Difference between revisions of "2021 JMPSC Sprint Problems/Problem 20"
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<cmath>\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},</cmath> | <cmath>\sqrt{\frac{(256+1)(256^2 + 256 + 1) + 1(256^2 + 257) + 256}{258}},</cmath> | ||
<cmath>\sqrt{\frac{258(256^2 + 257) + 256}{258}},</cmath> | <cmath>\sqrt{\frac{258(256^2 + 257) + 256}{258}},</cmath> | ||
− | <cmath>\sqrt{256^2 + 256 + 256 + 1} =</cmath> <cmath>\sqrt{256^2 + 2\cdot256 + 1} =</cmath> <cmath>\sqrt{(256+1)^2 = (257^2)}</cmath> | + | <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> |
which equals <math>\boxed{257}</math>. | which equals <math>\boxed{257}</math>. | ||
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~~abhinavg0627 | ~~abhinavg0627 | ||
+ | |||
+ | == Note: == | ||
+ | |||
+ | <math>257^3 = 16974593</math>, <math>256^2 = 65536</math>, and <math>257^2 = 66049</math>. | ||
+ | |||
+ | == 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 - | ||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 10:00, 12 July 2021
Problem
For all integers and
, define the operation
as
Find
Solution
Let . Then,
and
. We substitute these values into expression
to get
Recall the definition for the operation
; using this, we simplify our expression to
We have
and
, so we can expand the numerator of the fraction within the square root as
to get
~samrocksnature
Solution 2
Basically the same as above, but instead we can let . Then we have
which equals .
~~abhinavg0627
Note:
,
, and
.
Solution 3
Notice that , substituting this in, we get
. Therefore,
- kante314 -
See also
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