Digit Problems

Chapter 1 Digit problems 1.1 When can you cancel illegitimately and yet get the correct answer? Let ab and bc be 2-digit numbers. When do such illegitimate cancellations as ab ab a bc = bc6 = c , 6 a allowing perhaps further simplifications of c ? 16 1 19 1 26 2 49 4 Answer. 64 = 4 , 95 = 5 , 65 = 5 , 98 = 8 . Solution. We may assume a, b, c not all equal. 10a+b a Suppose a, b, c are positive integers 9 such that 10b+c = c . (10a + b)c = a(10b + c), or (9a + b≤)c = 10ab. If any two of a, b, c are equal, then all three are equal. We shall therefore assume a, b, c all distinct. 9ac = b(10a c). If b is not divisible− by 3, then 9 divides 10a c = 9a +(a c). It follows that a = c, a case we need not consider. − − It remains to consider b =3, 6, 9. Rewriting (*) as (9a + b)c = 10ab. If c is divisible by 5, it must be 5, and we have 9a + b = 2ab. The only possibilities are (b, a)=(6, 2), (9, 1), giving distinct (a, b, c)=(1, 9, 5), (2, 6, 5). 102 Digit problems If c is not divisible by 5, then 9a + b is divisible by 5. The only possibilities of distinct (a, b) are (b, a) = (3, 8), (6, 1), (9, 4). Only the latter two yield (a, b, c)=(1, 6, 4), (4, 9, 8). Exercise 1. Find all possibilities of illegitimate cancellations of each of the following types, leading to correct results, allowing perhaps further simplifications. abc c (a) b6 ad6 = d , 6 6 cab c (b) d6ba6 = d , 6 6 abc a (c) bcd6 6 = d . 6 6 2. Find all 4-digit numbers like 1805 = 192 5, which, when divided by the its last two digits, gives the square× of the number one more than its first two digits. 1.2 Repdigits 103 1.2 Repdigits A repdigit is a number whose decimal representation consists of a rep- etition of the same decimal digit. Let a be an integer between 0 and 9. For a positive integer n, the repdigit an consists of a string of n digits each equal to a. Thus, a a = (10n 1). n 9 − Exercise 1. Show that 16 1 19 1 26 2 49 4 n = , n = , n = , n = . 6n4 4 9n5 5 6n5 5 9n8 8 abn a Solution. More generally, we seek equalities of the form bnc = c for distinct integer digits a, b, c. Here, abn is digit a followed by n digits each equal to b. To avoid confusion, we shall indicate multiplication with the sign . The condition× (ab ) c =(b c) a is equivalent to n × n × b 10b 10na + (10n 1) c = (10n 1)+ c a, 9 − 9 − b 10b (10n 1)a + (10n 1) c = (10n 1) a. − 9 − 9 − 10n 1 Cancelling a common divisor 9− , we obtain (9a + b)c = 10ab, which ab a is the same condition for bc = c . 104 Digit problems Exercise 1. Prove that for 1 a, b 9, a b = b a . ≤ ≤ × n × n 2. Complete the following multiplication table of repdigits. 1n 2n 3n 4n 5n 6n 7n 8n 9n 1 1n 2n 3n 4n 5n 6n 7n 8n 9n 2 4n 6n 8n 1n0 13n 12 15n 14 17n 16 19n 18 − − − − 3 9n 13n 12 16n 15 19n 18 23n 11 26n 14 29n 17 − − − − − − 4 17n 16 2n0 26n 14 31n 208 35n 12 39n 16 − − − − − 5 27n 15 3n0 38n 15 4n0 49n 15 − − − 6 39n 16 46n 12 53n 228 59n 14 − − − − 7 54n 239 62n 216 69n 13 − − − 8 71n 204 79n 12 − − 9 98n 201 − 3. Verify 26n 2 . 6n5 = 5 Solution. It isenoughto verify 5 (26 )=2 (6 5). × n × n 5 (26 )= 5(20 +6 )=10 +3 0=13 0; × n n n n+1 n n 2 (6n5)= 2(6n0+5)=13n 120+10=13n0. × − 4. Simplify (1n)(10n 15). − Answer. (1n)(10n 15)=1n5n. − 1.3Sortednumberswithsortedsquares 105 1.3 Sorted numbers with sorted squares A number is sorted if its digits are nondecreasing from left to right. It is strongly sorted if its square is also sorted. It is known that the only strongly sorted integers are given in the table below. 1 1, 2, 3, 6, 12, 13, 15, 16, 38, 116, 117. • 16 7. • n 3 4. • n 3 5. • n 3 6 7. • m n (3 5 )2 =(10 3 + 5)2 n 1 · n =100 (3 )2 + 100 (3 )+25 · n · n =1n 108n 19102 +3n25 − − =1n 112n 1225 − − =1n2n+15. If x =3m6n7, then 3x = 10m 110n1, and it is easy to find its square. − 2 1m3m4n m+16m8n9, if n +1 m, (3m6n7) = − ≥ 1m3n+15m n 16n+18n9, if n +1 < m. ( − − More generally, the product of any two numbers of the form 3m6n7 is sorted. 1Problem 1234, Math. Mag., 59 (1986) 1, solution, 60 (1987)1. See also R. Blecksmith and C. Nicol, Monotonic numbers, Math. Mag., 66 (1993) 257–262. 106 Digit problems Exercise 1. Find all natural numbers whose square (in base 10) is represented by odd digits only. 2. Find the three 3-digit numbers each of which is equal to the product of the sum of its digits by the sum of the squares of its digits. Answer. 133, 315, 803. 3. Find all 4-digit numbers abcd such that √3 abcd = a + b + c + d. Answer. 4913 and 5832. Solution. There are only twelve 4-digit numbers which are cubes. For only two of them is the cube root equal to the sum of digits. n 10 11 12 13 14 17 16 17 18 19 20 21 n3 1000 1331 1728 2197 2744 3375 4096 4913 5832 6859 8000 9261 ∗∗∗ ∗∗∗ 4. Use each digit 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once to form prime numbers whose sum is smallest possible. What if we also include the digit 0? 5. There are exactly four 3-digit numbers each equal to the sum of the cubes of its own digits. Three of them are 153, 371, and 407. What is the remaining one? 6. Find all possibilities of a 3-digit number such that the three numbers obtained by cyclic permutations of its digits are in arithmetic progression. Answer. 148, 185, 259, 296. Solution. Let abc be one such 3-digit numbers, with a smallest among the digits (which are not all equal). The other two numbers are bca and cab. Their sum abc + bca + cab = 111 (a + b + c). Therefore the middle number = 37 (a + b + c). × × We need therefore look for numbers of the form abc = 37 k with digit sum equal to s, and check if 37 s = bca or cab.× We may ignore multiples of 3 for k (giving repdigits× for 37 k). Note that 3k < 27. We need only consider k =4, 5, 7, 8. × 1.3Sortednumberswithsortedsquares 107 k 37 k s 37 s arithmetic progression 4 148× 13 13 37=481× 148, 481, 814 5 185 14 14 × 37=518 185, 518, 851 7 259 16 16 × 37=592 259, 592, 925 8 296 17 17 × 37=629 296, 629, 962 × 7. A 10-digit number is called pandigital if it contains each of the digits 0, 1,..., 9 exactly once. For example, 5643907128 is pandigital. We regard a 9-digit number containing each of 1,..., 9 exactly once as pandigital (with 0 as the leftmost digit). In particular, the number A := 123456789 is pandigital. There are exactly 33 positive integers n for which nA are pandigital as shown below. n nA n nA n nA 1 123456789 2 246913578 4 493827156 5 617283945 7 864197523 8 987654312 10 1234567890 11 1358024679 13 1604938257 14 1728395046 16 1975308624 17 2098765413 20 2469135780 22 2716049358 23 2839506147 25 3086419725 26 3209876514 31 3827160459 32 3950617248 34 4197530826 35 4320987615 40 4938271560 41 5061728349 43 5308641927 44 5432098716 50 6172839450 52 6419753028 53 6543209817 61 7530864129 62 7654320918 70 8641975230 71 8765432019 80 9876543120 How would you characterize these values of n? 8. Find the smallest natural number N, such that, in the decimal nota- tion, N and 2N together use all the ten digits 0, 1,..., 9. Answer. N = 13485 and 2N = 26970. 108 Digit problems 1.4 Sums of squares of digits Given a number N = a1a2 an of n decimal digits, consider the “sum of digits” function ··· s(N)= a2 + a2 + + a2 . 1 2 ··· n For example s(11) = 2, s(56) = 41, s(85) = 89, s(99) = 162. For a positive integer N, consider the sequence S(N) : N, s(N), s2(N), ...,sk(N), ..., where sk(N) is obtained from N by k applications of s. Theorem 1.1. For every positive integer N, the sequence S(N) is either eventually constant at or periodic. The period has length 8 and form a cycle Exercise n 1 1. Prove by mathematical induction that 10 − > 81n for n 4. ≥ Solution. Clearly this is true for n =4: 103 > 81 4. n 1 · Assume 10 − > 81n. Then n n 1 10 = 10 10 − > 10 81n> 81(n + 1). · · n 1 Therefore, 10 − > 81n for n 4. ≥ 1.4 Sums of squares of digits 109 2. Prove that if N has 4 or more digits, then s(N) < N. Solution. If N has n digits, then n 1 (i) N 10 − , (ii) s(N≥) 81n.

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