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

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~Geometry285 | ~Geometry285 | ||

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+ | == 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>. | ||

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+ | - kante314 - | ||

==See also== | ==See also== |

## Revision as of 09:57, 12 July 2021

## Problem

If , , and each represent a single digit and they satisfy the equation find .

## Solution

Notice that can only be and . However, is not divisible by , so Thus,

~Bradygho

## Solution 2

Clearly we see does not work, but works with simple guess-and-check. We have , so and . The answer is

~Geometry285

## Solution 3

Easily, we can see that . Therefore,We can see that must be or . If , thenThis doesn't work because isn't divisible by . If , thenTherefore, . So, we have .

- kante314 -

## See also

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