Difference between revisions of "SANSKAR'S OG PROBLEMS"
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==Solution 1 by ddk001 (Casework)== | ==Solution 1 by ddk001 (Casework)== | ||
'''Case 1: <math>a>b</math>''' | '''Case 1: <math>a>b</math>''' | ||
+ | |||
In this case, we have | In this case, we have | ||
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, but this contradicts the original assumption of <math>b \ge 5</math>, so hence we must have <math>b \le 4</math>. | , but this contradicts the original assumption of <math>b \ge 5</math>, so hence we must have <math>b \le 4</math>. | ||
+ | |||
+ | With this in mind, we consider the unit digit of <math>\overline{ab}^2</math>. | ||
+ | |||
+ | '''Subcase 1.1: <math>a>b=1</math>''' | ||
+ | |||
+ | In this case, we have that | ||
+ | |||
+ | <cmath>a! \equiv \overline{ab}^2-b! \equiv (10a+1)^2-1 \equiv 0 \pmod{10} \implies 10|a! \implies a \ge 5</cmath>. | ||
+ | |||
+ | There is no apparent contradiction here, so we leave this as it is. | ||
+ | |||
+ | '''Subcase 1.2: <math>a>b=2</math>''' | ||
+ | |||
+ | In this case, we have that | ||
+ | |||
+ | <cmath>a! \equiv \overline{ab}^2-b! \equiv (10a+2)^2-2 \equiv 2 \pmod{10} \implies a! \equiv 2 \pmod{10} \implies a=2</cmath>. | ||
+ | |||
+ | This contradicts with the fact that <math>a>b</math>, so this is impossible. | ||
+ | |||
+ | '''Subcase 1.3: <math>a>b=3</math>''' | ||
+ | |||
+ | In this case, we have that | ||
+ | |||
+ | <cmath>a! \equiv \overline{ab}^2-b! \equiv (10a+3)^2-6 \equiv 3 \pmod{10}</cmath>. | ||
+ | |||
+ | However, this is impossible for all <math>a</math>. | ||
+ | |||
+ | '''Subcase 1.4: <math>a>b=4</math>''' | ||
+ | |||
+ | In this case, we have that | ||
+ | |||
+ | <cmath>a! \equiv \overline{ab}^2-b! \equiv (10a+4)^2-24 \equiv 2 \pmod{10}</cmath>. | ||
+ | |||
+ | Again, this yields <math>a=2</math>, which, again, contradicts <math>a>b</math>. <math>\square</math> | ||
+ | |||
+ | Hence, we must have <math>b=1</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 22:19, 28 January 2024
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Problem 1
Let be a 2-digit positive integer satisfying
. Find
.
Solution 1 by ddk001 (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 .
Problem2
For any positive integer ,
>1 can
be a perfect square? If yes, give one such
. If no, then prove it.