Difference between revisions of "SANSKAR'S OG PROBLEMS"
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==Problem1 == | ==Problem1 == | ||
Let <math>\overline{ab}</math> be a 2-digit [[positive integer]] satisfying <math>\overline{ab}^2</math> = <math>a! +b!</math>. Find <math>a+b</math> . | Let <math>\overline{ab}</math> be a 2-digit [[positive integer]] satisfying <math>\overline{ab}^2</math> = <math>a! +b!</math>. Find <math>a+b</math> . | ||
+ | ==Solution 1 by ddk001== | ||
+ | we have | ||
+ | |||
+ | <cmath>10a+b=a!+b!</cmath> | ||
+ | |||
+ | Obviously, the left increase much slower for big <math>a</math> and <math>b</math> so <math>a</math> and <math>b</math> must be small. (Namely, <math>a!+b!<100 \implies a,b<5</math>) <math>a=b=4</math> don't work so <math>a!+b! \le 3!+4!=30</math>, so since <math>a,b<5</math>, the max of <math>a!+b!</math> is <math>24</math>, which didn't work so <math>a,b<4</math>. since the max now would be <math>12=3!+3!</math>, and that doesn't work, the max is now <math>2!+3!=8</math>, which is not a 2-digit integer. Hence, no such ordered pair <math>(a,b)</math> exist. ~[[Ddk001]] | ||
==Problem2 == | ==Problem2 == | ||
For any [[positive integer]] <math>n</math>, <math>n</math>>1 can <math>n!</math> be a [[perfect square]]? If yes, give one such <math>n</math>. If no, then prove it. | For any [[positive integer]] <math>n</math>, <math>n</math>>1 can <math>n!</math> be a [[perfect square]]? If yes, give one such <math>n</math>. If no, then prove it. |
Revision as of 11:58, 24 January 2024
Hi, this page is created by ...~ SANSGANKRSNGUPTA This page contains exclusive problems made by me myself. I am the creator of these OG problems. What OG stands for is a secret! Please post your solutions with your name. If you view this page please increment the below number by one:
Problem1
Let be a 2-digit positive integer satisfying
=
. Find
.
Solution 1 by ddk001
we have
Obviously, the left increase much slower for big and
so
and
must be small. (Namely,
)
don't work so
, so since
, the max of
is
, which didn't work so
. since the max now would be
, and that doesn't work, the max is now
, which is not a 2-digit integer. Hence, no such ordered pair
exist. ~Ddk001
Problem2
For any positive integer ,
>1 can
be a perfect square? If yes, give one such
. If no, then prove it.