Difference between revisions of "SANSKAR'S OG PROBLEMS"

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       <math>02</math>
 
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==Problem1 ==
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==Problem 1==
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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Let <math>\overline{ab}</math> be a 2-digit [[positive integer]] satisfying <math>\overline{ab}^2=a! +b!</math>. Find <math>a+b</math> .
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==Solution 1 by ddk001 (Casework)==
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'''Case 1: <math>a>b</math>'''
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In this case, we have
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<cmath>\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2</cmath>.
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If
  
 
==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 22:03, 28 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:

      $02$

Problem 1

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

Solution 1 by ddk001 (Casework)

Case 1: $a>b$ In this case, we have \[\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2\]. If

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.