Difference between revisions of "2007 AMC 8 Problems/Problem 10"
MRENTHUSIASM (talk | contribs) (Made the solution more direct and formatted.) |
|||
| Line 1: | Line 1: | ||
== 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== | ==Video Solution by WhyMath== | ||
Latest revision as of 15:04, 15 July 2021
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
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. 
