Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
(→Solution) |
(→Solution) |
||
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
| − | Notice that <math>C</math> can only be <math>0</math> and <math>5</math>. However, 790 is not divisible by <math>3</math>, so the number <math>ABC = 265</math>. Thus, <math>3A + 2B + C = \boxed{23}</math> | + | 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 number <math>ABC = 265</math>. Thus, <math>3A + 2B + C = \boxed{23}</math> |
~Bradygho | ~Bradygho | ||
Revision as of 22:53, 10 July 2021
Problem
If 
, 
, and 
each represent a single digit and they satisfy the equation ![\[\begin{array}{cccc}& A & B & C \\ \times & & &3 \\ \hline & 7 & 9 & C\end{array},\]](http://latex.artofproblemsolving.com/d/9/2/d92d1fb62e2b18f640a996cd6d7f7f0ed2cf655b.png)
find 
.
Solution
Notice that can only be 
and 
. However, 
is not divisible by 
, so the number 
. Thus, 
~Bradygho
