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

Revision as of 08:57, 12 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

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 -

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 ~Geometry285
+== Solution 3 ==
+Easily, we can see that <math>A=2</math>. Therefore,<cmath>\overline{BC} \cdot 3 = \overline{19C}.</cmath>We can see that <math>C</math> must be <math>1</math> or <math>5</math>. If <math>C=1</math>, then<cmath>\overline{B1} \cdot 3 = 191.</cmath>This doesn't work because <math>191</math> isn't divisible by <math>3</math>. If <math>C=5</math>, then<cmath>\overline{B5} \cdot 3 = 195.</cmath>Therefore, <math>B=6</math>. So, we have <math>3(2) + 2(6) + 5=6+12+5=18+5=\boxed{23}</math>.
+- kante314 -
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