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Expert replies
by [email protected] » Thu Jul 03, 2014 10:03 pm
Hello


I am not sure why cant y = (2^16 - 1) be (2^4-1)(2^4-1)?

Pls explain-Thanks
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Source: — Problem Solving |

by GMATinsight » Fri Jul 04, 2014 12:16 am
Hi Shibsriz,

Please check the explanation of first statement

Statement 1)

y = 2^16-1 = (2)^16 - (1)^16 = (2^8 +1) x (2^8 -1)
= (2^8 +1) x (2^4 +1) x (2^4 -1) = (2^8 +1) x (2^4 +1) x (2^2 +1) x (2^2 -1)
= (2^8 +1) x (2^4 +1) x (2^2 +1) x (2 +1) x (2-1)

Which means y is a multiple of 3 therefore xy will be divisible by 3 SUFFICIENT
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by GMATinsight » Fri Jul 04, 2014 12:19 am
Second statement is not sufficient because if k=0 then x=1 which means x is not a multiple of 3 therefore nothing can be deduced with certainty about xy whether it's a multiple of 3 or not

but if k = 1 then x will have digit sum 6 which will make x as multiple of 3 because of the divisibility test of 3.

Inconsistent answer therefore second statement is not sufficient

Answer Option [spoiler]A[/spoiler]
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by GMATGuruNY » Fri Jul 04, 2014 2:51 am
If x, y and k are integers, is xy divisible by 3?

1. y = 2¹� - 1
2. The sum of the digits of x = 6^k
Information every test-taker should know:
1) a² - b² = (a+b)(a-b).
2) The powers of 2 up to 2¹�.
3) If the sum of the digits of an integer is a multiple of 3, then the integer itself is a multiple of 3.

Statement 1: 1. y = 2¹� - 1
y = 2¹� - 1 = (2� + 1)(2� - 1) = (256 + 1)(256 - 1) = (257)(255)
Since 2+5+5 = 12, 255 is a multiple of 3.
Thus, y is multiple of 3, implying that xy is divisible by 3.
SUFFICIENT.

Statement 2: The sum of the digits of x = 6^k
If k=0, then the sum of the digits of x = 6� = 1.
It's possible that x=1 and y=3, in which case xy is divisible by 3.
It's possible that x=1 and y=2, in which case xy is NOT divisible by 3.
INSUFFICIENT.

The correct answer is A.
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