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
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==See also==
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 17:22, 11 July 2021

Problem

Find the sum of all positive multiples of $3$ that are factors of $27.$

Solution

We use the fact that $3 = 3^1$ and $27 = 3^3$ to conclude that the only multiples of $3$ that are factors of $27$ are $3$, $9$, and $27$. Thus, our answer is $3 + 9 + 27 = \boxed{39}$.

~Bradygho

Solution 2

The factors of $27$ are $1$, $3$, $9$ and $27$. Out of these, only $3$, $9$ and $27$ are multiples of $3$, so the answer is $3 + 9 + 27 = \boxed{39}$.

~Mathdreams

See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

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