Difference between revisions of "2007 AMC 8 Problems/Problem 10"
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== Problem == | == Problem == | ||
− | For any positive integer <math>n</math>, <math>\boxed{n}</math> to be the sum of the positive factors of <math>n</math>. | + | For any positive integer <math>n</math>, define <math>\boxed{n}</math> to be the sum of the positive factors of <math>n</math>. |
For example, <math>\boxed{6} = 1 + 2 + 3 + 6 = 12</math>. Find <math>\boxed{\boxed{11}}</math> . | For example, <math>\boxed{6} = 1 + 2 + 3 + 6 = 12</math>. Find <math>\boxed{\boxed{11}}</math> . | ||
− | <math>\ | + | <math>\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30</math> |
== Solution == | == Solution == | ||
+ | We have | ||
+ | <cmath>\begin{align*} | ||
+ | \boxed{\boxed{11}}&=\boxed{1+11} \\ | ||
+ | &=\boxed{12} \\ | ||
+ | &=1+2+3+4+6+12 \\ | ||
+ | &=28, | ||
+ | \end{align*}</cmath> | ||
+ | from which the answer is <math>\boxed{\textbf{(D)}\ 28}.</math> | ||
− | + | ~Aplus95 (Fundamental Logic) | |
− | + | ~MRENTHUSIASM (Reconstruction) | |
− | + | ==Video Solution by WhyMath== | |
+ | https://youtu.be/Ih8lEBwPqEY | ||
− | + | ~savannahsolver | |
− | + | ==Video Solution by SpreadTheMathLove== | |
− | + | https://www.youtube.com/watch?v=omFpSGMWhFc | |
+ | ==See Also== | ||
+ | {{AMC8 box|year=2007|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:54, 2 July 2024
Contents
Problem
For any positive integer , define
to be the sum of the positive factors of
.
For example,
. Find
.
Solution
We have
from which the answer is
~Aplus95 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Video Solution by WhyMath
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.