Difference between revisions of "SANSKAR'S OG PROBLEMS"
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'''Case 1: <math>a>b</math>''' | '''Case 1: <math>a>b</math>''' | ||
In this case, we have | In this case, we have | ||
| + | |||
<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>. | <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>. | ||
| − | If | + | |
| + | If <math>b \ge 5</math>, we must have | ||
| + | |||
| + | <cmath>10|b!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0</cmath> | ||
| + | |||
| + | , but this contradicts the original assumption of <math>b \ge 5</math>, so hence we must have <math>b \le 4</math>. | ||
==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 21:06, 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:
Problem 1
Let 
be a 2-digit positive integer satisfying 
. Find 
.
Solution 1 by ddk001 (Casework)
Case 1: 
In this case, we have
![\[\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2\]](http://latex.artofproblemsolving.com/d/6/6/d66bc68298f08f74f02e1d5c7a55f96f1385d389.png)
.
If 
, we must have
![\[10|b!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0\]](http://latex.artofproblemsolving.com/0/a/9/0a96a133a9772e6ffdcdee8ff54fdbbb2a585060.png)
, but this contradicts the original assumption of , so hence we must have 
.
Problem2
For any positive integer 
, >1 can 
be a perfect square? If yes, give one such . If no, then prove it.
