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Perfect squares

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by lucas211 » Thu Jun 16, 2016 1:50 pm
Hello BTG

Would appreciate a little help on the attached question.
It seems a bit abstract to me, so would appreciate a little explanation in connection to it - if possible.

Thanks in advance
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by GMATGuruNY » Fri Jun 17, 2016 4:26 am
"  and Â¥ represent nonzero digits, and (" Â¥)² - (Â¥" )² is a perfect square. What is that perfect square?

121

361

576

961

1089
a² - b² = (a+b)(a-b).
Thus:
(" Â¥)² - (Â¥" )² = (" Â¥ + Â¥" )(" Â¥ - Â¥" ).

One approach is to TEST cases and look for a PATTERN.

Case 1: " Â¥=21, Â¥" =12
21² - 12² = (21+12)(21-12) = 33*9 = 3*3*3*11.

Case 2: x=52, Â¥" =25
52² - 25² = (52+25)(52-25) = 77*27 = 3*3*3*7*11.

Case 3: x=73, Â¥" =37
73² - 37² = (73+37)(73-37) = 110*36 = 2*2*2*3*3*5*11.

In every case, the resulting prime-factorization includes 3*3*11.
Implication:
The correct answer choice must be divisible by both 3*3=9 and 11.

An integer is divisible by 9 only if the sum of its digits is a multiple of 9.
A: 1+2+1 = 4, which is not a multiple of 9. Eliminate A.
B: 3+6+1 = 11, which is not a multiple of 9. Eliminate B.
C: 5+7+6 = 18, which is a multiple of 9. Hold onto C.
D: 9+6+1 = 16, which is not a multiple of 9. Eliminate D.
E: 1+0+8+9 = 18, which is a multiple of 9. Hold onto E.

Check whether answer choice C is divisible by 11.
C: 576 = 9*69 = 3*3*3*23, which is not a multiple of 11. Eliminate C.

The correct answer is E.
Last edited by GMATGuruNY on Fri Jun 17, 2016 8:15 am, edited 1 time in total.
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by lucas211 » Fri Jun 17, 2016 8:13 am
GMATGuruNY wrote:
"  and Â¥ represent nonzero digits, and (" Â¥)² - (Â¥" )² is a perfect square. What is that perfect square?

121

361

576

961

1089
a² - b² = (a+b)(a-b).
Thus:
(" Â¥)² - (Â¥" )² = (" Â¥ + Â¥" )(" Â¥ - Â¥" ).

One approach is to TEST cases and look for a PATTERN.

Case 1: " Â¥=21, Â¥" =12
21² - 12² = (21+12)(21-12) = 33*9 = 3*3*3*11.

Case 2: x=52, Â¥" =25
52² - 25² = (52+25)(52-25) = 77*27 = 3*3*3*7*11.

Case 3: x=73, Â¥" =37
73² - 37² = (73+37)(73-37) = 110*36 = 2*2*2*3*3*5*11.

In every case, the resulting prime-factorization includes 3*3*11.
Implication:
The correct answer choice must be divisible by both 3*3=9 and 11.

An integer is divisible by 9 only if the sum of its digit is a multiple of 9.
A: 1+2+1 = 4, which is not a multiple of 9. Eliminate A.
B: 3+6+1 = 11, which is not a multiple of 9. Eliminate B.
C: 5+7+6 = 18, which is a multiple of 9. Hold onto C.
D: 9+6+1 = 16, which is not a multiple of 9. Eliminate D.
E: 1+0+8+9 = 18, which is a multiple of 9. Hold onto E.

Check whether answer choice C is divisible by 11.
C: 576 = 9*69 = 3*3*3*23, which is not a multiple of 11. Eliminate C.

The correct answer is E.
Thanks a lot Mitch!
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by DavidG@VeritasPrep » Fri Jun 17, 2016 8:57 am
lucas211 wrote:Hello BTG

Would appreciate a little help on the attached question.
It seems a bit abstract to me, so would appreciate a little explanation in connection to it - if possible.

Thanks in advance
You can also think about this algebraically. Note that the relevant numbers are two-digit integers in which the tens and units digits are swapped. Say the two numbers are 'a and 'b.' We'll say a = 10x + y, so b = 10y + x. Notice that if you take the difference of these two numbers (a- b) you'll get 10x + y - (10y + x) = 9x - 9y = 9(x-y). This is a multiple of 9.

And if you add these two numbers (a+b) you'll get 10x + y + 10y + x = 11x + 11y = 11(x+y). This is a multiple of 11.

As Mitch noted, we can think of our expression here as a^2 - b^2 = (a-b)(a + b) so in this problem, you're multiplying a multiple of 9 and a multiple of 11; clearly the product must be a multiple of both 9 and 11.

In these types of questions, I like to start with E and work up. 1089 is clearly a multiple of 9 as the digits sum to 18. And 1100 is clearly a multiple of 11, and 1089 is simply 1100 - 11, so this, too, must be a multiple of 11. So we're done. The answer is E.
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by Matt@VeritasPrep » Thu Jun 23, 2016 4:31 pm
(" Â¥)² - (Â¥" )² = n²

(10"  + Â¥)² - (10Â¥ + " Â²) = n²

(100" Â² + 20" Â¥ + ¥²) - (100¥² + 20" Â¥ + " Â²) = n²

99" Â² - 99¥² = n²

Since the left side divides by 99, the right side must also divide by 99. So n² is a multiple of 99.

Only E fits, so we're set!
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