Difference between revisions of "SANSKAR'S OG PROBLEMS"

Revision as of 21:30, 28 January 2024

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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 $b \ge 5$ , we must have

$10|b!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0$

, but this contradicts the original assumption of $b \ge 5$ , so hence we must have $b \le 4$ .

With this in mind, we consider the unit digit of $\overline{ab}^2$ .

Subcase 1.1: $a>b=1$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+1)^2-1 \equiv 0 \pmod{10} \implies 10|a! \implies a \ge 5$ .

There is no apparent contradiction here, so we leave this as it is.

Subcase 1.2: $a>b=2$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+2)^2-2 \equiv 2 \pmod{10} \implies a! \equiv 2 \pmod{10} \implies a=2$ .

This contradicts with the fact that $a>b$ , so this is impossible.

Subcase 1.3: $a>b=3$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+3)^2-6 \equiv 3 \pmod{10}$ .

However, this is impossible for all $a$ .

Subcase 1.4: $a>b=4$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+4)^2-24 \equiv 2 \pmod{10}$ .

Again, this yields $a=2$ , which, again, contradicts $a>b$ . $\square$

Hence, we must have $b=1$ .

Now, with $b$ determined by modular arithmetic, we actually plug in the values.

To simplify future calculations, note that

$a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(a+2)$ .

For $a=5$ , this does not hold.

For $a=6$ , this does not hold.

For $a=7$ , this does not hold.

For $a=8$ , this does not hold.

For $a=9$ , this does not hold.

Hence, there is no positive integers $a$ and $b$ between $1$ and $9$ inclusive such that $a!+b!=\overline{ab}^2$ .

Case 2: $a=b$

For this case, we must have

$(11a)^2=\overline{ab}^2=a!+b!=2a! \implies 11|a!$

which is impossible if a is a integer and $1 \le a \le 9$ .

Case 3: $a<b$

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.

@@ Line 72: / Line 72: @@
 Hence, there is no positive integers <math>a</math> and <math>b</math> between <math>1</math> and <math>9</math> inclusive such that <math>a!+b!=\overline{ab}^2</math>.
-'''Case 2:<math>a=b</math>'''
+'''Case 2: <math>a=b</math>'''
+For this case, we must have
+<cmath>(11a)^2=\overline{ab}^2=a!+b!=2a! \implies 11|a!</cmath>
+which is impossible if a is a integer and <math>1 \le a \le 9</math>.
+'''Case 3: <math>a<b</math>'''
 ==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.

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Difference between revisions of "SANSKAR'S OG PROBLEMS"

Revision as of 21:30, 28 January 2024

Problem 1

Solution 1 by ddk001 (Casework)

Problem2