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

(Created page with "==Problem== For all integers <math>x</math> and <math>y</math>, define the operation <math>\Delta</math> as <cmath>x \Delta y = x^3+y^2+x+y.</cmath> Find <cmath>\sqrt{\dfrac{2...")
 
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==Solution==
 
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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

Revision as of 22:20, 10 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

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