Tough PS - Number Properties & Exponents

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 11
Joined: 15 Oct 2015

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 ?

User avatar
GMAT Instructor
Posts: 3650
Joined: 21 Jan 2009
Location: India
Thanked: 267 times
Followed by:81 members
GMAT Score:760

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]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

User avatar
Legendary Member
Posts: 2073
Joined: 03 Feb 2014
Location: New York City Metro Area and Worldwide Online
Thanked: 955 times
Followed by:136 members
GMAT Score:800

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.

2¹ = 2 ends in 2
2² = 4 ends in 4
2³ = 8 ends in 8
2� = 16 ends in 6
2� = 32 ends in 2

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

The correct answer is B.

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.

2¹� = 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 2� = 128

So 131,072 = 1024 x 128 = 2¹� x 2�

10 + 7 = 17

The correct answer is B.
Marty Murray
GMAT Coach
m.w.murray@hotmail.com
https://infinitemindprep.com/
In Person in the New York Area and Online Worldwide

GMAT Instructor
Posts: 2630
Joined: 12 Sep 2012
Location: East Bay all the way
Thanked: 625 times
Followed by:117 members
GMAT Score:780

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

2ˣʸᶻ = 131072 = 2¹�

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.

User avatar
GMAT Instructor
Posts: 3650
Joined: 21 Jan 2009
Location: India
Thanked: 267 times
Followed by:81 members
GMAT Score:760

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.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 15003
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1265 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Feb 20, 2016 2:01 pm
Related Resources
The following free videos provide information that's useful for answering this question: Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
If you enjoy my solutions, I think you'll like my GMAT prep course
Image
Watch these video reviews of my course
And check out these free resources