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by sanju09 » Thu May 22, 2014 2:57 am
What digit will appear in the tens place when a two-digit positive integer, X, raised to the power of 329, is expanded and solved?
(1) The tens digit in X is even.
(2) The ones digit in X is 5.



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by theCodeToGMAT » Thu May 22, 2014 9:42 am
TO find: Tens digit of X^329

Statement 1:
A B
A is EVEN = 2,4,6,8
It can be anything.
Say 20 ==> Tens digit will be "0"
Say 21 ==> Tens digit will not be "0"
INSUFFICIENT

Statement 2:
Unit digit = 5
If 15 then Tens digit will be either 2 or 7
INSUFFICIENT


Combining
Tens=2,4,6,8
Unit=5
For Even Digit, the last two digit will always be 25
So, 2
SUFFICIENT

[spoiler]{C}?[/spoiler]
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by theCodeToGMAT » Mon May 26, 2014 8:27 am
Sanjeev, what's your answer?
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by sanju09 » Tue May 27, 2014 1:27 am
theCodeToGMAT wrote:Sanjeev, what's your answer?
Mine is same as yours. How can it be different? :)
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by theCodeToGMAT » Tue May 27, 2014 1:53 am
sanju09 wrote:
theCodeToGMAT wrote:Sanjeev, what's your answer?
Mine is same as yours. How can it be different? :)
I really like your replies.. So lively!
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