# Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"

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asdf | asdf | ||

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+ | Easily, we can see that <math>C=3,5</math>. <math>C</math> must be <math>5</math> because <math>795</math> is divisible by <math>3</math> and <math>793</math> is not divisible by <math>3</math>. Therefore <math>\frac{795}{3}=265</math>. So, <math>3A+2B+C=6+12+5=23</math> | ||

+ | |||

+ | 13. | ||

+ | Case 1: <math>x+y=4,3</math>. | ||

+ | There are no possible answer when <math>x+y=3</math>, but when <math>x+y=4</math>, <math>x</math> can equal <math>3</math> or <math>1</math>. | ||

+ | Case 2: <math>x+y=5,0</math> | ||

+ | This works when <math>x=0,5</math>. | ||

+ | Therefore, the answer is <math>9</math>. ~ kante314 |

## Revision as of 00:19, 11 July 2021

## Problem

Let and be nonnegative integers such that Find the sum of all possible values of

## Solution

asdf

Easily, we can see that . must be because is divisible by and is not divisible by . Therefore . So,

13. Case 1: . There are no possible answer when , but when , can equal or . Case 2: This works when . Therefore, the answer is . ~ kante314