Question

Positive integer divisors $a$ and $b$ of $N$ are called complementary if $ab = N.$ Given that $N$ has a pair of complementary divisors that differ by $20$ and a pair of complementary divisors that differ by $23,$ find the sum of the digits of $N.$

$11$

$13$

$15$

$17$

$19$

Solution:

Complementary divisors differing by $20$ are $b$ and $b + 20$ with product $N,$ so $N = b^2 + 20b$ and $N + 100 = (b + 10)^2.$ A pair differing by $23$ gives $4N + 529 = (2d + 23)^2.$ Set $N + 100 = k^2.$ Then $4k^2 + 129 = m^2,$ so $(m - 2k)(m + 2k) = 129 = 3 \cdot 43.$ Take the factorization $1 \cdot 129:$ it gives $m = 65,$ $k = 32,$ hence $N = 32^2 - 100 = 924.$ Check it: $924 = 22 \cdot 42 = 21 \cdot 44,$ and the digit sum is $9 + 2 + 4 = 15.$ Thus, C is the correct answer.

2023 AMC 10A Problem 23

Solution:

Problem 23 in Other Years