PUMaC 2021 · 数论（B 组） · 第 1 题

PUMaC 2021 — Number Theory (Division B) — Problem 1

Andrew has a four-digit number whose last digit is 2. Given that this number is divisible by 9 , determine the number of possible values for this number that Andrew could have. Note that leading zeros are not allowed.

解析

Andrew has a four-digit number whose last digit is 2 . Given that this number is divisible by 9 , determine the number of possible values for this number that Andrew could have. Proposed by: Frank Lu Answer: 100 It suffices to find the smallest and largest four-digit numbers that satisfy these conditions, because any two such numbers differ by a multiple of 90 . We recall that an integer is divisible by 9 if and only if the sum of the digits is divisible by 9 . So for the smallest integer, this is 1062 . For the largest, we note that the largest possible sum of digits is 27 . This yields the integer 9972 . Their difference is 8910 , and dividing by 90 yields 99 . Hence, there are 100 such numbers.