# Difference between revisions of "2021 JMPSC Sprint Problems/Problem 20"

Line 24: | Line 24: | ||

== Solution 3 == | == Solution 3 == | ||

Notice that <math>x=y+1</math>, substituting this in, we get <math>x^2(x+1)</math>. Therefore, <math>\sqrt{\frac{257^2(258)}{258}}=\boxed{257}</math> | Notice that <math>x=y+1</math>, substituting this in, we get <math>x^2(x+1)</math>. Therefore, <math>\sqrt{\frac{257^2(258)}{258}}=\boxed{257}</math> | ||

+ | |||

+ | - kante314 - | ||

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

## Latest revision as of 09:00, 12 July 2021

## Problem

For all integers and , define the operation as Find

## Solution

Let . Then, and . We substitute these values into expression to get Recall the definition for the operation ; using this, we simplify our expression to We have and , so we can expand the numerator of the fraction within the square root as to get ~samrocksnature

## Solution 2

Basically the same as above, but instead we can let . Then we have

which equals .

~~abhinavg0627

## Note:

, , and .

## Solution 3

Notice that , substituting this in, we get . Therefore,

- kante314 -

## See also

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