Difference between revisions of "2021 JMPSC Sprint Problems/Problem 20"
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== Solution 3 == | == 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> | 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 - | ||
==See also== | ==See also== |
Latest revision as of 09:00, 12 July 2021
Contents
[hide]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
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.