Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"

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~Bradygho
 
~Bradygho
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==Solution 2==
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Clearly we see <math>C=1</math> does not work, but <math>C=5</math> works with simple guess-and-check. We have <math>AB5=\frac{795}{3}=265</math>, so <math>A=2</math> and <math>B=6</math>. The answer is <math>3(2)+6(2)+1(5)=\boxed{23}</math>
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~Geometry285

Revision as of 11:25, 11 July 2021

Problem

If $A$, $B$, and $C$ each represent a single digit and they satisfy the equation \[\begin{array}{cccc}& A & B & C \\ \times & &  &3 \\ \hline  & 7 & 9 & C\end{array},\] find $3A+2B+C$.

Solution

Notice that $C$ can only be $0$ and $5$. However, $790$ is not divisible by $3$, so the number $ABC = 265$. Thus, $3A + 2B + C = \boxed{23}$

~Bradygho

Solution 2

Clearly we see $C=1$ does not work, but $C=5$ works with simple guess-and-check. We have $AB5=\frac{795}{3}=265$, so $A=2$ and $B=6$. The answer is $3(2)+6(2)+1(5)=\boxed{23}$

~Geometry285