Difference between revisions of "2019 Mock AMC 10B Problems/Problem 22"
(Created page with "==Problem== Let <math>S = \{r_1, r_2, r_3, ..., r_{\mu}\}</math> be the set of all possible remainders when <math>15^{n} - 7^{n}</math> is divided by <math>256</math>, where...") |
|||
| Line 5: | Line 5: | ||
==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 = \boxed{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, so the answer is <math>2 + 7 + 31 = \boxed{\text{(A)} 40} | + | 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 = \boxed{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 19:24, 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![\[p^qr,\]](http://latex.artofproblemsolving.com/5/7/e/57ef78a97f88f2c6b8fe060205ea30c4927b9520.png)
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)8
n
S = \boxed{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}$.
