# Difference between revisions of "2019 Mock AMC 10B Problems/Problem 22"

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==Solution== | ==Solution== | ||

− | + | <math>15^n - 7^n \equiv 7^n - 7^n \equiv 0</math> <math>\text{mod}</math> <math>8</math> for all integer <math>n</math>. Therefore, <math>S = \{0, 8, 16, 24,...,248\}</math>. Since the sum of the elements in <math>S</math> is <math>8 + 16 + 24 + ... + 248 = 8(1 + 2 + 3 + ... + 31) = 8 \cdot \frac{31(31 + 1)}{2} = 2^7 \cdot 31</math>, the answer is <math>2 + 7 + 31 = \boxed{\text{(A)} 40}</math>. |

## Revision as of 19:25, 3 November 2019

## Problem

Let be the set of all possible remainders when is divided by , where is a positive integer and is the number of elements in . The sum can be expressed aswhere are positive integers and and are as small as possible. Find .

## Solution

for all integer . Therefore, . Since the sum of the elements in is , the answer is .