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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> .
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==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 10:58, 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: 

      $02$

Problem1

Let $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2$ = $a! +b!$. Find $a+b$ .

Solution 1 by ddk001

we have

\[10a+b=a!+b!\]

Obviously, the left increase much slower for big $a$ and $b$ so $a$ and $b$ must be small. (Namely, $a!+b!<100 \implies a,b<5$) $a=b=4$ don't work so $a!+b! \le 3!+4!=30$, so since $a,b<5$, the max of $a!+b!$ is $24$, which didn't work so $a,b<4$. since the max now would be $12=3!+3!$, and that doesn't work, the max is now $2!+3!=8$, which is not a 2-digit integer. Hence, no such ordered pair $(a,b)$ exist. ~Ddk001

Problem2

For any positive integer $n$, $n$>1 can $n!$ be a perfect square? If yes, give one such $n$. If no, then prove it.

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