FOR KIDS WHO LIKE to SOLVE PROBLEMS #2 MAGAZINE Mpower! Our STAFF

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FOR KIDS WHO LIKE TO SOLVE PROBLEMS #2 MAGAZINE MPower! Our STAFF

Editor-in-Chief Laura Gallus CONTENTS Contributing Editor Nina Dubinsky

Contributing Author Andrey Polin The Three Princesses Art Direction by 1

Factors & Faulty Lights 3 KEY Grades 2 & up Grades 4 & up Amusements 5 Grades 6 & up Master Solver The Secret Hideout 8 Note from the EDITOR MChallenge 10 There is no such thing as a “math person.” However, a child’s early learning experiences directly impact their perception of their Pythagorean Means mathematics ability. When children 12 are taught mathematics through fun and engaging problems, they not only build fluency in and an intuitive understanding of mathematics, but they also develop critical problem- solving skills.

MPower! was created as a resource for children who want to flex their mathematical muscles and show off their problem-solving ingenuity. We hope you enjoy the stories and problems inside.

The RSM Foundation is committed to the promotion of STEM activities for K-12 students. With the belief that Mathematics is the queen of the sciences, the Foundation advocates for ANSWERS the adoption of world-class standards and supports the Want to know if your implementation of innovative mathematical programs in a diverse context across the United States and abroad. The RSM answers are correct? Foundation strives to develop world leading intellectuals in All answers can be found online at: STEM fields, who will compete and collaborate across the www.rsmfoundation.org/mpower country. Connect with us at: ©2016 RSM Foundation. All Rights Reserved. No part of this work may be [email protected] reproduced or distributed in any form or by any means. The Three Princesses 1

The Three Princesses King Albert and King Edward wanted to unite to Prince John. However, he did not tell the their kingdoms through the marriage of King prince who is the eldest, middle, and youngest. Albert’s son, Prince John, and one of King Edward’s three daughters. King Edward invited Now, the prince intended to marry a princess he King Albert and Prince John to his castle could trust. Therefore, he decided that it must so that Prince John could choose one of the be either the eldest princess or the youngest princesses to marry. princess (he figures he can simply negate all of her responses). The prince was very clever, and It was common knowledge that the eldest he knew he could make his choice by simply princess always tells the truth, the youngest asking one of the princesses a single yes or no princess never tells the truth, and the middle question. With the three princesses standing princess, who is quite mischievous, tells the before him, he asked his question. He then truth only some of the time. King Edward made his choice. presented his daughters Princess Catherine, Princess Regina, and Princess Elizabeth What question did he ask?

Faulty Lights Andrey Polin

The holiday season was fast approaching, and Proof Miss Snow, who loved the season more than Let p be a factor of n. Then r = n/p is also factor anyone, decorated the beautiful tree in her of n. For this and each next factor we have the yard with a string of 300 lights. She turned pair (p, r). the lights on and marveled at the sight. At midnight, however, she noticed something 1. If the number of the factors is an odd number, strange happen—bulbs started blinking on it means the factor a does not pair with another or off every second. In the st1 second, every factor. However, that happens if n = a2. bulb was on; in the 2nd second every second 2. Otherwise, if n = a2, then for the factor a, the bulb, starting with bulb #2, blinked off; in pair is (a, n/a = a). So the two factors in this pair the 3rd second, every third bulb, starting from are the same and counted as one factor. Other bulb #3, blinked off or on (some bulbs that were off blinked on again—for example, bulb pairs have two different factors. It means the #6 blinked on again); in the 4th second, every number of factors is an odd number. fourth bulb, starting with bulb #4 blinked off or on; and so on. This continued for a Back to the faulty lights while, and then stopped. At what time did the In the pattern, each second corresponds to a blinking stop? How many bulbs on the string bulb number. Since there are 300 bulbs, the of lights were still on? pattern continues for 300 seconds, or 5 minutes. So, the lights stop blinking at 12:05 A.M. A little math on the side How many factors does a natural number We choose, for example, bulb #52. When does have? Every natural number, except 1, has at this bulb blink on or off? It happens in the st1 least two factors: 1 and the number itself. For second (bulb is on), in the 2nd second (bulb is example, 2 has two factors (1 and 2), 6 has th th four factors (1, 2, 3, 6), 9 has 3 factors (1, 3, 9), off), in the 4 second (bulb is on), in the 13 th 24 has 8 factors (1, 2, 3, 4, 6, 8, 12, 24). Note second (bulb is off), in the 26 second (bulb is that the numbers 2, 6, and 24 each have an on), and, finally, in the 52nd second (bulb is off). even number of factors and the number 9 has So we get 1, 2, 4, 13, 26, and 52—all the factors an odd number of factors. Without factoring, of 52. For each bulb number we get the same how can we determine if a given number has result in terms of factors. That is, the bulb will an even or an odd number of factors? only blink on or off when the number of seconds corresponds to a factor of the bulb number. We see that, for the numbers 2, 6, and 24, each factor of these numbers pairs with So, how many bulbs on the string of lights are another factor. However, the factor 3 of the still on? If the bulb number has an even number number 9 does not pair with another factor. of factors, the bulb is off. If the bulb number has an odd number of factors, the bulb is on. We A natural number n has an odd number of know that only square numbers have an odd factors if and only if n is a square number 2 2 (i.e. n = a2, where a is a natural number). number of factors. Therefore, since 1 = 1 , 4 = 2 , 9 = 32,…, 225 = 152, 256 = 162, 289 = 172, we have

What if the string of lights has 290 bulbs? 320 bulbs? How many of the bulbs will still be on at the end of the pattern? According to our solution, a bulb will be on if the bulb number is a square number. Since both numbers are greater than or equal to 289, but less than 324 (the next square number), we know that only 17 bulbs will still be on for both of these strings of lights.

Alternate blinking patterns In each second of the original pattern some bulbs blinked on or off. We can think of each second as a step of the pattern. That is, in each step of the pattern some bulbs blink on or off. What happens if we restrict when the bulbs blink on or off? The bulbs will still blink on or off when the number of seconds corresponds to a factor of the bulb number.

Pattern 1 In pattern 1, bulbs blink on or off only in even seconds. So, Step 1 is the 2nd second, Step 2 is the 4th second, and so on. How many of the 300 bulbs will be on at the end of pattern 1?

Solution Note that when the number of seconds is an odd number, nothing happens. So the odd-numbered bulbs will never blink on. Now we look closely at the numbers of seconds at which the steps happen:

Step 1 2nd second Step 2 4th second Step 3 6th second

Step 150 300th second

We can see that a step number is a factor of the corresponding number of seconds. It means that, for each step, the bulb will blink on or off if the step number corresponds to a factor of the bulb number. That’s because a factor of a factor of a number is also a factor of the number.

From our solution of the original problem, a bulb From the solution of the original problem and the will be on at the end of the new pattern if and only factorization above we can see that the answer is if the step number is a square number. We have obtained by finding alln = 2k × d, where k ≥ 0 and 150 steps and only 12 square numbers within the d is the square of an odd number (1 = 12, 9 = 32, step numbers. So, only 12 bulbs are still on at the 25 = 52, …, 169 = 132, 225 = 152, 289 = 172). So for end of the new pattern. Can you find which bulbs 300 bulbs we have: are still on? 1: 20×1=1, 2×1=2, 22×1=4, …, 28×1=256 Pattern 2 9: 20×9=9, 21×9=18, 22×9=36, …, 25×9 = 288 0 1 3 In pattern 2, bulbs blink on or off only in odd 25: 2 ×25=25, 2 ×25=50, … , 2 ×25=200 seconds. So, Step 1 is the 1st second, Step 2 is the 0 3rd second, and so on. How many of the 300 bulbs 169: 2 ×169=169 0 will be on at the end of pattern 2? 225: 2 ×225=225 289: 20×289=289 (Note that in pattern 2, a step number is not a factor of the corresponding number of seconds, So, the answer is 29 bulbs. except in the first step. So we cannot apply a solution similar to the solution of pattern 1.) PROBLEM 1 Suppose the faulty string of lights Miss Snow put Solution Factor any bulb number. For example, on her tree had 1000 bulbs. At what time would let’s choose the bulb #30. The factors of 30 are: the blinking have stopped? How many bulbs 1, 2, 3, 5, 6, 10, 15, and 30. According to the new would have still been on? pattern, the bulb blinks on or off at 1, 3, 5, and 15 seconds only. Note that 30 = 2 × 15 and we can check that 1, 3, 5 and 15 are all the possible factors PROBLEM 2 of 15. Let’s factor another bulb number—for Suppose 15 bulbs are still on at the end of the example, the bulb #112. The factors of 112 are: 1, pattern in our original problem. How many 2, 4, 7, 16, 28, 56, and 112. This bulb blinks on or bulbs are on the string of lights? Find all possible off at 1 second and 7 seconds only. We have 112 = answers. 24 × 7 and 1 and 7 are all the factors of 7.

In the examples above, we wrote each bulb PROBLEM 3 number as the product of a power of 2 and an odd Suppose we have a new pattern in which a step number. For each bulb, we see that it blinks on corresponds to a number of seconds that is a or off in the number of seconds that corresponds power of 2. That is, Step 1 is at the 20 = 1st second, to a factor of the odd number factor. This means Step 2 is at the 21 = 2nd second, Step 3 is at the 22 that for every natural number n, we can factor n = 4th second, and so on. How many of the 300 as n = 2k × d (where k ≥ 0 and d is an odd number) bulbs will be on at the end of this pattern? and take the factors of d to find the step numbers for the bulb number n. Examples: 4 = 22 × 1, 111 = (Hint: How many steps does the new pattern have? 20 × 111, 234 = 2 × 117 and so on. Note that d can If the number of steps is a relatively small number— be either prime or composite odd number. All the for example, less than 10—we can find the answer factors of this odd number d are all the steps at just by applying this pattern step by step.) which this bulb blinks on or off.

Amusements

1) My birthday is 2) Find the digits A, B, C, 45 days from today. and D in the multiplication My sister’s birthday problem. was the day before A B C D yesterday. If my x 4 sister’s birthday was on D C B A Monday, then on what day of the week will my birthday be?

3) How many three-digit multiples of 3 can you make using only the digits 0, 1, 2, 3, and 4, if no digits are repeated in a number?

3) If A:B=2:3 and A:C = 4) How many 3:2, then what is B:C? squares are on a checkerboard?

5) A snail is climbing up a wall. Each day it climbs 3 feet up, but each night it slides 1 foot down. If the wall is 12 feet high, then how many days does it take the snail to climb to the top of the wall? © 2015 Foundation RSM Amusements 7

7) On a coordinate plane, how many points with integer coordinates are exactly 5 units from 6) At 9:00 AM Julia told her the origin? friend Beth a secret. By 9:05 AM, Beth had told two other students Julia’s secret. By 9:10 AM, each of those students had told two more students the secret. This continued so that every five minutes each student who was told the secret had told it to two other people. All the students in the school knew the secret by 9:35 am. Assuming that each student told exactly two other students the secret and each student was told only once, then how many students attend this school? © 2015 Foundation RSM 8 The Secret Hideout

The Secret Hideout A well-known secret math society had a special hideout. Only those who knew the password could enter the hideout.

Oliver was a curious boy. Soon a girl came along and He wondered what the secret knocked on the door. The voice math society did and why they said, “Six.” The girl replied, needed a hideout. So, one day, “Three.” The door opened and he decided he would find out for the girl went inside. Then a boy himself. came along and knocked on the door. The voice on the other side He knocked on the door to the said, “Twelve.” The boy answered, secret hideout. A voice on the “Six.” The door opened and the other side said, “Eight.” boy went inside.

“Eight?” he replied. “Oh! Now I know the password!” exclaimed Oliver. He quickly “That’s not the password! You walked to the door and knocked. can’t get in unless you know the The voice inside said, “Eight.” password,” said the voice on the Oliver confidently replied, “Four.” other side. The voice said, “I’m sorry, that is Oliver scratched his head. He was not the password.” confused and didn’t know what to do. So he went home to think What password should Oliver about it. have given to the voice behind the door? The next day, he returned to the hideout. However, this time he waited, watched, and listened.

QUESTION 1 QUESTION 2 Anna, Joe, and Ezra have some pencils. If Ezra How many blocks were used to make the gives 7 pencils to Anna, then Anna and Ezra tower below? will have the same number of pencils. If Anna gives 9 of her pencils to Joe, then Joe and Anna will have the same number of pencils. How many more pencils than Joe does Ezra have?

QUESTION 3 QUESTION 4 x+y What is the difference between the largest and If x ∆ y = xy , then what is the value of the smallest three-digit numbers that leave a 1 ∆ (3 ∆ 6)? remainder of 1 when divided by 2, 3, or 4? © 2015 Foundation RSM MChallenge 11

QUESTION 5 QUESTION 6 Two congruent equilateral triangles are If 4 is the first number in a pattern and 24 is overlapped to form a regular hexagon as shown the fifth number in the pattern, then what is in the figure. If the area of the hexagon is 24 the 100th number in this pattern? square centimeters, then what is the area of one of the overlapping triangles?

QUESTION 7 QUESTION 8 Determine the area of the region enclosed by Find all real values of x such that the graphs of y = a|x| – b and y =–a|x| + b, (x2 + x - 1)x2-2x-3 = 1 where a and b are real numbers (a ≠ 0). © 2015 Foundation RSM 12 Pythagorean Means

Pythagorean Means Definition For two numbers a and b*: Andrey Polin and Laura Gallus Arithmetic mean A is a − A = A − b

The word “average” is widely known. When we use Geometric mean G is a = G G b it, we usually have in mind the arithmetic mean. Harmonic mean H is a − H H − b But, did you know that the arithmetic mean is only = a b one of three classical Pythagorean means? The These formulas are more commonly written as other two are the mean proportional, or geometric a + b 2ab mean, and the harmonic mean. A = ; G = ab ; and H = . 2 a + b

Pythagoras, Harmony, and Means For example, the Pythagorean means For ancient Greeks, the notions of beauty, proportion, and for a = 4 and b = 9 are: 4 + 9 13 1 harmony were closely related. An object’s beauty resulted A = = = 6 from a harmony (or wholeness) created from its parts being 2 2 2 = ⋅ = = joined together in proper proportion. Pythagoras of Samos G 4 9 36 6 2 ⋅ 4 ⋅ 9 72 7 (569-475 BC), a Greek philosopher and mathematician H = = = 5 + most known for the theorem named for him, described 4 9 13 13 musical harmony through proportions. In fact, Pythagoras Comparing the means discovered the music intervals of the fourth, the fifth, and the In the example above, we see that the arithmetic mean octave. According to legend, while passing by a blacksmith’s A has the greatest value and the harmonic mean H has shop, Pythagoras stopped to listen to the workmen pound the least value. metal against an anvil with hammers of different weights. He noticed that the pounding of the hammers produced In fact, for any two numbers, H ≤ G ≤ A. harmonious sounds. He entered the shop and examined the hammers, noting that the weights were in the ratio of 6, 8, 9, We will prove this statement after we prove the and 12 pounds. He experimented further, investigating the following theorem about right triangles: relationship between the ratio of the length of a vibrating string and the musical tone it produced. When a string of Right Triangle Altitude Theorem length 12 was shortened to length 9, a fourth was heard (9:12 In a right triangle, the measure of the altitude to the = 3:4); if it was shortened to length 8, a fifth was heard (8:12 = hypotenuse is the geometric mean between the 2:3); and if it was shortened to length 6, an octave was heard measures of the two segments into which the altitude (6:12 = 1:2). divides the hypotenuse. That is, CD = AD⋅ DB .

In addition to the musical relationship, the numbers 6, 8, 9, and 12 have an arithmetic relationship. The number 9 is the arithmetic mean of 12 and 6, and the number 8 is the harmonic mean of 12 and 6. (See Problem 2.)

Later, Archytas (428–347 BC), a follower of Pythagoras’ teachings, defined the Pythagorean means.

*The Pythagorean means can be found for more than two numbers. 13

Proof: 2) Geometric mean: From the right triangle altitude

In ∆ABC, m∠C = 90° and CD ⊥ AB. theorem, we see that for the right ∆ABC: Let’s take right triangles ∆ACD and ∆CBD. AD CD = or CD = AC ⋅CB = a ⋅ b . CD DB 1) ; m∠ADC = m∠CDB = 90° 3) Harmonic mean: Finally, from the similar right 2) ∠ = ∠ = °− ∠ ; and CE CD m CAD m BAC 90 m ABC triangles, ∆CDO and ∆CED, = , or m∠BCD = 90°− m∠CBD = 90°− m∠ABC . So CD OC 2 m∠CAD = m∠BCD ; CD2 ( a ⋅ b) 2ab CE = = = 3) Similarly, we can show m∠ACD = m∠CBD . OC a + b a + b . 2 From 1-3 we get that ∆ACD is similar to ∆CBD. Since We now complete our proof. In a right triangle, corresponding sides are proportional, we have the length of a leg is less than the length of the hypotenuse. Therefore, CE < CD (for DEC) and CD < AD CD ∆ = , or = ⋅ OC (for OCD). Combining both inequalities, we have CD DB CD AD DB . ∆ DE < DC < OD, or H < G < A.* (See Problem 1.) Now, we will give a geometric proof of our previous statement: Problem 1 We promised to prove that H ≤ G ≤ A. However, we For any two numbers, H ≤ G ≤ A. proved that H < G < A. Finish the proof by stating the condition for which H = G = A. Proof: Let’s take two numbers a and b. Draw a semi- circle with center O and diameter AB = AD + DB, Problem 2 where AD = a and DB = b. Next, draw line segment CD (a) Show that 9 is the arithmetic mean of 12 and 6. perpendicular to diameter AB. Finally, draw radius OC and line segment DE perpendicular to OC. (b) Show that 8 is the harmonic mean of 12 and 6.

Before we continue, see if you can find the lengths that Problem 3 are the arithmetic mean, the geometric mean, and the Can you find a triangle where the length of one side is harmonic mean of a and b in the figure below. the arithmetic mean, geometric mean and harmonic mean of the other two side of the triangle?

Problem 4 Prove H ≤ G ≤ A algebraically. (Hint: First show a + b 2ab ≥ ab , and then show ab ÷ ≥1 .) 2 a + b

1) Arithmetic mean: Since the line segment OC is the AB a + b radius, OC = = 2 2 At the RSM Foundation, we believe that there is no such thing as a “math person.” When introduced logically, slowly, and with excitement, all children can learn math and appreciate its beauty and importance in the world around them.

But what is math? It’s more than numbers and calculation. From identifying the fastest line in the supermarket to helping a knight slay a dragon with instantly regenerating heads, math is a pathway to higher-level thinking and reasoning. It provides the tools you need to analyze, understand, and solve complex problems. It plays a critical role in the development of logical thinking in young minds.

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