Difference between revisions of "SANSKAR'S OG PROBLEMS"
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Testing cases, we can see that there is no such <math>b</math>. | Testing cases, we can see that there is no such <math>b</math>. | ||
| − | '''Subcase 3.2: <math>a=2</math>''' | + | '''Subcase 3.2: <math>b>a=2</math>''' |
<cmath>(20+b)^2=2+b!</cmath> | <cmath>(20+b)^2=2+b!</cmath> | ||
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Testing cases, we can see that there is no such <math>b</math>. | Testing cases, we can see that there is no such <math>b</math>. | ||
| − | '''Subcase 3.3: <math>a=3</math>''' | + | '''Subcase 3.3: <math>b>a=3</math>''' |
<cmath>(30+b)^2=3+b!</cmath> | <cmath>(30+b)^2=3+b!</cmath> | ||
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Testing cases, we can see that there is no such <math>b</math>. | Testing cases, we can see that there is no such <math>b</math>. | ||
| − | '''Subcase 3.4: <math>a=4</math>''' | + | '''Subcase 3.4: <math>b>a=4</math>''' |
<cmath>(40+b)^2=4+b!</cmath> | <cmath>(40+b)^2=4+b!</cmath> | ||
Revision as of 21:50, 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 (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 
.
With this in mind, we consider the unit digit of 
.
Subcase 1.1: 
In this case, we have that
![\[a! \equiv \overline{ab}^2-b! \equiv (10a+1)^2-1 \equiv 0 \pmod{10} \implies 10|a! \implies a \ge 5\]](http://latex.artofproblemsolving.com/d/a/9/da936f2fd27de182ddc41454512bb74d8c5df17e.png)
.
There is no apparent contradiction here, so we leave this as it is.
Subcase 1.2: 
In this case, we have that
![\[a! \equiv \overline{ab}^2-b! \equiv (10a+2)^2-2 \equiv 2 \pmod{10} \implies a! \equiv 2 \pmod{10} \implies a=2\]](http://latex.artofproblemsolving.com/8/c/9/8c93a979cb35e8204216892a0f16c67c1a91bd3a.png)
.
This contradicts with the fact that , so this is impossible.
Subcase 1.3: 
In this case, we have that
![\[a! \equiv \overline{ab}^2-b! \equiv (10a+3)^2-6 \equiv 3 \pmod{10}\]](http://latex.artofproblemsolving.com/5/d/8/5d89b14f3c045404e4f4430d25969e6b653fa49b.png)
.
However, this is impossible for all 
.
Subcase 1.4: 
In this case, we have that
![\[a! \equiv \overline{ab}^2-b! \equiv (10a+4)^2-24 \equiv 2 \pmod{10}\]](http://latex.artofproblemsolving.com/a/6/5/a6592f984dcfc69cf9d41ceff23490e5679cf091.png)
.
Again, this yields 
, which, again, contradicts . 
Hence, we must have 
.
Now, with 
determined by modular arithmetic, we actually plug in the values.
To simplify future calculations, note that
![\[a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(a+2)\]](http://latex.artofproblemsolving.com/0/2/b/02b4fda89a4800ba937ad4f3597fdd2f1e31f827.png)
.
For 
, this does not hold.
For 
, this does not hold.
For 
, this does not hold.
For 
, this does not hold.
For 
, this does not hold.
Hence, there is no positive integers and between 
and 
inclusive such that 
.
Case 2: 
For this case, we must have
![\[(11a)^2=\overline{ab}^2=a!+b!=2a! \implies 11|a!\]](http://latex.artofproblemsolving.com/9/b/3/9b389fcce14b2e05053d66793af4c945d55a81c0.png)
which is impossible if a is a integer and 
.
Case 3: 
In this case, we have
![\[\overline{ab}^2=a! +b!=(1+b \cdot (b-1) \cdot \dots \cdot (a+1)) \cdot a! \implies a!|\overline{ab}^2=(10a+b)^2\]](http://latex.artofproblemsolving.com/5/1/f/51f22c600d52858d10f0937ccdfc61f2f6fe6b44.png)
.
If 
, we must have
![\[10|a!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0 \implies 10|a!+b!\]](http://latex.artofproblemsolving.com/b/5/b/b5bbffa4588b0a77e97d51a4b55035410946c473.png)
which is impossible since 
and 
.
Hence, 
.
Subcase 3.1: 
![\[(10+b)^2=1+b!\]](http://latex.artofproblemsolving.com/5/d/f/5df4393248891a8de37b645bad69f2823c04d435.png)
Testing cases, we can see that there is no such .
Subcase 3.2: 
![\[(20+b)^2=2+b!\]](http://latex.artofproblemsolving.com/6/8/a/68a82d4da3381dacd97e81ee5e351c11e4bb1247.png)
Testing cases, we can see that there is no such .
Subcase 3.3: 
![\[(30+b)^2=3+b!\]](http://latex.artofproblemsolving.com/0/9/2/092c66729544056e54c7602a09ee172e0e050809.png)
Testing cases, we can see that there is no such .
Subcase 3.4: 
![\[(40+b)^2=4+b!\]](http://latex.artofproblemsolving.com/0/a/7/0a7dd24e932289c2baff754e9980e03d82cde380.png)
Testing cases, we can see that there is no such .
We see there is no and that satisfy the given equation. 
~Ddk001
Problem2
For any positive integer 
, >1 can 
be a perfect square? If yes, give one such . If no, then prove it.
