Question

For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9,$ and $10.$ For example, $R(15) = 1 + 0 + 3 + 0 + 3 + 1 + 7 + 6 + 5 = 26.$ How many two-digit positive integers $n$ satisfy $R(n) = R(n + 1)?$

$0$

$1$

$2$

$3$

$4$

Solution:

Going from $n$ to $n + 1,$ each remainder $n \bmod m$ increases by $1$ unless $m \mid n + 1,$ in which case it drops from $m - 1$ to $0.$ So $R(n+1) - R(n) = 9 - \sum_{\substack{2 \le m \le 10 \\ m \mid n+1}} m.$

We need those divisors to sum to $9.$ If $n + 1$ is divisible by $3, 4, 5, 6, 8, 9,$ or $10,$ it picks up additional small divisors that push the sum past $9,$ so the only workable case is $n + 1$ divisible by $2$ and $7$ but no other value in $\{2, \ldots, 10\},$ giving $2 + 7 = 9.$

Among the two-digit $n,$ this means $n + 1 = 14$ or $n + 1 = 98,$ so $n = 13$ or $n = 97.$ That is $2$ values.

Thus, the correct answer is C.

2021 AMC 12B Fall Problem 25

Solution:

Problem 25 in Other Years