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 cancella- tions 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 fol- lowing types, leading to correct results, allowing perhaps further simplifications.
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