Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 Mock AMC 10B Problems/Problem 22"
(→‎Problem)
Line 2: Line 2:
  
 
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 <math>n</math> is a positive integer and <math>\mu</math> is the number of elements in <math>S</math>. The sum <math>r_1 + r_2 + r_3 + ... + r_{\mu}</math> can be expressed as<cmath>p^qr,</cmath>where <math>p, q, r</math> are positive integers and <math>p</math> and <math>r</math> are as small as possible. Find <math>p+q+r</math>.
 
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 <math>n</math> is a positive integer and <math>\mu</math> is the number of elements in <math>S</math>. The sum <math>r_1 + r_2 + r_3 + ... + r_{\mu}</math> can be expressed as<cmath>p^qr,</cmath>where <math>p, q, r</math> are positive integers and <math>p</math> and <math>r</math> are as small as possible. Find <math>p+q+r</math>.
 +
 +
<math>\textbf{(A)}\ 40\qquad\textbf{(B)}\ 41\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 43\qquad\textbf{(E)}\ 44</math>
  
 
==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 > 0</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>.
 
<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 > 0</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 19:30, 3 November 2019

Problem

Let $S = \{r_1, r_2, r_3, ..., r_{\mu}\}$ be the set of all possible remainders when $15^{n} - 7^{n}$ is divided by $256$, where $n$ is a positive integer and $\mu$ is the number of elements in $S$. The sum $r_1 + r_2 + r_3 + ... + r_{\mu}$ can be expressed as\[p^qr,\]where $p, q, r$ are positive integers and $p$ and $r$ are as small as possible. Find $p+q+r$.

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 41\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 43\qquad\textbf{(E)}\ 44$

Solution

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_22&oldid=110854"