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2021 JMPSC Sprint Problems/Problem 20

Revision as of 00:15, 11 July 2021 by Abhinavg0627 (talk | contribs) (→‎Solution 2)

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 \[\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},\] \[\sqrt{\frac{258(256^2 + 257) + 256}{258}},\] \[\sqrt{256^2 + 256 + 256 + 1} =\] \[\sqrt{256^2 + 2\cdot256 + 1} =\] \[\sqrt{(256+1)^2} =\] \[\sqrt{(257^2)}\]

which equals $\boxed{257}$.


~~abhinavg0627

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