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If x and y are positive integers, what is the remainder when

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If x and y are positive integers, what is the remainder when [#permalink] New post 10 Jan 2005, 11:26
If x and y are positive integers, what is the remainder when 3^(4+4x)+9^y is divided by 10?
(1) x=25
(2) y=1


I can do this with "C" but OA is "B"
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 [#permalink] New post 10 Jan 2005, 11:36
B is right.

Ones digit of 3 ^ m is always 1 if m is a multiple of 4.

So digit at ones place for 3 ^ (4x + 4) is 1.

(b) says y = 1. So 9 ^ y = 9

So 3^(4+4x)+9^y has the last digit as 0;
therefore remainder is 0.

i.e. (b) is sufficient to find hte remainder.
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 [#permalink] New post 10 Jan 2005, 13:12
ssumitsh, can you be more brief
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 [#permalink] New post 10 Jan 2005, 16:09
3^(4+4x)+9^y ===> 81^(1+x) + 9^y

Statement 1: For any x last digit of 81^(1+x) is always 1, but we don't know abt y so not suff.

Statement 2: Again for any x last digit of 81^(1+x) is always 1, for y = 1 , 9^y = 9...so last digit will be 0, divisible by 10. Suff.


Plz...don't post the OA in the ques....kills the problem solving process right away.
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Re: integers [#permalink] New post 10 Jan 2005, 16:56
pb_india wrote:
If x and y are positive integers, what is the remainder when 3^(4+4x)+9^y is divided by 10?
(1) x=25
(2) y=1


I can do this with "C" but OA is "B"


:arrow: As a rule of thumb with remainder type questions, most likely we are only concerned with the unit digit....so check for unit digits for each number.

this is interesting....we know that the 3^x changes its unit digit every 4 times i.e.

3^1=3
3^2=9
3^3=7
3^4=1
3^5=3<---repeats...
so we know that 3^4x will always give 1 as unit digit, but we dont know what 9^y is if y is odd it will be 9, if it is even then it will be 9. In this problem we only need to know y to determine if it divisible by 10.
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 [#permalink] New post 11 Jan 2005, 09:48
banerjeea_98 wrote:
3^(4+4x)+9^y ===> 81^(1+x) + 9^y

Statement 1: For any x last digit of 81^(1+x) is always 1, but we don't know abt y so not suff.

Statement 2: Again for any x last digit of 81^(1+x) is always 1, for y = 1 , 9^y = 9...so last digit will be 0, divisible by 10. Suff.


Plz...don't post the OA in the ques....kills the problem solving process right away.


Can you show me how to get 81^(1+x) from 3^(4+4x)?
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 [#permalink] New post 11 Jan 2005, 11:21
jinino wrote:
banerjeea_98 wrote:
3^(4+4x)+9^y ===> 81^(1+x) + 9^y

Statement 1: For any x last digit of 81^(1+x) is always 1, but we don't know abt y so not suff.

Statement 2: Again for any x last digit of 81^(1+x) is always 1, for y = 1 , 9^y = 9...so last digit will be 0, divisible by 10. Suff.


Plz...don't post the OA in the ques....kills the problem solving process right away.


Can you show me how to get 81^(1+x) from 3^(4+4x)?



3^(4+4x) = 3^4 * 3^(4x) = 81 * 81^x = 81^(1+x).
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 [#permalink] New post 11 Jan 2005, 12:38
B is the answer to a great question!

Any number substituted for x will give you 3^(multiple of 4) which will always give you 1 in the units digit. So if you add that with 9^y and y = 1 then addition of the two gives you a units digit of 0 which is divisible by 10 with no remainder.


I'm surprised they didn't just say and y is an odd integer as choice B. That would have made a hard question even harder...something I'm sure ETS wouldn't mind...bastards!
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  [#permalink] 11 Jan 2005, 12:38
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