## If r, s, and t are all positive integers

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### If r, s, and t are all positive integers

by ddm » Tue Aug 26, 2008 8:30 pm
If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t

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### Re: If r, s, and t are all positive integers

by sudhir3127 » Wed Aug 27, 2008 1:04 am
ddm wrote:If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t
i would got for B for this

P= rst

statement 1. s is even .

thus from this we know rst will so will be even

but then

assume rst =4
2^4 /10 = 6 as remainder

2^6 /10 = 4 as remainder

hence insufficient

Statement B
p =4t

thus we know P is a factor of 4

2^ any factor of 4 divided by 10 will always leave a remainder of 6.

hence B is sufficient.

thus B.

hope that helps..

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### Re: If r, s, and t are all positive integers

by mehravikas » Wed Aug 27, 2008 8:16 pm
I would go for 'B' too....nice explanation
sudhir3127 wrote:
ddm wrote:If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t
i would got for B for this

P= rst

statement 1. s is even .

thus from this we know rst will so will be even

but then

assume rst =4
2^4 /10 = 6 as remainder

2^6 /10 = 4 as remainder

hence insufficient

Statement B
p =4t

thus we know P is a factor of 4

2^ any factor of 4 divided by 10 will always leave a remainder of 6.

hence B is sufficient.

thus B.

hope that helps..

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### Re: If r, s, and t are all positive integers

by stern » Thu Aug 28, 2008 11:28 am
ddm wrote:If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t

IMO B. sinct p = 4t

(2^4t)/10 = (16^t)10. Since t is positive the last digit for 16 to the power of anything should be 6. Hence the remainder is going to be 6.

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### Re: If r, s, and t are all positive integers

by stern » Thu Aug 28, 2008 11:30 am
ddm wrote:If r, s, and t are all positive integers, what is the remainder of 2^p/10, if p = rst?

(1) s is even

(2) p = 4t

IMO B. sinct p = 4t

(2^4t)/10 = (16^t)10. Since t is positive the last digit for 16 to the power of anything should be 6. Hence the remainder is going to be 6.

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by indiheats » Tue Jan 01, 2013 7:58 pm
Why can T not be zero ? Making this 1/10 - and therefore a different remainder ... ?

O is a positive integer, is it not ?

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by ceilidh.erickson » Sat Jan 05, 2013 8:25 am
The key to understanding this problem is to think about units digits. Whenever a question asks "what is the remainder when divided by 10?", it's really asking "what is the units digit?"

In this problem, we're asking about the units digit of 2 raised to some power. The units digits of powers of 2 form the following pattern:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6 (only looking at the units digit here)
2^5 = 2
2^6 = 4
etc.

You can see that the units digit repeats every 4 powers. So, if we know that p is a multiple of 4, we'll know that the units digit is 6. Otherwise, we won't know. Statement (1) tells us that s (and therefore p) is a multiple of 2, but that's not enough. The units digit could be 4 or 6. Statement (2) tells us that p is a multiple of 4, though, so it's sufficient.

For more info, check out these posts on patterns of units digits:
https://www.beatthegmat.com/if-n-and-m-a ... tml#544266
https://www.beatthegmat.com/what-is-the- ... tml#544267
Ceilidh Erickson
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by ceilidh.erickson » Sat Jan 05, 2013 8:28 am
indiheats wrote:Why can T not be zero ? Making this 1/10 - and therefore a different remainder ... ?

O is a positive integer, is it not ?
No, 0 is not a positive integer - it's the only integer that's neither positive nor negative! Without that positive constraint, you're right, the answer here would have been E. But with it, statement (2) is sufficient.
Ceilidh Erickson
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