Difference between revisions of "2019 Mock AMC 10B Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | S15^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 = \ | + | S15^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>, so the answer is </math>2 + 7 + 31 = \boxed{\text{(A)} 40}$. |
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 as
where
are positive integers and
and
are as small as possible. Find
.
Solution
S15^n - 7^n =\equiv 7^n - 7^n \equiv 0$$ (Error compiling LaTeX. Unknown error_msg)\text{mod}$$ (Error compiling LaTeX. Unknown error_msg)8n
S = \{0, 8, 16, 24,...,248\}
S
8 + 16 + 24 + ... + 248 = 8(1 + 2 + 3 + ... + 31) = 8 \cdot \frac{31(31 + 1)}{2} = 2^7 \cdot 31
2 + 7 + 31 = \boxed{\text{(A)} 40}$.