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

Abhinavg0627 (talk | contribs) (→Solution) |
Abhinavg0627 (talk | contribs) (→Solution 2) |
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<cmath>\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},</cmath> | <cmath>\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},</cmath> | ||

<cmath>\sqrt{\frac{258(256^2 + 257) + 256}{258}},</cmath> | <cmath>\sqrt{\frac{258(256^2 + 257) + 256}{258}},</cmath> | ||

− | <cmath>\sqrt{256^2 + 256 + 256 + 1} =</cmath> <cmath>\sqrt{256^2 + 2\cdot256 + 1} =</cmath> <cmath>\sqrt{(256+1)^2 = (257^2)}</cmath> | + | <cmath>\sqrt{256^2 + 256 + 256 + 1} =</cmath> <cmath>\sqrt{256^2 + 2\cdot256 + 1} =</cmath> <cmath>\sqrt{(256+1)^2} =</cmath> <cmath>\sqrt{(257^2)}</cmath> |

which equals <math>\boxed{257}</math>. | which equals <math>\boxed{257}</math>. |

## Revision as of 00:15, 11 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