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If r, s, and t are all positive integers, what is the remainder of 2^p
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23 Oct 2014, 00:40
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Tough and Tricky questions: Remainders. 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, what is the remainder of 2^p
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24 Oct 2014, 00:52
Bunuel wrote: Tough and Tricky questions: Remainders. 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 Sol: The questions becomes what is the remainder when 2^(rst)/10 ? St 1 says: s is even that means r*s*t=even Now 2^(Even number) can end is 4 or 6..So we have to remainders possible. A and D ruled out St 2 says p=4*t or p is a multiple of 4 so we will have expression of the form 2^4 or 2^8 or 2^12...all end up in 6.. So remainder will be 6.. Ans is B
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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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23 Oct 2014, 00:55
Bunuel wrote: Tough and Tricky questions: Remainders. 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 Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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25 Sep 2017, 17:41
Bunuel wrote: Tough and Tricky questions: Remainders. 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 Bunuel how do we know whether or not the question reads 2^(p/10) pr (2^p) /10?



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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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25 Sep 2017, 21:11
Nunuboy1994 wrote: Bunuel wrote: Tough and Tricky questions: Remainders. 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 Bunuel how do we know whether or not the question reads 2^(p/10) pr (2^p) /10? 2^p/10 can mean nothing but (2^p)/10. If it were 2^(p/10), it would have been written that way.
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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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30 Jul 2019, 02:43
WoundedTiger wrote: Bunuel wrote: Tough and Tricky questions: Remainders. 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 Sol: The questions becomes what is the remainder when 2^(rst)/10 ? St 1 says: s is even that means r*s*t=even Now 2^(Even number) can end is 4 or 6..So we have to remainders possible. A and D ruled out St 2 says p=4*t or p is a multiple of 4 so we will have expression of the form 2^4 or 2^8 or 2^12...all end up in 6.. So remainder will be 6.. Ans is B Hi, I think I have missed some concept . can you help me with how 2^4 or 2^8 or 2^12...all end up in 6..?



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If r, s, and t are all positive integers, what is the remainder of 2^p
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30 Jul 2019, 03:12
p=rst
Statement2: p=4t —> \(\frac{2^p}{10}\)=\(\frac{2^{4t}}{10}\)= \(\frac{16^t}{10}\)= \(\frac{(10+6)^t}{10}\)
—> \(\frac{(10^t+10^{t1}*6^1+...+6^t)}{10}\)
Each term is divided by 10, except the last one, \(6^t.\)
If you divide \(6^t\) by 10, —> no matter what kind of positive integer t is, the last digit of it will be 6. The remainder will be always 6. Sufficient
The answer choice is B.



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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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30 Jul 2019, 04:00
lacktutor wrote: p=rst
Statement2: p=4t —> \(\frac{2^p}{10}\)=\(\frac{2^{4t}}{10}\)= \(\frac{16^t}{10}\)= \(\frac{(10+6)^t}{10}\)
—> \(\frac{(10^t+10^{t1}*6^1+...+6^t)}{10}\)
Each term is divided by 10, except the last one, \(6^t.\)
If you divide \(6^t\) by 10, —> no matter what kind of positive integer t is, the last digit of it will be 6. The remainder will be always 6. Sufficient
The answer choice is B. I also observed that the unit digit of the numerator (cyclicity of numerator) is deciding the Remainder when we divide the numerator by 10. FOR EG 2^even no. would give us 4/10 or 16/10 or 64/10 so REMAINDER will be 4,6,4 and so on . i.e unit digits . In option B 2^4t /10 = 16^t/10 and 6 being the unit digit has only 6 in cycle and hence the remainder will be 6 always .



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If r, s, and t are all positive integers, what is the remainder of 2^p
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08 Aug 2019, 05:43
Bunuel wrote: Tough and Tricky questions: Remainders. 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 Concept Tested Cyclicity 2 has a cyclicity of 4: 2^1 => 2, 2^2 => 4, 2^3 => 8, 2^4 => 6 (Focus on the unit digit) 1) Remainder could be 4 or 6 => Not sufficient 2) Remainder would always be 6. => Sufficient => B



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Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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12 Aug 2019, 09:36
Any number when divided by 10, remainder will be unit's digit. For example, 51 divided by 10 has a remainder of 1. This question asks for the remainder when an integer power of 2 is divided by 10. Powers of 2 (2, 4, 8, 16, 32, 64…), we see that the units digit alternates in a consecutive pattern of 2, 4, 8, 6. Thus, we need to know which of the above mentioned possible units digits we have with 2p.
(1) INSUFFICIENT: If s is even, we know that the product rst is even and so is p. Then units digit will be either 4 or 6 (22 = 4, 24 = 16....).
(2) SUFFICIENT: If p = 4t and t is an integer, p must be a multiple of 4. Since every fourth power of 2 ends with 6 (4 = 16, 8 = 256...), we know that the remainder when 2p is divided by 10 is 6. The correct answer is B.
Kudos if you like the explanation.




Re: If r, s, and t are all positive integers, what is the remainder of 2^p
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12 Aug 2019, 09:36






