Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"
(Created page with "==Problem== Let <math>x</math> and <math>y</math> be nonnegative integers such that <math>(x+y)^2+(xy)^2=25.</math> Find the sum of all possible values of <math>x.</math> ==S...") |
(→Solution) |
||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
asdf | asdf | ||
| + | |||
| + | 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 23:19, 10 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
