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Theory Theory and questions for topic based enrichment activities/teaching

Contents Prerequisites ...... 1 Questions ...... 2 Investigative ...... 2 General Questions ...... 2 Senior Maths Challenge Questions ...... 3 British Maths Olympiad Questions ...... 8 Key Theory ...... 9 Solutions...... 13 Theory Questions ...... 13 General Questions ...... 15 SMC Questions ...... 20 British Maths Olympiad ...... 33

Prerequisites  For some questions, appreciation of the function n!  For one of the questions, knowledge of the difference between a convergent and divergent infinite .  For one of the questions, knowledge of logarithms and determining a turning point using calculus.  Appreciation of big-sigma  to indicate a over a range. Questions

Investigative These questions are intended to highlight key theory in number theory or proofs you should be aware of (the latter of which you should research, as you would not be expected to prove them). For more general questions, see those in the next section or the SMC/BMO questions.

1. If x is divisible by a, then is x2 divisible by a? 2. If x2 is divisible by a, then is x divisible by a? More generally, if xn is divisible by a for some n, then is x divisible by a? 3. Prove that √2 is irrational.

4. Prove that is divergent.

5. Show that there are an infinite number of prime . 6. Why is 1 not considered to be prime?

General Questions 1. Show that the 2, 5, 8, 11, 14, 17, … can never be a number. 2. Show that 2n = n3 has no integer solution for n. 3. Does n have an integer solution to 4. Find integer solutions to xy + x + y = 2004 where x>0, y>0.

5. How many positive integer values of n is also an integer?

6. Suppose an integer has d distinct prime factors, where each is repeated ai times. How many does the number have? How many factors does 1010 have? 7. Find all integer solution to the equation ab = ba that are non-trivial (ab), and show that these are the only solutions. 8. (a) Suppose n > 2 is a positive integer and that 2k is the highest power of 2 that divides any term in the sequence 2, 3, 4, ... , n. Show that only one term of the sequence is divisible by 2k. Deduce that lcm(2, 3, 4, ... , n) = 2kc where c is an odd integer.

(b) By expressing the number in the form a/b, show that x cannot be an integer.

9. Show that if a number is divisible by 3, its digits add up to a multiple of 3. 10. Determine the largest that is a factor of the following: (a) 50! (b) 1000! (c) In general, n! Senior Maths Challenge Questions Level 1 Level 3 1. How many prime numbers are there less than 20? 1. Which of these five expressions represents the largest A 6 B 7 C 8 number? 9 D 9 E 10 A 9(9 ) B 999 C 999 D (99)9 E 999

2. What is the remainder when the number 743589 × 301647 2. What is the value of is divided by 5? 1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + A 0 B 1 C 2 2))))))))))? D 3 E 4 A 210 +1 B 211 −1 C 211 +1 D 212 −1 E 212 + 1

3. Exactly one of the following numbers is divisible by 11. 3. I am trying to do a rectangular jigsaw puzzle. The puzzle was Which is it? made by starting with a rectangular picture and then cutting 7 7 7 A 10 − 11 B 10 −1 C 10 it into 1000 pieces by sawing along the lines of a (wiggly!) 7 7 D 10 +1 E 10 + 11 rectangular grid. I start by separating out all the edge and corner pieces. Which of the following could not possibly be Level 2 the number of corner and edge pieces of such a jigsaw? 1. The mean of seven consecutive odd numbers is 21. What is A 126 B 136 C 216 the sum of the first, third, fifth and seventh of these D 316 E A-D are all possible numbers? A 16 B 21 C 84 4. Observe that 18 = 42 + 12 + 12 + 02. How many of the first D 147 E more information needed fifteen positive can be written as the sum of the of four integers? 2. The value of the product wxyz is 2002 and w, x, y and z are A 11 B 12 C 13 2 2 2 2 prime numbers. What is the value of w + x + y + z ? D 14 E 15 A 66 B 203 C 260 D 285 E 343 5. The probability of a single ticket winning the jackpot in the National Lottery is 6 5 4 3 2 1 A 2016 B 2025 C 2040      . 49 48 47 46 45 44 D 2048 E 2050 If I buy one ticket every week, approximately how often might I expect to win the jackpot? 10. Which of the following numbers n gives a counter-example A once every hundred years for the statement: ‘If n is a then n2 + 2 is also B once every twenty thousand years a prime number’? C once every hundred thousand years A 3 B 5 C 6 D once every quarter of a million years D 9 E none of them E once every million years n 20 n 6. The factorial of n, written n!, is defined by n! = 1 × 2 × 11. When n = 81, what is the value of 3 ? 3×…×(n − 2) × (n − 1) × n 1 1 e.g. 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720. What is the smallest A less than 100 B 3 positive integer which is not a factor of 50! ? C 1 D 3 E more than 100 A 51 B 52 C 53 D 54 E 55 12. The number of the year 2003 is prime. How many square numbers are factors of 20032003? 2 2 2 2 7. What is the value of (61 − 39 ) ÷ (51 − 49 )? A 0 B 1 C 44 A 10·5 B 11 C 12 D 1002 E 2003 D 21 E 22 13. A is divided by 6. Which of the following 8. Given that a and b are integers greater than zero, which of could not be the remainder? the following equations could be true? A 0 B 1 C 2 A a – b = a  b B a + b = a  b D 3 E 4 C a – b = a  b D a + b = a – b E a+ b = a + b

9. Observe that 2000 = 24 × 53. What is the number of the next year after the year 2000 which can be written ab × cd where a, b, c, d are 2, 3, 4, 5 in some order? Level 4 6. What is the value of 22003 − 22002 − 22001 − 22000? A −22002 B 0 C 2−4000 1. What is the maximum possible value of the product of a list D 64 E 22000 of positive integers, which are not necessarily all different, given that the sum of these numbers is 100? 7. The value of 12004 + 32004 + 52004 + 72004 + 92004 is calculated 10 50 20 20 A 10 B 2 C 2 × 3 using a powerful computer. What is the units digit of the 20 2 32 D 5 E 2 × 3 correct answer? A 9 B 7 C 5 2. Sam correctly calculates the value of 58 × 85. How many D 3 E 1 digits does her answer contain? A 11 B 12 C 13 D 14 E 15 8. Which is the lowest positive integer by which 396 must be multiplied to make a perfect ? 3. Note that 2001 = 3 × 23 × 29. What is the number of the A 11 B 66 C 99 next year which can be written as the product of three D 121 E 726 distinct primes? A 2002 B 2004 C 2006 9. How many two-digit numbers N have the property that the D 2007 E none of these sum of N and the number formed by reversing the digits of N is a square? 4. For how many integer values of n does the equation x2 + nx A 2 B 5 C 6 − 16 = 0 have integer solutions? D 7 E 8 A 2 B 3 C 4 D 5 E 6 10. Damien wishes to find out if 457 is a prime number. In order to do this he needs to check whether it is exactly divisible by 5. Which of the following is divisible by 3 for every whole some prime numbers. What is the smallest number of number x? possible prime number divisors that Damien needs to check A x3 − x B x3 − 1 C x3 before he can be sure that 457 is a prime number? D x3 + 1 E x3 + x A 8 B 9 C 10 D 11 E 12

Level 5 6. Let N be a positive integer less than 102002. When the digit 1 is placed after the last digit of N, the number formed is 1. Three people each think of a number which is the product three times the number formed when the digit 1 is placed in of two different primes. Which of the following could be the front of the first digit of N. How many different values of N product of the three numbers which are thought of? are there? A 120 B 144 C 240 A 1 B 42 C 333 D 3000 E 12100 D 667 E 2002

2. Just one of the following is a prime number. Which one is it? 7. Given an unlimited supply of 50p, £1 and £2 coins, in how A 10002 + 1112 B 5552 + 6662 many different ways is it possible to make a sum of £100? C 20002 – 9992 D 10012 + 10022 A 1326 B 2500 C 2601 E 10012 + 10032 D 5050 E 10 000

3. The statement “There are exactly four integer values of n 8. How many pairs of positive integers (x, y) are solutions of 1 2 3 for which (2n + y)/ (n − 2) is itself an integer” is true for the equation + = ? certain values of y only. For how many values of y in the x y 19 range 1 ≤ y ≤ 20 is the statement true? A 0 B 1 C 2 A 0 B 7 C 8 D 3 E more than 3 D 10 E 20 9. The number N is exactly divisible by 7. It has 4008 digits. 4. How many pairs of positive integers (x, y) satisfy the Reading from left to right, the first 2003 digits are all 2s, the equation x – 17  y ? next digit is n and the last 2004 digits are all 8s. What is the A 0 B 1 C 2 value of n? D 17 E infinitely many A 4 B 5 C 0 or 3 D 2 or 9 E 1 or 8  n1   5. For how many values of n are both n and 4 n1 integers? A 1 B 2 C 3 D 4 E 5

10. Positive integers x and y satisfy the equation: 12. Given that S = (x + 20) + (x + 21) + (x + 22) +…+ (x + 100), where x is a positive integer, what is the smallest value of x x + 1 y – x – 1 y = 1. 2 2 such that S is a perfect square? Which of the following is a possible value of y? A 1 B 2 C 4 A 5 B 6 C 7 D 8 E 64 D 8 E 9 13. The factorial of n, written n!, is defined by n! = 1 × 2 × 3 ×…× 11. What is the sum of the values of n for which both n and (n − 2) × (n − 1) × n. For how many positive integer values of n 2 – 9 k less than 50 is it impossible to find a value of n such that n! are integers? n – 1 ends in exactly k zeros? A −8 B −4 C 0 A 0 B 5 C 8 D 4 E 8 D 9 E 10

British Maths Olympiad Questions 1. One number is removed from the set of integers from 1 to 7. Determine the least possible value of the largest term in an n. The average of the remaining numbers is 40.75. Which of seven distinct primes. integer was removed? 8. Let s be an integer greater than 6. A solid cube of side s has 2. Determine the a0, a1, a2, … which satisfy all of the a square hole of side x<6 drilled directly through from one following conditions: face to the opposite face (so the drill removes a cuboid).

 for every integer , The volume of the remaining solid is numerically equal to  is a and the total surface of the remaining solid. Determine all  for some I, j with possible integer values of x. 3. Find all positive integers n such that both n+8000 divides n2 9. Determine the least n for which the + 2008 and n+2009 divides n2 + 2009. following result holds: No matter how the elements of the 4. Find four prime numbers less than 100 which are factors of set {1, 2, … , n} are coloured red or blue, there are integers, 332 – 232. x, y, z, w in the set (not necessarily distinct) of the same 5. Let n be an integer. Show that, if is an colour such that x + y + z = w. integer, then it is a perfect square. 6. Each of Paul and Jenny has a whole number of pounds. He 10. Find, showing your method, a six-digit integer n with the says to her: “If you give me £3, I will have n times as much following properties: (i) n is a perfect square, (ii) the number as you.” She says to him: “If you give me £n, I will have 3 formed by the last three digits of n is exactly one greater times as much as you”. Given that all these statements are than the number formed by the first three digits of n. (Thus true and that n is a positive integer, what are the possible n might look like 123124, although this is not a square.) values for n?

Key Theory Explanation What kind of problems do we find in Number theory is concerned solely with the study of ‘integers’, i.e. whole numbers, and various properties number theory? concerning them. It’s interested in problems like the following:  Integer solutions to equations.  Various properties of prime numbers: Are there an infinite number of them? Is there any way to enumerate all the prime numbers? What’s the probability that a given number is prime? Can all integers be expressed as the sum of two primes?  Divisibility problems: Properties involving factors of numbers and expressions and remainders.  Properties of rational numbers: Rational numbers are simply those that can be expressed as a fraction involving integers, so number theory typically asks questions about rational numbers as well. Is 2 rational? Is e rational?  Types of numbers and associated questions. e.g. Perfect numbers (where the factors of the number add up to itself, e.g. 6): do they all end in 6 or 8? Are there an infinite number of them?

If a and b are integers, then ab is divisible This might seem obvious, but is important to keep in mind. If we have an expression like 3x2 say and we by both a and b. know that x is an integer, then 3x2 is divisible by 3. Key term: Modulo Arithmetic Modulo arithmetic is when we do arithmetic on numbers, but where the numbers ‘wrap around’ when reaching a certain number known as the modulus. We do modulo arithmetic with angles and time on a clock for example.

Formally, we write:

to mean that the remainder when we divide both a and b by c is the same. So for example and . Sometimes you might see the mod used more in a functional way like say addition or subtraction, for example 11 mod 2 = 1, which means “the remainder when we divide 11 by 2”. But usually the ‘mod’ comes at the end and in brackets. I use the equality symbol = from here on for convenience. When we find the remainder of a sum of This is one of the features of modulo arithmetic. For example, suppose we were to find the remainder of 7 + numbers when dividing by a, we can find 4 when dividing by 3. We could find the remainder when dividing 7 by 3 (i.e. 1), and similarly with 4 (i.e. 1 the remainders of each term in the sum, again), then add these remainders together (i.e. 2). This gives us the same result than if we were to directly add them together, and then find the divide 11 by 3. remainder. An implication of this is that we can tell 3n + 2 gives us a remainder of 2 when dividing by 3. Because 3n gives us a remainder of 0, and 2 gives us a remainder of 2. So the overall remainder is 2.

Modulo arithmetic also works with If 5 = 2 (mod 3) and 8 = 2 (mod 3), then we can work out the remainder when we divide 5 x 8 = 40 by 3. We . multiply the remainders, giving 2 x 2 = 4. But since this is modulo arithmetic, we actually get 1 (since the 4 wraps around). If a = b (mod c), then an = bn (mod c) for This follows from the above. If a gives a remainder of b when dividing by c, then gives a remainder of any positive integer n. since we earlier said that we could multiply the remainders. Repeating this n times results in the stated law.

Examples: 4 = 1 (mod 3), so 16 = 1 (mod 3), and 64 = 1 (mod 3). 7 = 2 (mod 5), so 49 = 4 (mod 5) 5 = 2 (mod 3), so 25 = 1 (mod 3), because the 22 = 4 ‘wraps’ to 1 in modulo 3.

We’ll see how this law can be used in solving various problems. If a = b (mod c), then ka = kb (mod kc) For example 24 = 3 (mod 7). Then 48 = 6 (mod 14). If a = b (mod c), then a-b is divisible by c. If for example 103 gives a remainder of 3 when divided by 10, then clearly 103-3 = 100 is divisible by 10. This gives an effective means of dealing with many divisibility-themed questions, as will be seen. The factorial function n! Number theory type problems frequently involve the factorial function. It is defined as such: 0! = 1 n! = n x (n-1) x (n-2) x …. x 2 x 1 So 4! = 24 for example. See the Combinatorics set of problems for lots of examples of its use. Diophantine Equations Diophantine equations are polynomial equations where the variables are restricted to integers. For example for constants a, b and c we might be interested in integer solutions for ax + by = c (known as a Linear Diophantine Equations), or solutions to x2 + y2 = z2 (known as Pythagorean triples).

A famous historical problem known as Hilbert’s 10th Problem (posed in 1900) asks whether there is some standard procedure by which we can either find all solutions to a Diophantine Equations or show that none exist (for example in the case x3 + y3 = z3, due to Fermat’s Last Theorem). It’s since been proven that no such procedure exists. There are 4 well-known examples of Diophantine Equation:

Form Info ax + by = c Linear Diophantine Equation: This has infinitely many solutions if c is a multiple of the greatest common of a and b, and no solutions otherwise. It may be worth learning how to solve such equations. The Erdos-Staus Conjecture states that for any integer n>1, 4/n can be

expressed as the sum of 3 positive unit fractions. It’s been verified for values up to very large n, but not yet proven. See ‘Famous Theorems’ below. Known as Pell’s Equation. We’re interested in integer solutions for x and y (apart from the trivial x=1 and y=0) when n is non-square. Historical interest in the equation was for its use in finding approximations to square roots of numbers. For example, in finding solutions for x2 – 2y2 = 1, x/y gives an approximation for 2. Lagrange proved that there are infinitely many solutions for any Pell Equation.

If we find the factors of some number x, For example, 30 = 5 x 3 x 2. Then 3 x 2 divides 30, 5 x 2 divides 30, 2 divides 30, etc. then the product of any subset of factors This might seem obvious, but is a principle used in some SMC questions. of x divides x. Key Theorem: Fundamental Theorem of All integers greater than 1 can be uniquely factorised as a product of primes. If 12 = 2 x 2 x 3, then there is Arithmetic (also known as the Unique no other product of primes which yields 12 (except by reordering the factors). Factorisation Theorem) When we find the prime factorisation of a number, we tend to group identical factors using powers, i.e. 300 = 22 x 3 x 52. This makes it easier to both count the total prime factors and the distinct factors. Key Term: Lowest Common Multiple The smallest number which divides a given set of integers. (LCM) Fermat’s Little Theorem Not to be confused with ‘Fermat’s Last Theorem. If p is a prime number, then for any integer a:

or equivalently:

So 612 = 1 (mod 13) for example, and 612 – 1 is divisible by 13. The guidance document for the British Maths Olympiad recommends that you know this for Round 2. Famous Theorems in Number Theory Reading about some of these problems will broaden your general mathematical knowledge (and if applicable, hopefully provide some good fodder for your university personal statement!)

Name Explanation Solved? Goldbach Conjecture Are all positive integers are the sum of two primes? No. Fermat’s Last an + bn = cn has no integer solutions for a, b and c when n>2. Yes, by Andrew Theorem Wiles in 1995. Prime Number The probability that a randomly chosen integer n is prime is NA Theorem approximately 1/ln(n) (the reciprocal of the natural log of n). Euclid’s Theorem There are an infinite number of prime numbers (see Theory Yes (by Euclid!) Questions).

Euler Product Euler in 1737.

Formula

i.e. The formula relates a involving the natural numbers with a product involving all the prime numbers.

e.g.

Riemann Hypothesis The summation on the LHS of the equation above is known as No. the .

The domain of the function can be extended to complex numbers. The Riemann Hypothesis states that all roots of the Riemann Zeta function, i.e. s for which , have the

form for any a (except for ‘obvious’ roots known as

‘trivial roots’). Recommended book: “Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in ” by John Derbyshire.

Solutions

Investigative Questions Question Solution If x is divisible by a, then is x2 To formally prove: If a is a factor of x, then x = ka for some k. Then x2 = k2 a2, where the RHS is clearly divisible by a. divisible by a? But the answer is clear just from observation. If x is divisible by a, then any whole number times x is still visible by a.

Teaching points: The idea here is to get students to appreciate that multiplying an integer by another doesn’t ‘lose’ any factors from the original number, only potentially adds them. If x2 is divisible by a, then is x Only if a is prime. The key here is prime factorisation. Whatever prime factors we have in x will appear twice in x2, divisible by a? More generally, if xn is but ultimately we don’t add any new prime factors when squaring (or taking to any integer power) other than divisible by a for some integer n, repeating them. If we have the same unique prime factors in x2 and x, then clearly a will be a factor of x. Exactly then is x divisible by a? the same argument applies for xn.

There’s simple counter examples if a is not prime. 32 is divisible by 9, but 3 is obviously not divisible by 9. This is because 9 is not prime.

Teaching points: It’s a fact needed in the proof of root 2’s irrationality amongst others. Prove that √2 is irrational We do a ‘proof by contradiction’, i.e. asserting the opposite of the theorem, and showing this leads to a contradiction.

If √2 is rational, then suppose there are some a and b such that √2 = a/b and a/b is a fraction in its simplest form. By multiplying by sides by b and squaring both sides, we get 2b2 = a2. We can see 2b2 is divisible by 2. So a2 must also be divisible by 2. By the theorem above (i.e. if xn is divisible by a and a is prime, then is x divisible by a), a must be divisible by 2 since 2 is prime. Since a is divisible by 2 (i.e. is even), then our fraction a/b is not in the simplest form. This contradicts what we initially asserted, so √2 must be irrational.

This reasoning applies for the root of any prime number.

Teaching points: Appreciating how to solve this problem is much more important than this particular proof itself, otherwise students are likely to quickly forget the proof. The emphasis in presenting the solution should probably be the strategy in getting to the each subsequent step of the proof. The hardest part of the proof perhaps (in terms of knowing what step you would take next) is making the assertion that you need the fraction in its simplest form – it’s not obvious at that particular point why it will be useful to us later in the proof.

Prove that is This particular series is known as the harmonic series. A series is divergent when the sum of an infinite number of

terms is . This contrasts to a such as , which gradually approaches 2. divergent.

There’s two proofs for this on http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29

Teaching points: Note that the harmonic series is not a geometric series. Show that there are an infinite This is known as Euclid’s Theorem, published in the book ‘Elements’ in 300BC. number of prime numbers. The proof goes as such:

 Suppose we have a list of all the known primes. Multiply them together, and add 1. i.e. n = (p1p2....pk) + 1.

 We know that if we divide n by any of our known primes pi, then the remainder will be 1. Thus none of these primes can be divisors of n.  Therefore either n is prime not in our original list, or n is composite, but its prime factors are not in our original list.

Teaching points: This proof is important as part of a broad number theory knowledge. To teach it, you could potentially give different sets of cards, just give them the first bullet point of the proof, and see what they notice for different sets, so as to distinguish between the cases when n is prime and composite. One point to emphasise is that the method doesn’t enumerate all prime factors; it only shows we can always find a new prime given a finite list. If we start with just 2 for example, 2 + 1 = 3, then (2 x 3) + 1 = 7, then (2 x 3 x 7) + 1 = 43, (2 x 3 x 7 x 43) + 1 = 1807 = 13 x 39. We so far have generated the primes {2, 3, 7, 13, 43, 139}, which is clearly not a complete list of the first few prime numbers. Why is 1 not considered to be prime? Here’s a nice explanation of the various reasons 1 is not considered a prime: http://primes.utm.edu/notes/faq/one.html

The most convincing reason is due to the definition of the Fundamental Law of Arithmetic. If 1 is prime, then a number 10 say can be prime factorised as 5 x 2, or 5 x 2 x 1, or 5 x 2 x 1 x 1 and so on. This violates the condition that there must be a unique prime factorisation of any integer.

General Questions Question Solution Source 1 Show that the sequence 2, 5, 8, 11, 14, This is equivalent to saying “Show that the square numbers are always a multiple of 3 or one ? 17, … can never be a square number. more than a multiple of 3.

JAF’s solution: We can use modulo arithmetic to get the solution. For any number x, then x = 0, 1 or 2 in modulo 3 arithmetic (cycling through these as x increases). So then x2 = 02, 12 or 22 (mod 3). Thus x2 = 0, 1 or 1 (mod 3), since 4 is 1 in modulo 3 arithmetic. This tells us that the remainder of the squares when dividing by 3 oscillates between 0, 1 and 1, i.e. being a multiple of 3, followed by two square numbers one more than a multiple of 3. But the numbers in the sequence are all one less than a multiple of 3 (i.e. are all 2 in modulo 3).

Teaching points: This law is really incredibly useful in a number of problems involving divisibility. Students ought to learn it and appreciates its usefulness. 2 Show that 2n = n3 has no integer This is one of many questions where we wish to appreciate the prime factors on each side of solution. for n. the equation. On the LHS of the equation, we only have prime factors of 2. Then n3 can only have factors of 2, and thus by our usual theorem n can only have factors of 2. If this is the case, we can put n in the form 2k for some integer k. Substituting this in, we get:

Then 2k = 3k by comparing the powers. The RHS of the equation is divisible by 3, but the LHS is not. So k cannot be an integer, which contradicts what we asserted about k. 3 Find integer solutions to xy + x + y = This problem is in the British Maths Olympiad guidance document under ‘key things you BMO 2004 where x>0, y>0. should know about Number Theory’. So take heed! guidance doc The key here is to spot potential factorisations. xy + x + y + 1 = (x+1)(y+1). Therefore (x+1)(y+1) = 2005. The only factors pairs of 2005 are 1 and 2005 (giving 0 and 2004, which is not allowed), and 5 and 401 (giving 4 and 400 as solutions). 4 Does n have an integer solution to We can use the same method as above. Firstly note we can factorise the RHS as n2(1 + n). If again the RHS has to only contain prime factors of 2, then each of its factors must also only contain prime factors of two. So the n term we factored out must again only contain prime factors of two. By again substituting n = 2k, we obtain:

We can see the RHS must be odd unless k = 0 (which yields n = 1 as a solution). But otherwise, the RHS is odd and LHS even, so there’s no other solution. 5 How many positive integer values of n JAF’s solution: The first thing that came to me was “Wouldn’t it be easier if the subtraction SMC

is also an integer? was in the numerator instead of the denominator – then we could simplify”. So I created a

variable n = 100 – x. Substituting this in, you get (100 – x) / x, which conveniently simplifies to (100/x) – 1. If you choose factors of 100 for x, then the whole expression is clearly an integer. So x = 1, 2, 4, 5, 10, 20, 25, 50, 100. Since n = 100 – x, we obtain n = 99, 98, 96, 95, 90, 80, 75, 50, 0. 6 Suppose an integer has d distinct This is more of a combinatorics problem, since we’re concerned with the number of ways of JAF prime factors, where each is repeated combining the prime factors to form different divisors. ai times. How many divisors does the If we look at 20 for example (2 x 2 x 5), our divisors are obtained using 2, (2 x 2), 5, (2 x 5) and number have? How many factors does (2 x 2 x 5), as well as using none of the prime factors (yielding the divisor 1). The factor of 2 1010 have? can be seen zero, one or two times, and the factor of 5 zero or one time. So the number of divisors is just:

where the +1 is because we need to include the ‘zero’ number of times a prime factor can appear in a factor. For the second part of the question, 1010 = (2 x 5)10 = 210 x 510. Using the formula, this gives 11 x 11 = 121 factors. 7 Find all integer solution to the This is an incredibly interesting problem! Older students might initially be inclined to take logs Classic equation ab = ba that are non-trivial of both sides, giving b log(a) = a log(b). Rearranging, we get log(a) / a = log(b) / b. We can’t problem (ab), and show that these are the simplify further (i.e. we can’t make a the subject of the formula in a = log(a)). However, if we only solutions. define a function f(x) = log(x) / x, then we have a solution if f(a) = f(b).

At this point, we need to sketch the function f(x) = log(x) / x. (The key to sketching it is identifying the domain, the range, roots and the turning points)

The turning point (using calculus) is at (e, 1/e), and the root is x = 1. By observation of the shape of the graph, we can see that if f(a) = f(b), then if , then . If however a < 1, then the only b for which f(a) = f(b) is when a = b (i.e. our trivial solution). So one of a or b must be in the range (1,e). 2 is the only integer in this range, and thus is the only integer solution. So 24 = 42 is the only integer solution. 8 (a) Suppose n > 2 is a positive integer We had an earlier question on the proof of the harmonic series up to infinity being divergent. Oxford and that 2k is the highest power of 2 Now we’re trying to show (in part (b)) that it’s not an integer either for any n. Interview that divides any term in the sequence 2, 3, 4, ... , n. Show that only one term The first part of (a) is rather confusing until we use a concrete example. Suppose we have the of the sequence is divisible by 2k. number 2, 3, 4, 5, 6, 7, 8, 9, 10. Then 8 is the highest power of 2. Clearly nothing less than 8 is Deduce that divisible by 8, and the lowest number greater than 8 that is divisible by 8 is twice it, i.e. the lcm(2, 3, 4, ... , n) = 2kc next power of 2. But we know this must be greater than n because we defined 2k to be the where c is an odd integer. highest power of 2 less than n. So 2k itself is the only term in the sequence divisible by 2k.

(b) By expressing the number The next part of (a) is a little tricky to understand. All integers can be expressed in the form 2r

in the form a/b, c, where c is odd. So the key here is showing that r = k when we have the lowest common k show that x cannot be an integer. multiple of 2 to n. We previously showed that 2 is the highest power of 2 that will divide one of the terms in the sequence 2 to n. By finding the LCM, we’ve found a number which divides all of 2 to n. Now, because we need to ensure this LCM can be divided by the number 2k in the sequence. But we also don’t need r to be any greater than k, because no term in the sequence 2 to n has more the k twos as factors, and thus the LCM need not have more than k twos to be divisible by all the numbers.

(b) is a bit more fun. In order to add the numerators, we need to make the denominators the same, and finding the LCM of 2 to n is sufficient. Then:

for some natural number a. To make the sum of fractions a sum of integers, we can multiply both sides by 2kc, since we know 2kc / 2 is an integer, 2kc / 3 is an integer, etc. (because of the fact 2kc is our LCM of 2 to n). So:

The LHS is even (because it has a factor of 2). And all of the RHS terms are even, except for the one where the denominator is 2k, because we’re dividing by a number that contains less than k twos (by definition of k). However when the denominator is 2k, we end up with c, which by definition is odd. Therefore the LHS is even but the RHS is odd. This is a contradiction, and therefore x cannot be an integer.

9 Show that if a number is divisible by 3, Suppose the digits of n are nmnm-1...n1 (where there’s m digits). Then in JAF

its digits add up to a multiple of 3. general. If its digits add up to a multiple of 3, then for some integer k. We have two simultaneous equations! By subtracting the second from the first, we get:

So we just need to show the RHS is divisible by 3. Certainly 3k is divisible by 3. And is divisible by 3 because for i=1, 2, 3, etc. we get 0, 9, 99, 999, ... (or we could more formally show is divisible by 3, since 10 = 1 (mod 3), and thus 10i-1 = 1i-1 = 1 (mod 3), and thus = 0 (mod 3)).

Teaching points: For younger students, get them to think about 2 digit numbers (say using a and b for the two digits to avoid confusing subscript notation), then 3 digit numbers, and so on. Hopefully they’ll identify the pattern that emerges in order to give an informal argument for why the trick works for any number of digits. A related problem is showing that if the digits add up to a multiple of 9, the number is divisible by 9. 10 Determine the largest power of 10 Once again, consider the prime factors of 50! and 1000! Classic that is a factor of the following: We note that 10 = 2 x 5. So each time we find a pair of 2 and 5 in the prime factorisation of Problem (a) 50! some number, we obtain a factor of 10. In both 50! and 1000!, the factor of 2 is going to (b) 1000! appear and awful lot more than 5, so it’s sufficient to count the number of factors of 5, (c) In general, n! knowing that we’ll definitely have enough 2s floating about to pair them up with. a) 50! = 50 x 49 x 48 x ... x 1. How many factors of 5 appear in these 50 numbers? Well, each multiple of 5 has one. But we also get additional factors of 5 from 25 (= 52) and 50 (= 52 x 2). So that’s 10 plus the additional 2, giving us 12. b) We spotted from the first part that multiples of 5 give us one factor of 5 each, and that multiples of 52 give us a bonus 5. Similarly multiples of 53 = 125 will give us a further 5, as will multiples of 54 = 625. That gives us 200 + 40 + 8 + 1 = 249.

c) Using the pattern from above, we get: , where is a function which

rounds down its value (we’re looking for the whole number of times 5k goes into n). Although I’ve summed to infinity, the floor function is going to give 0 once 5k exceed n (e.g. in part (ii), none of the numbers between 1 and 1000 have a factor of 55, and so

gives us 0 when rounded down, and is not contributing to the summation). We

could use logs to limit the summation (by finding the maximum power of 5 that

divides k), so a better answer would be

SMC Questions Level 1 Question Solution Ref 1 How many prime numbers are there less Answer C 1999 than 20? The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Q1 A 6 B 7 C 8 D 9 E 10

2 What is the remainder when the number Answer D 2000 743589 × 301647 is divided by 5? The units digit of the product is 3; therefore the remainder when that number is divided by 5 is Q1 A 0 B 1 also 3. C 2 D 3 E 4 3 Exactly one of the following numbers is Answer D 2001 divisible by 11. Since 107 is not divisible by 11, then neither is 107 − 11 nor 107 + 11. Therefore, the correct answer Q2 Which is it? is either B or D. 7 7 A 10 − 11 B 10 −1 107 − 1 = 9 999 999 = 11 × 909090 + 9. Therefore 7 7 C 10 D 10 +1 107 + 1 = 11 × 909090 + 11. Hence D is the correct answer. E 107 + 11

Level 2 Question Solution Ref 1 The mean of seven consecutive odd Answer C 2001 numbers is 21. What is the sum of the first, Since the four numbers are symmetrical about the mean of the group, they must have the same Q6 third, fifth and seventh of these numbers? mean, 21, giving their total as 84. Alternatively, it is possible to state that the original numbers A 16 B 21 must be 15, 17, 19, 21, 23, 25, 27 (since the numbers are given as consecutive and odd). Hence C 84 D 147 the required sum is 15 + 19 + 23 + 27 = 84. E more information needed

2 The value of the product wxyz is 2002 and Answer E 2002 w, x, y and z are prime numbers. 2002 = 2 × 7 × 11 × 13; 22 + 72 + 112 + 132 = 343 Q5 What is the value of w2 + x2 + y2 + z2? A 66 B 203 C 260 D 285 E 343

Level 3 Question Solution Ref 1 Which of these five expressions represents Answer: E 1997 9 the largest number? 999 = 9 × 111 < 9 × 93 = 94 < 999 < (93)9 = 927 < 981 = (99)9 < 999 < 99 Q11 9 A 9(9 ) B 999 C 999 D (99)9 E 999 2 What is the value of Answer: D 1997 1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211. Q12 2(1 + 2(1 + 2(1 + 2))))))))))? Now (xn − 1 + xn − 2 + … + x2 + x + 1) (x − 1) = (xn − 1) 10 11 A 2 +1 B 2 −1  211 + 210 + 29 + … + 22 + 2 + 1) = (212 − 1)/ (2 − 1) = 212 – 1 11 12 C 2 +1 D 2 −1 E 212 + 1 (JAF: Know your formula for the sum of a geometric series!)

3 I am trying to do a rectangular jigsaw Answer: D 1997 puzzle. The puzzle was made by starting Suppose there are h pieces along each horizontal edge and v pieces along each vertical edge. Q14 with a rectangular picture and then cutting Then h.v = 1000 = 23.53. it into 1000 pieces by sawing along the Thus the only possibilities for the pair {h, v} are {1, 1000}, {2, 500}, {4, 250}, {5, 200}, lines of a (wiggly!) rectangular grid. I start {8, 125}, {10, 100}, {20, 50}, {25, 40}. by separating out all the edge and corner The total number of edge pieces is 2h + 2v − 4 = 2 (h + v − 2) (since 2h + 2v counts each pieces. of the four corners twice). Which of the following could not possibly 126 (= 2 (25 + 40 − 2)), 136 (= 2 (20 + 50 − 2)), be the number of corner and edge pieces 216 (= 2 (10 + 100 − 2)) are all possible; but 316 is not. of such a jigsaw? A 126 B 136 C 216 D 316 E A-D are all possible 4 Observe that 18 = 42 + 12 + 12 + 02. Answer E 1998 How many of the first fifteen positive The question really asks how many of the first 15 positive integers can be expressed as the sum Q6 integers can be written as the sum of the of 4 or fewer perfect squares. These can be listed as follows:− squares of four integers? 1 = 12; 2 = 12 + 12; 3 = 12 + 12 + 12; 4 = 22. Expressions for 5, 6, 7 and 8 can be obtained by A 11 B 12 adding 22 to each of the ones already given. 9 = 32 and expressions for 10, 11, 12, 13, 14, and C 13 D 14 15 can be obtained by adding to the ones for 1, 2, 3, 4, 5 and 6. E 15 Note: In fact, the French mathematician Lagrange proved in 1770 that every positive integer can be expressed as the sum of four squares. 5 The probability of a single ticket winning Answer D 1998 the jackpot in the National Lottery is 49 48 47 46 45 44 Q16 Number of years =      52 6 5 4 3 2 1 6 5 4 3 2 1      . 49 48 47 46 45 44 = 49 × 1 × 47 × 46 × 3 × 44 ÷ 52 If I buy one ticket every week,  503 3  375000. approximately how often might I expect to Note: 375000 is clearly an overestimate since 49 × 1 × 47 × 46 is less than 503 and 3 × 44 ÷ 52 win the jackpot? is less than 3. A once every hundred years B once every twenty thousand years C once every hundred thousand years D once every quarter of a million years E once every million years 6 The factorial of n, written n!, is defined by Answer C 1999 n! = 1 × 2 × 3×…×(n − 2) × (n − 1) × n All of the positive integers from 1 to 50 inclusive must be factors of 50! Q9 e.g. 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720. 51 = 3 × 17 and 52 = 2 × 26 which means that these two numbers are also factors of 50!. 53 is What is the smallest positive integer which prime and is not a factor of 50!. is not a factor of 50! ? A 51 B 52 C 53 D 54 E 55 7 What is the value of (612 − 392) ÷ (512 − Answer B 2000 2 Q9 49 )? 612 – 392  100 22 A 10·5 B 11 a2 − b2 = (a + b) (a − b). Thus  11. 512 – 492 100 2 C 12 D 21   E 22 JAF note: Questions that require you to spot the difference of two squares are quite common.

8 Given that a and b are integers greater Answer A 2000 than zero, which of the following equations B cannot be true since a + b > a but a ÷ b ≤ a; C cannot be true since a − b < a but a × b ≥ a; Q11 2 could be true? similarly, D cannot be true. Furthermore, E cannot be true since  a  b  a +b. However, A A a – b = a  b is true if a = 4 and b = 2. B a + b = a  b

C a – b = a  b D a + b = a – b E a+ b = a + b

9 Observe that 2000 = 24 × 53. What is the Answer B 2000 number of the next year after the year If the number of the year is even then it is either a power of 2, namely 25 × 43 or 23 × 45, both of Q12 2000 which are greater than 2025, or it is 2a × 3b or 2a × 5b , but no number between 2001 and 2024 b d which can be written a × c where a, b, c, is of this form. If the number is odd then it is either 34 × 52, which is 2025, or 32 × 54, which is d are 2, 3, 4, 5 in some order? much greater. Therefore 2025 is the next number of the given form. A 2016 B 2025 C 2040 D 2048 E 2050

10 Which of the following numbers n gives a Answer B 2001 counter-example for the statement: Since 6 and 9 are not prime, C and D cannot provide counter-examples. For 3, both parts are Q9 ‘If n is a prime number then n2 + 2 is also a true (since 11 is prime) but since 52 + 2 = 27 is not prime, B does provide a counter-example. prime number’? A 3 B 5 C 6 D 9 E none of them 11 n 20 Answer B 2002 n 8120 380 1 Q13 When n = 81, what is the value of 3 ? 4   Note that 81 = 3 . Therefore 81 81 . 1 1 3 3 3 A less than 100 B 3 C 1 D 3 E more than 100

12 The number of the year 2003 is prime. Answer D 2003 How many square numbers are factors of The only square numbers which are factors of 20032003 have the form 20032n for a non- Q15 20032003? negative integer n. But 2n ≤ 2003 so n = 0, 1, … , 1001. A 0 B 1 C 44 D 1002 E 2003

13 A square number is divided by 6. When divided by 6, a whole number leaves remainder 0, 1, 2, 3, 4 or 5. So the possible 2005 Which of the following could not be the remainders when a square number is divided by 6 are the remainders when 0, 1, 4, 9, 16 and 25 Q14 remainder? are divided by 6. These are 0, 1, 4, 3, 4 and 1 respectively, so a square number cannot leave A 0 B 1 remainder 2 (or remainder 5) when divided by 6. C 2 D 3 E 4

Level 4 Question Solution Ref 1 What is the maximum possible value of the Answer E 2000 product of a list of positive integers, which Any list whose product is a maximum does not contain any 1s since one 1 and one n could be Q20 are not necessarily all different, given that replaced by one n+ 1 and this would contribute n + 1 to the product rather than n. We can also the sum of these numbers is 100? say that any list of maximum product would contain a maximum of two 2s, since any three 2s A 1010 B 250 could be replaced by two 3s which would contribute 9 to the product rather than 8. C 220 × 320 D 520 Furthermore, it would contain a maximum of one 4 since two 4s could be replaced by one 2 and two 3s thereby contributing 18 to the product rather than 16. It would not contain any E 22 × 332 number greater than 4 since 2 (n − 2) > n when n > 4 and therefore the product would be

increased by replacing one n with one 2 and one n − 2. Any whose product is a maximum, therefore, consists of two 2s and thirty-two 3s or one 4 and thirty-two 3s and the maximum product is 22 × 332. 2 Sam correctly calculates the value of 58 × 58 × 85 = 58 × 215 = (5 × 2)8 × 27 = 128 × 100 000 000 = 12 800 000 000 which has 11 digits. 2001 85. Q15 How many digits does her answer contain? JAF: Whenever you see a question involving numbers of 0s or number of digits, then always A 11 B 12 remember the prime factors of 10 are 2 and 5. C 13 D 14 E 15

3 Note that 2001 = 3 × 23 × 29. Answer C 2001 What is the number of the next year which 2002 = 2 × 1001 = 2 × 7 × 11 × 13 which is the product of four primes. Q19 can be written as the product of three Since 2003 < 13 × 13 × 13, for 2003 to be a product of three distinct primes, distinct primes? at least one of them must be less than 13. However, none of 2, 3, 5, 7 or 11 is A 2002 B 2004 a factor of 2003. 2004 = 2 × 2 × 3 × 167 which has a repeated prime factor. C 2006 D 2007 2005 = 5 × 401 which is the product of two primes. E none of these 2006 = 2 × 1003 = 2 × 17 × 59 and is therefore the next year which is the product of three distinct primes.

4 For how many integer values of n does the Answer D 2002 equation x2 + nx − 16 = 0 have integer For the equation to have integer solutions, it must be possible to write x2 + nx − 16 in the form Q15 solutions? (x − α) (x − β), where α and β are integers. Therefore x2 + nx − 16 = x2 − (α + β)x + αβ and we A 2 B 3 require that αβ = −16. The possible integer values of α, β are 1,−16; −1, 16; 2, −8; −2, 8; 4, –4 C 4 D 5 (we do not count −16, 1 as being distinct from 1, −16, for instance). E 6 As n = −(α + β), the possible values of n are 15, −15, 6, −6 and 0.

5 Which of the following is divisible by 3 for Answer A 2003 every whole number x? Since x3 − x = x (x2 − 1) = (x − 1) × x × (x + 1), x3 − x is always the product of three consecutive Q13 3 3 A x − x B x − 1 whole numbers when x is a whole number. As one of these must be a multiple of 3, x3 − x will C x3 D x3 + 1 be divisible by 3. Substituting 2 for x in the expressions in B,C and E and substituting 3 for x in E x3 + x the expression in results in D numbers which are not divisible by 3.

JAF’s alternative answer: Use modulo arithmetic. If x = 0, 1, 2 (mod 3) oscillating between these as x increases, then: x3 = 0, 1, 8 = 0, 1, 2 (mod 3) Then x3 – x = 0-0, 1-1, 2-2 = 0, 0, 0 (mod 3) i.e. For numbers of x, x3 – x gives us a remainder of 0 when dividing by 3. 6 2003 2002 2001 2003 What is the value of 2 − 2 − 2 − 22003 − 22002 − 22001 − 22000 22000? Q18 2000 3 2 A −22002 B 0 = 2 (2 − 2 − 2 − 1) C 2−4000 D 64 = 22000 (8 − 4 − 2 − 1) = 22000. E 22000 7 The value of 12004 + 32004 + 52004 + 72004 + Answer A 2004 4 4 2 4 501 92004 is calculated using a powerful The last digit of 3 is 1, as is the last digit of 7 and the last digit of 9 . So the last digit of (3 ) , Q13 computer. that is of 32004, is 1. Similarly, the last digit of (74)501, that is of 72004, is 1 and the last digit of What is the units digit of the correct (92)1002, that is of 92004, is 1. Furthermore, 12004 = 1 and the last digit of 52004 is 5. So the units answer? digit of the expression is 1 + 1 + 5 + 1 + 1, that is 9. A 9 B 7

C 5 D 3 E 1 8 Which is the lowest positive integer by Answer E 2004 which 396 must be multiplied to make a Expressed as the product of prime factors, 396 = 22 × 32 × 11. Therefore the lowest positive Q20 perfect cube? integer by which it must be multiplied to make a perfect cube is 2 × 3 × 112, that is 726. A 11 B 66 C 99 D 121 E 726 9 How many two-digit numbers N have the Answer E 2007 property that the sum of N and the Let N be the two-digit number ‘ab’, that is N = 10a + b. So the sum of N and its ‘reverse’ is Q12 number formed by reversing the digits of N 10a + b + 10b + a = 11a + 11b = 11 (a + b). As 11 is prime and a and b are both single digits, 11 is a square? (a + b) is a square if, and only if, a + b = 11. So the possible values of N are 29, 38, 47, 56, 65, 74, A 2 B 5 83, 92. C 6 D 7 E 8

10 Damien wishes to find out if 457 is a prime The smallest number of possible prime divisors of 457 that Damien needs to check is the number. In order to do this he needs to number of prime numbers less than or equal to the of 457. Since 212 < 457 < 222, check whether it is exactly divisible by he needs to check only primes less than 22. These primes are 2, 3, 5, 7, 11, 13, 17 and 19. some prime numbers. What is the smallest number of possible prime number divisors JAF note: Such efficiency tricks like this come into play when programming! that Damien needs to check before he can be sure that 457 is a prime number? A 8 B 9 C 10 D 11 E 12

Level 5 Question Solution Ref 1 Three people each think of a number which Answer E 1999 is the product of two different primes. 120 = 23 × 3 × 5; 144 = 24 × 32; 240 = 24 × 3 × 5; Q15 Which of the following could be the product 3000 = 23 × 3 × 53 12100 = 22 × 52 × 112. of the three numbers which are thought of? The product of the three numbers must have 6 prime factors, not necessarily all different, A 120 B 144 but with no prime factor repeated more than 3 times. Of these, only 12100 satisfies this C 240 D 3000 condition. The three numbers are 10 (2 × 5), 22 (2 × 11) and 55 (5 × 11). E 12100 Teaching Points (JAF): An interesting question is finding the form of all possible solutions. If we pick 6 primes, the choice of numbers is valid if no prime is repeated more than 3 times. If we have only two distinct prime factors, each prime must occur three times (i.e. a3 x b3), and therefore each person can pick an a and a b to end up with different primes. 2 Just one of the following is a prime number. Answer A 1999 Which one is it? B cannot be prime since 111 is clearly a factor of it; C cannot be prime since it equals 2999 × Q21 A 10002 + 1112 1001 (difference of two squares); the units digit of D will be 5 and therefore 5 must be a B 5552 + 6662 factor of it whilst the units digit of E is 0 and therefore it cannot be prime either since both 2 and 5 will be factors of it. C 20002 − 9992 2 2 D 1001 + 1002 JAF note: This question is essentially combining all the various mini number-theory tricks 2 2 E 1001 + 1003 from other questions: i.e. difference of two squares and observing the units digit. 3 The statement “There are exactly four Answer B 1999 integer values of n for which (2n + y)/ (n − 2) Q23 2n  y 2n – 4 y  4 y  4 is itself an integer” is true for certain values    2  (n  2) of y only. n – 2 n – 2 n – 2 n – 2 For how many values of y in the range 1 ≤ y ≤ Thus (2n + y)/ (n − 2) is an integer if and only if (y + 4)/ (n − 2) is an integer. As n varies, the 20 is the statement true? integer values taken by (y + 4) / (n − 2) are all the integers which divide exactly into y + 4. A 0 B 7 There are exactly 4 of these if and only if y + 4 is prime. They are {(y + 4), −(y + 4), 1, −1}. C 8 D 10 Thus the integer values of the expression will be 2 + y + 4 = y + 6; 2− (y + 4) = −(y + 2); 3 and E 20 1. For 1 ≤ y ≤ 20, the values of y for which y + 4 is prime are 1, 3, 7, 9, 13, 15 and 19.

4 How many pairs of positive integers (x, y) Answer E 2000 satisfy the equation x – 17  y ? Squaring both sides of the equation gives y = x − 2 17x 17. The righthand side will be a Q24 A 0 B 1 positive integer if x is of the form 17n2 for an integer n ≥ 2, and the equation then becomes C 2 D 17 y = 17n2 − 34n + 17 = 17 (n − 1)2. E infinitely many As there are infinitely many n ≥ 2, there are infinitely many pairs of positive integers (x, y) which satisfy the equation e.g. (68, 17), (153, 68), (272, 153) etc.

 n–1 5 For how many values of n are both n and   2001 n1 n –1 1 3 n –1 k  n1     4 is an integer if the value of is 0, ,1, ,... i.e.  where k = 0, 1, 2, 3, … Q22 n1 n 1 2 2 n 1 2 4  integers? k n –1 n 1– 2 2 4 A 1 B 2 Let   1– . Then k = 2 – . For k to be an integer, the only C 3 D 4 2 n 1 n 1 n 1 n 1 E 5 possible values of n + 1 are −4, −2, −1, 1, 2, 4; i.e. the possible values of n are −5, −3, −2, 0, 1, 3. The corresponding values of k are 3, 4, 6, −2, 0, 1. However, k cannot be negative and therefore n = 0 is not valid. Therefore the five possible values of n are −5, −3, −2, 1, 3.

6 Let N be a positive integer less than 102002. Answer C 2002 When the digit 1 is placed after the last digit Let N have x digits, so that x ≤ 2002. When the digit 1 is placed at its end, N becomes 10N + Q25 of N, the number formed is three times the 1. When 1 is placed in front of it, N becomes 10x + N. Therefore: 10N + 1 = 3 (10x + N) i.e. number formed when the digit 1 is placed in 7N = 3 × 10x − 1. So we need to find which of the numbers 2, 29, 299, 2999, 29999, …. are front of the first digit of N. divisible by 7. The first such number is 299999 (corresponding to x = 5), giving N = 42857 and How many different values of N are there? we check that 428571 = 3 × 142857. A 1 B 42 The next such numbers correspond to x = 11, x = 17, x = 23, x = 29 and the largest number in C 333 D 667 the given range corresponds to x = 1997. E 2002 The number of different values of N, therefore, is 1 + (1997 − 5) ÷ 6 = 333.

7 How many pairs of positive integers (x, y) are Answer D 2003 1 2 3 1 2 3 Q25 solutions of the equation + = ?   . x y 19 x y 19 A 0 B 1 i.e. 38x + 19y = 3xy. C 2 D 3 i.e. 9xy − 114x − 57y + 38 × 19 = 38 × 19. E more than 3 i.e. (3x − 19) (3y − 38) = 2 × 192. The factors of 2 × 192 are 1, 2, 19, 38, 361 and 722 and 3x − 19 has to be one of these. If 3x − 19 = 1, 19 or 361, then x is not an integer. If 3x − 19 = 2, then x = 7 and 3y − 38 = 361 giving y = 133. If 3x − 19 = 38, then x = 19 and 3y − 38 = 19 giving y = 19 as well. If 3x − 19 = 722, then x = 247 and 3y − 38 = 1 giving y = 13. 1 2 3 (The equation   always has exactly 3 solutions when x and y are positive integers x y p and p is a prime greater than or equal to 5. Can you prove this result?)

8 The number N is exactly divisible by 7. It has First note that 1, 11, 111, 1111, 11111 are not divisible by 7 but that 111111 = 15873 × 7. So 2004 4008 digits. Reading from left to right, the any number in which all the digits are the same is divisible by 7 if the number of digits is a Q23 first 2003 digits are all 2s, the next digit is n multiple of 6. Therefore the 2004-digit number 888…888 is a multiple of 7. Similarly, the and the last 2004 digits are all 8s. 2004-digit number 222…222 is a multiple of 7, as is the 2004-digit number 222…229 since it What is the value of n? differs from 222…222 by 7. Furthermore, the 2004-digit number 222…22n is not a multiple A 4 B 5 of 7 if n is any digit other than 2 or 9. Now N = 222…22n × 102004 + 888…888, so if N is C 0 or 3 D 2 or 9 divisible by 7 then 222…22n is divisible by 7 and we deduce that n = 2 or 9. E 1 or 8

9 Positive integers x and y satisfy the equation: Answer C 2004 Q25 x + 1 y – x – 1 y = 1. 1 1 2 2 x  y – x – y 1. 2 2 Which of the following is a possible value of 1 1 1 y? Therefore x + y – 2 x 2 – y  x – y 1, A 5 B 6 2 4 2 C 7 D 8 1 that is 2x − 1 = 2 x 2 – y . E 9 4 1 Therefore 4x2 − 4x + 1 = 4 (x2 − y ),that is y = 4x − 1. 4 So y must be 1 less than a multiple of 4. Of the values offered, the only possibility is 7. This gives x = 2, but it is necessary to check that these values are solutions of the original equation since they were derived from an argument which involved squaring equations. It is not difficult to show that x = 2, y = 7, satisfy the second equation in the above solution. We now need to confirm that 1 1 2  7 – 2 – 7 does indeed equal 1 rather than –1, which we are able to do since 2 2 1 1 it is clear that 2  7 is greater than 2 – 7 . 2 2

10 What is the sum of the values of n for which Answer E 2005 n 2 – 9 Note that n2 − 1 is divisible by n − 1. Thus: Q21 both n and are integers? n – 1 n2 – 9 n2 – 1 8 8  –  n 1 – (n 1). A −8 B −4 n –1 n – 1 n – 1 n – 1 C 0 D 4 n 2 – 9 E 8 So, if n is an integer, then is an integer if and only if n − 1 divides exactly into 8. The n – 1

possible values of n − 1 are −8, −4, −2, −1, 1, 2, 4, 8, so n is −7, −3, −1, 0, 2, 3, 5, 9. The sum of these values is 8. (Note that the sum of the 8 values of n – 1 is clearly 0, so the sum of the 8 values of n is 8.)

11 Given that S = (x + 20) + (x + 21) + (x + 22) Answer C 2005 +…+ (x + 100), where x is a positive integer, 1 Q22 There are 81 terms in the series, so, using the formula S  n(a  l) for an arithmetic what is the smallest value of x such that S is 2 a perfect square? series: A 1 B 2 81 C 4 D 8 S = (x + 20 + x + 100) = 81 (x + 60) . 2 E 64 Now 81 is a perfect square, so S is a perfect square if and only if x + 60 is a perfect square. As

x is a positive integer, the smallest possible value of x is 4.

12 The factorial of n, written n!, is defined by n! Answer D 2005 = 1 × 2 × 3 ×…× (n − 2) × (n − 1) × n. When n! is written in full, the number of zeros at the end of the number is equal to the Q24 For how many positive integer values of k power of 5 when n! is written as the product of prime factors, because there is at least that less than 50 is it impossible to find a value of high a power of 2 available. For example, 12! = 1 × 2 × 3 × … × 12 = 210 × 35 × 52 × 7 × 11. n such that n! ends in exactly k zeros? This may be written as 28 × 35 × 7 × 11 × 102, so 12! ends in 2 zeros, as 28 × 35 × 7 × 11 is not A 0 B 5 a multiple of 10. C 8 D 9 We see that 24! ends in 4 zeros as 5, 10, 15 and 20 all contribute one 5 when 24! is written E 10 as the product of prime factors, but 25! ends in 6 zeros because 25 = 5 × 5 and hence contributes two 5s. So there is no value of n for which n! ends in 5 zeros. Similarly, there is no value of n for which n! ends in 11 zeros since 49! ends in 10 zeros and 50! ends in 12 zeros. The full set of values of k less than 50 for which it is impossible to find a value of n such that n! ends in k zeros is 5, 11, 17, 23, 29, 30 (since 124! ends in 28 zeros and 125! ends in 31 zeros), 36, 42, 48.

British Maths Olympiad Legal note: While model solutions can be purchased online, I have not accessed these, and all solutions (at least what I’ve got round to so far!) are my own, unless their source is otherwise stated.

Question Solution Ref 1 One number is removed from the set of Yet to attempt. 2010 integers from 1 to n. The average of the remaining numbers is 40.75. Which integer was removed?

2 Determine the sequences a0, a1, a2, … which Yet to attempt. 2008 satisfy all of the following conditions:

11. for every integer , 12. is a rational number and 13. for some I, j with 3 Find all positive integers n such that both Yet to attempt. 2008 n+2008 divides n2 + 2008 and n+2009 divides n2 + 2009. 4 Find four prime numbers less than 100 JAF’s solution: (332 – 232) = (316 + 216)(316 – 216) 2006 which are factors of 332 – 232. And continuing to factorise the difference of two squares we get: (316 + 216) (38 + 28) (34 + 24) (32 + 22)(3 + 2)(3 – 2) That gives us our first prime factors of 5 and 13 immediately. 34 + 24 = 81+16 = 97 which is also prime. 38 + 28 is 812 + 162 = 6561 + 256 = 6817 = 17 x 401. So 17 is our fourth prime factor, and we’re done. 5 Let n be an integer. Show that, if Yet to attempt. 2006 is an integer, then it is a perfect square. 6 Each of Paul and Jenny has a whole number JAF’s solution: The equations are: p + 3 = nj and j + n = 3p. Rearranging we get: 2004 of pounds.

He says to her: “If you give me £3, I will have n times as much as you.” And we’re interested in integers n when j is an integer. Note that 3j-1 grows more rapidly She says to him: “If you give me £n, I will than j+9, so we’ll find eventually as j increases that 3j – 1 > j + 9 and we can no longer have have 3 times as much as you”. integer solutions (since the denominator is bigger than the numerator). Integer solutions Given that all these statements are true and can only exist when 3j – 1 ≤ j + 9, which gives j ≤ 5. This is quite handy as there’s now only 5 that n is a positive integer, what are the values of j we need to try! It works when j=1, giving us n=5 and p=2. It also works when j=5, possible values for n? giving us n=1 and p=2. And we’re done. 7 Determine the least possible value of the Yet to attempt. 2004 largest term in an arithmetic progression of seven distinct primes. 8 Let s be an integer greater than 6. A solid Yet to attempt. 2010 cube of side s has a square hole of side x<6 drilled directly through from one face to the opposite face (so the drill removes a cuboid). The volume of the remaining solid is numerically equal to the total surface area of the remaining solid. Determine all possible integer values of x. 9 Determine the least natural number n for Yet to attempt. 2004 which the following result holds: No matter how the elements of the set {1, 2, … , n} are coloured red or blue, there are integers, x, y, z, w in the set (not necessarily distinct) of the same colour such that x + y + z = w. 10 Find, showing your method, a six-digit JAF’s solution: Suppose the first 3 digits are a, b and c. Then the number is 100,000a + 1993 integer n with the following properties: (i) n 10,000b + 1,000c + 100a + 10b + c + 1 = 1001(100a + b + c) + 1, which needs to be a square is a perfect square, (ii) the number formed number. by the last three digits of n is exactly one Thus for some integer n, 1001(100a + 10b + c) = n2 – 1 = (n+1)(n-1). greater than the number formed by the first To finish writeup, but I found the numbers to be 183184, 328329, 528529 and 715716. The three digits of n. (Thus n might look like number squared in each case is 428, 573, 727 and 846. 123124, although this is not a square.)

11 Let m = (4p − 1)/3, where p is a prime Yet to attempt. 1993 number exceeding 3. Prove that 2m−1 has Round remainder 1 when divided by m. 2 12 Starting with any three digit number n (such Yet to attempt. 1994 as n = 625) we obtain a new number f(n) which is equal to the sum of the three digits of n, their three products in pairs, and the product of all three digits. (i) Find the value of n/f(n) when n = 625. (The answer is an integer!) (ii)Find all three digit numbers such that the ratio n/f(n)=1. 13 How many perfect squares are there (mod Solution from http://schoolexercisebooks.blogspot.co.uk/2010/09/30th-british- 1994 2n)? mathematical-olympiad-1994.html Round Answer: n odd (2n-1 + 5)/3, n even (2n-1 + 4)/3. 2 We find the first few: 2: 0, 1 4: 0, 1 8: 0, 1, 4 16: 0, 1, 4, 9 32: 0, 1, 4, 9, 16, 17, 25 64: 0, 1, 4, 9, 16, 17, 25, 33, 36, 41, 49, 57

a2 = b2 mod 2n iff (2a)2 = (2b)2 mod 2n+2 (JAF note: see Key Theory), so the number of even squares mod 2n+2 equals the number of squares mod 2n. This suggests looking separately at n the odd and even squares. Let un be the number of odd squares mod 2 and vn be the number of even squares mod 2n. (2a + 1)2 = (2b + 1)2 mod 2n is equivalent to 4a2 + 4a = 4b2 + 4b mod 2n, which is equivalent to a2 + a = b2 + b mod 2n-2 or (a - b)(a + b + 1) = 0 mod 2n-2. But a - b and a + b + 1 have sum 2a + 1 which is odd, so they have opposite , so either a - b = 0 mod 2n-2 or (a + b + 1) = 0 mod 2n-2. Thus just 3 other odd numbers have the same square as (2a + 1), namely (2a + 2n-1 + 1), -(2a + 1) and -(2a + 2n-1 + 1). So there are 2n-3 different odd squares mod 2n. In other n-3 words, vn = 2 . Now un = un-2 + vn-2 = un-4 + vn-4 + vn-2 etc. So the total number of squares u2m + v2m = v2m + v2m- 2m-3 2m-5 m-2 m-3 m-1 2 + v2m-4 + ... + v4 + u4 = 2 + 2 + ... + 2 + 2 = 2(4 + 4 + ... + 1) + 2 = 2(4 - 1)/3 + 2 = (2n-1 + 4)/3. Similarly, the total number of squares for n odd is (2n-1 + 5)/4. 14 Find the first integer n > 1 such that the Yet to attempt. 1994 average of 12, 22, 32, . . . , n2 Round is itself a perfect square. 2 15 Find all solutions in positive integers a, b, c Solution here: http://sciencecentral1234.blogspot.co.uk/2012/02/two-solutions-to- 2002/ to the equation problem-5-from-bmo-1.html (the unique solution is a=3, b=3, c=4). 3 a! b! = a! + b! + c!