## Tough PS - Number Properties & Exponents

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### Tough PS - Number Properties & Exponents

by late4thing » Wed Feb 17, 2016 8:43 pm
For integers x, y, and z, if ((2^x)^y)^z = 131072 which of the following must be true ?

A. The product xyz is even
B. The product xyz is odd
C. The product xy is even
D. The product yz is prime
E. The product yz is positive

I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ?

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by sanju09 » Thu Feb 18, 2016 12:26 am
late4thing wrote:For integers x, y, and z, if ((2^x)^y)^z = 131072 which of the following must be true ?

A. The product xyz is even
B. The product xyz is odd
C. The product xy is even
D. The product yz is prime
E. The product yz is positive

I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ?
As we know that no perfect square integer ends in either of the digits 2, 3, 7, or 8 and the result 131072 ends in 2; hence 131072 is not a perfect square. In other words, the exponent over 2 (i.e. xyz) is definitely not even, which means that [spoiler]xyz must be odd.

Pick B
[/spoiler]
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by Marty Murray » Thu Feb 18, 2016 3:58 am
late4thing wrote:For integers x, y, and z, if ((2^x)^y)^z = 131072 which of the following must be true ?

A. The product xyz is even
B. The product xyz is odd
C. The product xy is even
D. The product yz is prime
E. The product yz is positive
C, D and E can be eliminated because there is no way to determine these things about two out of the three exponents, because the third exponent can be positive, negative, or a fraction.

For instance, xy could be even or odd, because z could be anything. So if xyz were to have to be odd, and xy were even, then z could be a fraction having an odd number in the numerator and 2 in the denominator, making xyz odd.

So the choice becomes between A, the product xyz is even, and B, the product xyz is odd.

131072 ends in 2. So you can look for a pattern in the powers of 2.

2Â¹ = 2 ends in 2
2Â² = 4 ends in 4
2Â³ = 8 ends in 8
2â�´ = 16 ends in 6
2â�µ = 32 ends in 2

That a power of 2 ending in 2 is an odd power of 2 becomes clear.

Alternate Method

C, D and E can be eliminated because there is no way to determine these things about two out of the three exponents, because the third exponent can be positive, negative, or a fraction.

If xyz is even or odd, then 131,072 must be an integer power of 2.

2Â¹â�° = 1024

So 131,072 = 1024 x (some integer power of 2 around 100).

The only integer power of 2 that is close to 100 is 2â�· = 128

So 131,072 = 1024 x 128 = 2Â¹â�° x 2â�·

10 + 7 = 17

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by Matt@VeritasPrep » Thu Feb 18, 2016 6:04 pm
If we have

2Ë£Ê¸á¶» = 131072 = 2Â¹â�·

we know that x * y * z = 17.

Beyond that, we can't find x, y, and z individually. We could have x = 17, y = 1, z = 1, or x = 17/2, y = 2, z = 1, etc. etc.

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by sanju09 » Fri Feb 19, 2016 11:48 pm
Be careful friends and reread "For integers x, y, and z" to avoid consider fractional possibilities.
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by Brent@GMATPrepNow » Sat Feb 20, 2016 2:01 pm
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