Difference between revisions of "SANSKAR'S OG PROBLEMS"

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(Solution 1 by ddk001)
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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==
 
==Solution 1 by ddk001==
we have
+
oopies, read the question wrong
  
<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.

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Problem1

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

Solution 1 by ddk001

oopies, read the question wrong

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.