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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...")
 
 
(6 intermediate revisions by 4 users not shown)
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==Solution==
 
==Solution==
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asdf
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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
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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} =</cmath> <cmath>\sqrt{(257^2)}</cmath>
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which equals <math>\boxed{257}</math>.
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~~abhinavg0627
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== Note: ==
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<math>257^3 = 16974593</math>, <math>256^2 = 65536</math>, and <math>257^2 = 66049</math>.
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== Solution 3 ==
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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>
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- kante314 -
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==See also==
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

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 -

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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