I don't understand

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

Suppose p p is a two-digit number and q q has the same digits, but in reverse order. The number p2q2p^2 - q^2 is a non-zero perfect square. The sum of the digits of p p is:

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

Can you please solve this problem?

Note by Syed Hamza Khalid
1 month ago

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1 vote

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

Substituting these values, we get p2q2=99(a2b2)\displaystyle p^2-q^2=99\left(a^2-b^2\right). And thus, for this to be a perfect square, a2b2=11\displaystyle a^2-b^2=11.

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

Aaghaz Mahajan - 1 month ago

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

Syed Hamza Khalid - 1 month ago

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Well, we only need to figure out the values of aa and bb because then pp and qq will be determined. So for solving a2b2=11\displaystyle a^2-b^2=11 observe that this implies

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

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

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

Aaghaz Mahajan - 1 month ago

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@Aaghaz Mahajan Thanks... I understood this part now but can you tell me how you got from p2q2=99(a2b2)p^2 - q^2 = 99 (a^2 -b^2) to a2b2=11a^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 month ago

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

According the statement of the question, we need 3211(a2b2)\displaystyle 3^2\cdot11\left(a^2-b^2\right) to be a perfect square. That means a2b2=11m2a^2-b^2=11m^2 for some integer mm. Now I think you can proceed from here onwards, by checking the possible values of mm i.e. 1,2,31,2,3.

Aaghaz Mahajan - 1 month ago

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@Aaghaz Mahajan Thanks alot... :)

Syed Hamza Khalid - 1 month ago

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Let p=10a+bp = 10a + b where a,ba, b are non-zero digits. Then q=10b+aq = 10b + a and p2q2=(10a+b)2(10b+a)2=99a299b2=32×11×(ab)(a+b)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>ba \gt b and, since 1111 is prime, that (i) 11(a+b)11 | (a + b) and (ii) (ab)(a - b) is a perfect square. Condition (i) gives us the options for (a,b)(a,b) of (9,2),(8,3),(7,4),(6,5)(9,2), (8,3), (7,4), (6,5), but of these only (a,b)=(6,5)(a,b) = (6,5) satisfies condition (ii). Thus p=65p = 65 and indeed 652562=1089=33265^{2} - 56^{2} = 1089 = 33^{2}.

Brian Charlesworth - 1 month ago

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Let p=10x+yp=10x+y and q=10y+xq=10y+x. Then p2q2=(p+q)(pq)=(11x+11y)(9x9y)=(9)(11)(x+y)(xy)p^2-q^2=(p+q)(p-q)=(11x+11y)(9x-9y)=(9)(11)(x+y)(x-y) . Since xx and yy are non-zero digits, each of them may be 1,2,3,4,5,6,7,81,2,3,4,5,6,7,8 or 99. So x+y<19x+y<19 and xy<9x-y<9. Also, since 1111 is a prime number, x+yx+y can only be 1111, so that xyx-y can only be 11. Therefore x=6x=6 and y=5y=5. Hence x+y=11x+y=11.

Alak Bhattacharya - 1 month ago

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

Syed Hamza Khalid - 1 month ago

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See, x is at most 9 and so also y. So x+yx+y is at most 18. Similarly, both x and y can at least be 1. So xyx-y is at most 91=89-1=8.

Alak Bhattacharya - 1 month ago

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@Alak Bhattacharya Thanks, can you also tell why x+yx+ y can only be 11 ?

Syed Hamza Khalid - 1 month ago

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@Syed Hamza Khalid Because the next larger number is 1122=4411*2^2=44 which is more than 1818

Alak Bhattacharya - 1 month ago

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@Alak Bhattacharya okay thanks

Syed Hamza Khalid - 1 month ago

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