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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
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==See also==
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
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{{JMPSC Notice}}

Revision as of 16:23, 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

See also

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

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

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