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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
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<cmath>\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},</cmath>
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<cmath>\sqrt{\frac{258(256^2 + 257) + 256}{258}},</cmath>
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<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 00:15, 11 July 2021

Problem

For all integers $x$ and $y$, define the operation $\Delta$ as \[x \Delta y = x^3+y^2+x+y.\] Find \[\sqrt{\dfrac{257 \Delta 256}{258}}.\]

Solution

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


Solution 2

Basically the same as above, but instead we can let $257 = 256 + 1$. Then we have \[\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},\] \[\sqrt{\frac{258(256^2 + 257) + 256}{258}},\] \[\sqrt{256^2 + 256 + 256 + 1} =\] \[\sqrt{256^2 + 2\cdot256 + 1} =\] \[\sqrt{(256+1)^2 = (257^2)}\]

which equals $\boxed{257}$.


~~abhinavg0627

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