# 2021 JMPSC Accuracy Problems/Problem 7

## 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 $0$, $1$, and $5$. However, $790$ and $791$ are not divisible by $3$, so $$3 \times ABC = 795$$ $$ABC = 265$$ Thus, $3A + 2B + C = \boxed{23}$

## 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

## Solution 3

Easily, we can see that $A=2$. Therefore, $$\overline{BC} \cdot 3 = \overline{19C}.$$We can see that $C$ must be $1$ or $5$. If $C=1$, then $$\overline{B1} \cdot 3 = 191.$$This doesn't work because $191$ isn't divisible by $3$. If $C=5$, then $$\overline{B5} \cdot 3 = 195.$$Therefore, $B=6$. So, we have $3(2) + 2(6) + 5=6+12+5=18+5=\boxed{23}$.

- kante314 -

## Solution 4

Notice that the only values of $C$ that have $3C = 10n+C$ for some $n$ are $0$ and $5$. If $C=0$, then we have $AB0 \cdot 3 = 790$, and so $AB \cdot 3 = 79$. Notice that $79$ is not divisible by $3$, so $C=0$ is not a valid solution. Next, when $C=5$, we have that $AB5 \cdot 3 = 795$. Solving for $A$ and $B$ tells us that $A=2$ and $B=6$, so the answer is $3 \cdot 2 + 2 \cdot 6 + 5 = 6 + 12 + 5 = \boxed{23}$.

~Mathdreams

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