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07 Dec 2021, 05:00
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What smallest positive integer must be added to the number x32xy4232y3 so that the given number is divisible by 11?
(A) 1
(B) 3
(C) 4
(D) 7
(E) 8
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(A) 1
(B) 3
(C) 4
(D) 7
(E) 8
Are You Up For the Challenge: 700 Level Questions
07 Dec 2021, 05:32
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Bunuel
The question needs to make clear that x and y are digits, because as written the number could be interpreted as a product. This question is testing the divisibility test for 11, which the GMAT would certainly never test. But to use it, we sum the odd-placed digits (the first, third, fifth, etc), then subtract the sum of the even-placed digits, and if the result is a multiple of 11, the original number is a multiple of 11. Adding the odd placed digits here, we get x + 2 + y + 2 + 2 + 3 = x + y + 9, and adding the even-placed digits we get 3 + x + 4 + 3 + y = 10 + x + y, and when we subtract we get -1, so if we add 1 to the number, increasing the odd-placed sum by 1 (it's an 11-digit number), we'll get zero, and thus we'll get a multiple of 11.
General Discussion
12 Jun 2022, 06:30
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x32xy4232y3 = (3,200,423,203)+ (X*10^10+X*10^7+Y*10^6+Y*10)
(X*10^10+X*10^7+Y*10^6+Y*10)=10*(X*10^9+X*10^6)+10*(Y*10^5+Y)=10*X*10^6(10^3+1)+10*Y*(10^5+1)
From above we can find (10^3+1) and (10^5+1) are divisible by 11, so it depends on how much add to (3,200,423,203) to make the sum divisible by 11.
further we can find (3,200,423,203) = 290,947,560*11+43, so the solution is 1.
(X*10^10+X*10^7+Y*10^6+Y*10)=10*(X*10^9+X*10^6)+10*(Y*10^5+Y)=10*X*10^6(10^3+1)+10*Y*(10^5+1)
From above we can find (10^3+1) and (10^5+1) are divisible by 11, so it depends on how much add to (3,200,423,203) to make the sum divisible by 11.
further we can find (3,200,423,203) = 290,947,560*11+43, so the solution is 1.
