Difference between revisions of "SANSKAR'S OG PROBLEMS"

Revision as of 22:43, 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 $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2=a! +b!$ . Find $a+b$ .

Solution 1 (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: $a<b$

In this case, we have

$\overline{ab}^2=a! +b!=(1+b \cdot (b-1) \cdot \dots \cdot (a+1)) \cdot a! \implies a!|\overline{ab}^2=(10a+b)^2$ .

If $a \ge 5$ , we must have

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

which is impossible since $10|a!$ and $b!=1$ .

Hence, $a \le 4$ .

Subcase 3.1: $a=1$

Subcase 3.2: $a=2$

Subcase 3.3: $a=3$

Subcase 3.4: $a=4$

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 88: / Line 88: @@
 If <math>a \ge 5</math>, we must have
-<cmath>10|a!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0</cmath>
+<cmath>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!</cmath>
-We have 2 cases:
+which is impossible since <math>10|a!</math> and <math>b!=1</math>.
-'''Subcase 3.1: <math>a \ge 5</math>'''
+Hence, <math>a \le 4</math>.
-'''Subcase 3.2: <math>a \le 4</math>'''
+'''Subcase 3.1: <math>a=1</math>'''
+'''Subcase 3.2: <math>a=2</math>'''
+'''Subcase 3.3: <math>a=3</math>'''
+'''Subcase 3.4: <math>a=4</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.

Page

Toolbox

Search

Difference between revisions of "SANSKAR'S OG PROBLEMS"

Revision as of 22:43, 28 January 2024

Problem 1

Solution 1 (Casework)

Problem2