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

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==Solution== | ==Solution== | ||

Let <math>258=a</math>. Then, <math>257=a-1</math> and <math>256=a-2</math>. We substitute these values into expression <math>(1)</math> to get <cmath>\sqrt{\frac{(a-1) \Delta (a-2)}{a}}.</cmath> Recall the definition for the operation <math>\Delta</math>; using this, we simplify our expression to <cmath>\sqrt{\frac{(a-1)^3+(a-2)^2+(a-1)+(a-2)}{a}}.</cmath> We have <math>(a-1)^3=a^3-3a^2+3a-1</math> and <math>(a-2)^2=a^2-4a+4</math>, so we can expand the numerator of the fraction within the square root as <math>a^3-3a^2+3a-1+a^2-4a+4+a-1+a-2=a^3-2a^2+a</math> to get <cmath>\sqrt{\frac{a^3-2a^2+a}{a}}=\sqrt{a^2-2a+1}=\sqrt{(a-1)^2}=a-1=\boxed{257}.</cmath> ~samrocksnature | Let <math>258=a</math>. Then, <math>257=a-1</math> and <math>256=a-2</math>. We substitute these values into expression <math>(1)</math> to get <cmath>\sqrt{\frac{(a-1) \Delta (a-2)}{a}}.</cmath> Recall the definition for the operation <math>\Delta</math>; using this, we simplify our expression to <cmath>\sqrt{\frac{(a-1)^3+(a-2)^2+(a-1)+(a-2)}{a}}.</cmath> We have <math>(a-1)^3=a^3-3a^2+3a-1</math> and <math>(a-2)^2=a^2-4a+4</math>, so we can expand the numerator of the fraction within the square root as <math>a^3-3a^2+3a-1+a^2-4a+4+a-1+a-2=a^3-2a^2+a</math> to get <cmath>\sqrt{\frac{a^3-2a^2+a}{a}}=\sqrt{a^2-2a+1}=\sqrt{(a-1)^2}=a-1=\boxed{257}.</cmath> ~samrocksnature | ||

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+ | ==Solution 2== | ||

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+ | Basically the same as above, but instead we can let <math>257 = 256 + 1</math>. Then we have | ||

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

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

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+ | ~~abhinavg0627 |

## Revision as of 01:15, 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