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

Latest revision as of 09:00, 12 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 $\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},$ $\sqrt{\frac{258(256^2 + 257) + 256}{258}},$ $\sqrt{256^2 + 256 + 256 + 1} =$ $\sqrt{256^2 + 2\cdot256 + 1} =$ $\sqrt{(256+1)^2} =$ $\sqrt{(257^2)}$

which equals $\boxed{257}$ .

~~abhinavg0627

Note:

$257^3 = 16974593$ , $256^2 = 65536$ , and $257^2 = 66049$ .

Solution 3

Notice that $x=y+1$ , substituting this in, we get $x^2(x+1)$ . Therefore, $\sqrt{\frac{257^2(258)}{258}}=\boxed{257}$

- kante314 -

@@ Line 24: / Line 24: @@
 == Solution 3 ==
 Notice that <math>x=y+1</math>, substituting this in, we get <math>x^2(x+1)</math>. Therefore, <math>\sqrt{\frac{257^2(258)}{258}}=\boxed{257}</math>
+- kante314 -
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