Difference between revisions of "SANSKAR'S OG PROBLEMS"
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To simplify future calculations, note that | To simplify future calculations, note that | ||
− | <cmath>a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a( | + | <cmath>a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(10a+2)</cmath>. |
For <math>a=5</math>, this does not hold. | For <math>a=5</math>, this does not hold. | ||
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For <math>a=6</math>, this does not hold. | For <math>a=6</math>, this does not hold. | ||
− | For <math>a=7</math>, this does | + | For <math>a=7</math>, this does hold. Hence, <math>(a,b)=(7,1)</math> is a solution. |
For <math>a=8</math>, this does not hold. | For <math>a=8</math>, this does not hold. | ||
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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>. | ||
− | We see | + | We see that <math>(a,b)=(7,1) \implies a+b=\boxed{008}</math>. <math>\blacksquare</math> ~[[Ddk001]] |
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==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 16:37, 3 February 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
.
If , we must have
, 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
.
There is no apparent contradiction here, so we leave this as it is.
Subcase 1.2:
In this case, we have that
.
This contradicts with the fact that , so this is impossible.
Subcase 1.3:
In this case, we have that
.
However, this is impossible for all .
Subcase 1.4:
In this case, we have that
.
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
.
For , this does not hold.
For , this does not hold.
For , this does hold. Hence, is a solution.
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
which is impossible if a is a integer and .
Case 3:
In this case, we have
.
If , we must have
which is impossible since and .
Hence, .
Subcase 3.1:
Testing cases, we can see that there is no such .
Subcase 3.2:
Testing cases, we can see that there is no such .
Subcase 3.3:
Testing cases, we can see that there is no such .
Subcase 3.4:
Testing cases, we can see that there is no such .
We see that . ~Ddk001
Problem2
For any positive integer , >1 can be a perfect square? If yes, give one such . If no, then prove it.