2021 JMPSC Accuracy Problems/Problem 7

Revision as of 17:21, 11 July 2021 by Mathdreams (talk | contribs)

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

See also

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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png