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Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
IshanGre wrote:
I think option A should be the answer since we raise 13 to different first few positive powers I got different units therefore same when we are adding 13^x +12^x then we are ought to get more than 4 types of unit digit.

Please clarify?

Answer is B
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Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
Bunuel wrote:
\(13^x + 12^x = n\), where x is a positive integer

Quantity A
Quantity B
The number of distinct numbers that can be the units digit of n
4


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


Kudos for correct solution.

My Answer is B
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Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
Bunuel wrote:
\(13^x + 12^x = n\), where x is a positive integer

Quantity A
Quantity B
The number of distinct numbers that can be the units digit of n
4


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


Kudos for correct solution.

My Answer is B
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Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
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