Difference between revisions of "SANSKAR'S OG PROBLEMS"

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 $\overline{ab}$ 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

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

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

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

For $a=7$ , this does hold. Hence, $(a,b)=(7,1)$ is a solution.

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: $b>a=1$

$(10+b)^2=1+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.2: $b>a=2$

$(20+b)^2=2+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.3: $b>a=3$

$(30+b)^2=3+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.4: $b>a=4$

$(40+b)^2=4+b!$

Testing cases, we can see that there is no such $b$ .

We see that $(a,b)=(7,1) \implies a+b=\boxed{008}$ . $\blacksquare$ ~Ddk001

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 58: / Line 58: @@
 To simplify future calculations, note that
-<cmath>a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(a+2)</cmath>.
+<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.
@@ Line 64: / Line 64: @@
 For <math>a=6</math>, this does not hold.
-For <math>a=7</math>, this does not hold.
+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.
@@ Line 118: / Line 118: @@
 Testing cases, we can see that there is no such <math>b</math>.
-We see there is no <math>a</math> and <math>b</math> that satisfy the given equation. <math>\blacksquare</math> ~[[Ddk001]]
+We see that <math>(a,b)=(7,1) \implies a+b=\boxed{008}</math>. <math>\blacksquare</math> ~[[Ddk001]]
-'''FEEDBACK'''
-NICE TRY, KEEP IT UP BUT THERE EXIST SOME A AND B SO DO TRY AGAIN ~SANSGANKRSNGUPTA
-Which case in particular did the solution lie in?~[[Ddk001]]
-you have solved everything correctly but why did you skip subcase 1.1 if there wasn't any contradiction? ok now just don't delete the solution just solve subcase 1.1 below also, try looking for another shorter method. Thanks ~SANSGANKRSNGUPTA
 ==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 16:37, 3 February 2024

Problem 1

Solution 1 (Casework)

Problem2