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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
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==Solution==
 
==Solution==
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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 <cmath>3 \times ABC = 795</cmath> <cmath>ABC = 265</cmath> Thus, <math>3A + 2B + C = \boxed{23}</math>
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Notice that <math>C</math> can only be <math>0</math>, <math>1</math>, and <math>5</math>. However, <math>790</math> and <math>791</math> are not divisible by <math>3</math>, so <cmath>3 \times ABC = 795</cmath> <cmath>ABC = 265</cmath> Thus, <math>3A + 2B + C = \boxed{23}</math>
  
 
~Bradygho
 
~Bradygho
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- kante314 -
 
- kante314 -
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== Solution 4 ==
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Notice that the only values of <math>C</math> that have <math>3C = 10n+C</math> for some <math>n</math> are <math>0</math> and <math>5</math>. If <math>C=0</math>, then we have <math>AB0 \cdot 3 = 790</math>, and so <math>AB \cdot 3 = 79</math>. Notice that <math>79</math> is not divisible by <math>3</math>, so <math>C=0</math> is not a valid solution. Next, when <math>C=5</math>, we have that <math>AB5 \cdot 3 = 795</math>. Solving for <math>A</math> and <math>B</math> tells us that <math>A=2</math> and <math>B=6</math>, so the answer is <math>3 \cdot 2 + 2 \cdot 6 + 5 = 6 + 12 + 5 = \boxed{23}</math>.
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~Mathdreams
  
 
==See also==
 
==See also==

Latest revision as of 09:42, 18 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 $0$, $1$, and $5$. However, $790$ and $791$ are not divisible by $3$, so \[3 \times ABC = 795\] \[ABC = 265\] Thus, $3A + 2B + C = \boxed{23}$

~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 -

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

See also

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

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