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

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==Solution==
 
==Solution==
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Notice that <math>C</math> can only be <math>0</math> and <math>5</math>. However, <math>790</math> is not divisible by <math>3</math>, so the <math>3 \times ABC = 795 \Longrightarrow ABC = 265</math>. Thus, <math>3A + 2B + C = \boxed{23}</math>
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Notice that <math>C</math> can only be <math>0</math> and <math>5</math>. However, <math>790</math> is not divisible by <math>3</math>, so <cmath>3 \times ABC = 795</cmath> <cmath>ABC = 265</cmath> Thus, <math>3A + 2B + C = \boxed{23}</math>
  
 
~Bradygho
 
~Bradygho

Revision as of 13:18, 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 \[3 \times ABC = 795\] \[ABC = 265\] 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

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