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Exponents with exponents

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by Brent@GMATPrepNow » Sat Jan 31, 2009 3:47 pm
If x and y are positive integers and x+y>3, then what is the units digit of 87^(x^y)?

(1) x is a multiple of 6
(2) y is a multiple of 2

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
Brent Hanneson - Creator of GMATPrepNow.com
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Source: — Data Sufficiency |

by dmateer25 » Sat Jan 31, 2009 4:11 pm
Here is the pattern of the ones digit: 7 9 3 1 7 9 3 1 7 9 3 1

(1) x is a multiple of 6

Let's say x = 6 and y =1

then it would be 87^6 and the ones digit would be 9

Let's say x = 6 and y = 2

then it would be 87^12 and the ones digit would be 1

Insuff

(2) y is a multiple of 2
if y=2 and x=3

then it would be 87^6 and the ones digit would be 9

if y =4 and x=3

then it would be 87^12 and the ones digit would be 1

Insuff

Together

if x=6 and y=2

then it would be 87^12 and the ones digit would be 1

if x=6 and y=4

then it would be 87^24 and the ones digit would be 1

if x=12 and y=3

then it would be 87^36 and the ones digit would be 1

I will go with C
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by Brent@GMATPrepNow » Sat Jan 31, 2009 4:18 pm
Nice work, dmateer - the answer is C

The important point here is that if x is a multiple of 6, and y is greater than 1, then x^y will be divisible by 4. If x^y is divisible by 4, then 87^(x^y) will have a 1 as its units digit.
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by vikram_k51 » Mon Feb 02, 2009 7:53 am
If x and y are positive integers and x+y>3, then what is the units digit of 87^(x^y)?

(1) x is a multiple of 6
(2) y is a multiple of 2


Statements A and B together are sufficient,however,the statements on their own are not.

The digit 7 follows a cyclic pattern of 4.
Now 6=283 and hence 2 is a factor of x.now if y has 2 as a factor it means y is even.Thus the power of e87 will always be evenly distributed by4 and the last digit will be 1.

Hence C
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