Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 1"
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The factors of <math>27</math> are <math>1</math>, <math>3</math>, <math>9</math> and <math>27</math>. Out of these, only <math>3</math>, <math>9</math> and <math>27</math> are multiples of <math>3</math>, so the answer is <math>3 + 9 + 27 = \boxed{39}</math>. | The factors of <math>27</math> are <math>1</math>, <math>3</math>, <math>9</math> and <math>27</math>. Out of these, only <math>3</math>, <math>9</math> and <math>27</math> are multiples of <math>3</math>, so the answer is <math>3 + 9 + 27 = \boxed{39}</math>. | ||
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| + | ~Mathdreams | ||
==See also== | ==See also== | ||
| − | #[[2021 Accuracy | + | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] |
| − | #[[2021 Accuracy | + | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] |
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
{{JMPSC Notice}} | {{JMPSC Notice}} | ||
Latest revision as of 16:22, 11 July 2021
Contents
Problem
Find the sum of all positive multiples of 
that are factors of 
Solution
We use the fact that 
and 
to conclude that the only multiples of that are factors of 
are , 
, and . Thus, our answer is 
.
~Bradygho
Solution 2
The factors of are 
, , and . Out of these, only , and are multiples of , so the answer is .
~Mathdreams
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
