I don't understand

I got this question in a math competition; a multiple choice question:

Suppose $ p $ is a two-digit number and $ q $ has the same digits, but in reverse order. The number $p^2 - q^2 $ is a non-zero perfect square. The sum of the digits of $ p $ is:

A) 7
B) 9
C) 11
D) 12
E) 13

Can you please solve this problem?

#NumberTheory

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Comments

Let $a$ and $b$ be digits. Now, let $\displaystyle p=10a+b$ and $\displaystyle q=10b+a$ .

Substituting these values, we get $\displaystyle p^2-q^2=99\left(a^2-b^2\right)$ . And thus, for this to be a perfect square, $\displaystyle a^2-b^2=11$ .

Solving these equations give $p=65$ making the answer C.

Aaghaz Mahajan - 1 year, 9 months ago

What does "Solving these equations" mean? By trial and error? Can you further demonstrate the process?

Syed Hamza Khalid - 1 year, 9 months ago

Well, we only need to figure out the values of $a$ and $b$ because then $p$ and $q$ will be determined. So for solving $\displaystyle a^2-b^2=11$ observe that this implies

$\left(a-b\right)\left(a+b\right)=11$

Now, the LHS is an integer which implies that either $a-b=11$ and $a+b=1$ or $a-b=1$ and $a+b=11$

We see that only the second pair of equations holds true...….hence it is solved.

Aaghaz Mahajan - 1 year, 9 months ago

@Aaghaz Mahajan – Thanks... I understood this part now but can you tell me how you got from $p^2 - q^2 = 99 (a^2 -b^2)$ to $a^2 -b^2 = 11$ .

(I'm sorry to ask you such questions but I am not so good at competition math)

Syed Hamza Khalid - 1 year, 9 months ago

@Syed Hamza Khalid – No need for saying sorry!!! We are all here to learn new stuff!!

According the statement of the question, we need $\displaystyle 3^2\cdot11\left(a^2-b^2\right)$ to be a perfect square. That means $a^2-b^2=11m^2$ for some integer $m$ . Now I think you can proceed from here onwards, by checking the possible values of $m$ i.e. $1,2,3$ .

Aaghaz Mahajan - 1 year, 9 months ago

@Aaghaz Mahajan – Thanks alot... :)

Syed Hamza Khalid - 1 year, 9 months ago

Let $p = 10a + b$ where $a, b$ are non-zero digits. Then $q = 10b + a$ and $p^{2} - q^{2} = (10a + b)^{2} - (10b + a)^{2} = 99a^{2} - 99b^{2} = 3^{2} \times 11 \times (a - b)(a + b)$ .

For this to equal a non-zero perfect square we require $a \gt b$ and, since $11$ is prime, that (i) $11 | (a + b)$ and (ii) $(a - b)$ is a perfect square. Condition (i) gives us the options for $(a,b)$ of $(9,2), (8,3), (7,4), (6,5)$ , but of these only $(a,b) = (6,5)$ satisfies condition (ii). Thus $p = 65$ and indeed $65^{2} - 56^{2} = 1089 = 33^{2}$ .

Brian Charlesworth - 1 year, 9 months ago

Let $p=10x+y$ and $q=10y+x$ . Then $p^2-q^2=(p+q)(p-q)=(11x+11y)(9x-9y)=(9)(11)(x+y)(x-y)$ . Since $x$ and $y$ are non-zero digits, each of them may be $1,2,3,4,5,6,7,8$ or $9$ . So $x+y<19$ and $x-y<9$ . Also, since $11$ is a prime number, $x+y$ can only be $11$ , so that $x-y$ can only be $1$ . Therefore $x=6$ and $y=5$ . Hence $x+y=11$ .

A Former Brilliant Member - 1 year, 9 months ago

How did you set up the inequality $x + y < 19$ and $x - y < 9$ ?

Syed Hamza Khalid - 1 year, 9 months ago

See, x is at most 9 and so also y. So $x+y$ is at most 18. Similarly, both x and y can at least be 1. So $x-y$ is at most $9-1=8$ .

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – Thanks, can you also tell why $x+ y$ can only be 11 ?

Syed Hamza Khalid - 1 year, 9 months ago

@Syed Hamza Khalid – Because the next larger number is $11*2^2=44$ which is more than $18$

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – okay thanks

Syed Hamza Khalid - 1 year, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$