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What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 10:10
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What is the remainder when ((55)^15!)^188 is divided by 17? A) 1 B) 16 C) 3 D) 5 E) 15 PS: I am not sure whether this falls under 600700 difficulty level.
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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 10:23
What is the remainder when ((55)^15!)^188 is divided by 17? A) 1 B) 16 C) 3 D) 5 E) 15 ((55)^15!)^188 When 55 is divided by 17, the remainder is 4 because we are interested in remainder, we can reduce the previous question as(4^15!)^188 15! will be a even number, so we can safely divide 15! by 2 & raise the power of 4 by 2 i.e. ={16^(15!/2)}^188 ={[171]^(15!/2)}^188 ={[1]^(15!/2)}^188 Negative 1 raised to any even power will reduce to +1 ={1}^188 =1 Thus the remainder is 1 So the answer is A. Hope this explanation helps. Fame
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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 13:33
Why did you plug the remainder of 4 into the problem? I understand 55/17 gives remainder of 4 and I see where you went with the problem but I don't understand why you put the remainder back into the problem? Im lost!



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 14:46
When a number a is divided by another number b, the quotient would be some number q and we would get a remainder r. Since we need to find the remainder, we can ignore the quotient and feed in the remainder. On feeding in the remainder and raising it to the power, we can find the total remainder. Hope this helps! Let me know if you need any further notification. Richard0715 wrote: Why did you plug the remainder of 4 into the problem? I understand 55/17 gives remainder of 4 and I see where you went with the problem but I don't understand why you put the remainder back into the problem? Im lost!



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 17:21
Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again! Kris01 wrote: When a number a is divided by another number b, the quotient would be some number q and we would get a remainder r. Since we need to find the remainder, we can ignore the quotient and feed in the remainder. On feeding in the remainder and raising it to the power, we can find the total remainder. Hope this helps! Let me know if you need any further notification. Richard0715 wrote: Why did you plug the remainder of 4 into the problem? I understand 55/17 gives remainder of 4 and I see where you went with the problem but I don't understand why you put the remainder back into the problem? Im lost!



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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23 Feb 2013, 21:54
We need to find the remainder when ((17)^10)^123) is divided by 3 As you suggested let us divide, 17/3 and feed in the remainder back into the question.
((2)^10)^123)= (1024)^123/3 You can divide 1024 by 3. The remainder is 1. Hence 1^123=1
This was possible because the remainder of 17/3 was a small number,2
Another way of doing this is
((2)^10)^123)=((31)^10)^123 When you divide the data inside the bracket by 3, 3 would get completely divided by 3 and hence can be ignored.
((1)^10)^123 Any negative number raised to an even integer= positive number
1^123=1
Hope that helps!
[quote="Richard0715"]Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again!



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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27 Apr 2013, 01:30
Richard0715 wrote: Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again! Kris01 wrote: When a number a is divided by another number b, the quotient would be some number q and we would get a remainder r. Since we need to find the remainder, we can ignore the quotient and feed in the remainder. On feeding in the remainder and raising it to the power, we can find the total remainder. Hope this helps! Let me know if you need any further notification. Richard0715 wrote: Why did you plug the remainder of 4 into the problem? I understand 55/17 gives remainder of 4 and I see where you went with the problem but I don't understand why you put the remainder back into the problem? Im lost! I would always try to play with 1 wherever it is possible to reduce the time spent on calculations. In your question, 17 can be written as (181) where 18 is divisible by 3 and the rest about 1 is already explained by Kris01. Hope this helps



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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10 Aug 2014, 11:47
fameatop wrote: What is the remainder when ((55)^15!)^188 is divided by 17? A) 1 B) 16 C) 3 D) 5 E) 15
((55)^15!)^188
When 55 is divided by 17, the remainder is 4 because we are interested in remainder, we can reduce the previous question as(4^15!)^188
15! will be a even number, so we can safely divide 15! by 2 & raise the power of 4 by 2 i.e.
={16^(15!/2)}^188 ={[171]^(15!/2)}^188 ={[1]^(15!/2)}^188
Negative 1 raised to any even power will reduce to +1
={1}^188 =1
Thus the remainder is 1
So the answer is A.
Hope this explanation helps.
Fame Hello.. I understand till the point you plugged in 4 in the expression. But post that aren't you only simplifying the expression (4^15!)^188 ? instead of calculating the remainder when divided by 17. I hope I have expressed my doubt clearly. Basically I just lost the context after you plugged in 4 in the expression though I understood why it is done. But after this, I feel you have merely simplified the expression whereas our main task was to find the remainder. Kindly elaborate.



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What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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Updated on: 24 Aug 2014, 21:04
saratchandra wrote: What is the remainder when ((55)^15!)^188 is divided by 17?
A) 1 B) 16 C) 3 D) 5 E) 15
PS: I am not sure whether this falls under 600700 difficulty level. Remainder of 55/17=4 The problem reduces to ((4)^15!)^188 divided by 17 (4^(4*2*3*5*6*...15)^188) We see the remainders of 4^4 by 17 is 1. So the remainder for the whole expression is 1.
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What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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11 Aug 2014, 02:27
fameatop wrote: What is the remainder when ((55)^15!)^188 is divided by 17? A) 1 B) 16 C) 3 D) 5 E) 15
((55)^15!)^188
When 55 is divided by 17, the remainder is 4 because we are interested in remainder, we can reduce the previous question as(4^15!)^188
15! will be a even number, so we can safely divide 15! by 2 & raise the power of 4 by 2 i.e.
={16^(15!/2)}^188 ={[171]^(15!/2)}^188 ={[1]^(15!/2)}^188
Negative 1 raised to any even power will reduce to +1
={1}^188 =1
Thus the remainder is 1
So the answer is A.
Hope this explanation helps.
Fame This method is perfect; however requires some correction as done below. Hopefully, it will be helpful \((55^{15!})^{188} = 55^{15!*188} = (51+4)^{15!*188}\) Using formula \((a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{(n1)} b^n + ........... & so on\), we can conclude that all terms except the last is divisible by 17; so focus only on the last term \(4^{(15!*188)} = 4^{(2*15!*94)} = (4^2)^{(15!*94)} = 16^{15!*94}\) \(16^{15!*94} = (171)^{15!*94}\) By the same expansion concept; the last term \((1)^{(15!*94)}\) would provide a remainder 15!*94 would provide an even result & \((1)^{(Even No)} = +1\) So, Answer = 1 Answer = A
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What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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11 May 2015, 00:07
PareshGmat wrote: fameatop wrote: What is the remainder when ((55)^15!)^188 is divided by 17? A) 1 B) 16 C) 3 D) 5 E) 15
((55)^15!)^188
When 55 is divided by 17, the remainder is 4 because we are interested in remainder, we can reduce the previous question as(4^15!)^188
15! will be a even number, so we can safely divide 15! by 2 & raise the power of 4 by 2 i.e.
={16^(15!/2)}^188 ={[171]^(15!/2)}^188 ={[1]^(15!/2)}^188
Negative 1 raised to any even power will reduce to +1
={1}^188 =1
Thus the remainder is 1
So the answer is A.
Hope this explanation helps.
Fame This method is perfect; however requires some correction as done below. Hopefully, it will be helpful \((55^{15!})^{188} = 55^{15!*188} = (51+4)^{15!*188}\) Using formula \((a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{(n1)} b^n + ........... & so on\), we can conclude that all terms except the last is divisible by 17; so focus only on the last term \(4^{(15!*188)} = 4^{(2*15!*94)} = (4^2)^{(15!*94)} = 16^{15!*94}\) \(16^{15!*94} = (171)^{15!*94}\) By the same expansion concept; the last term \((1)^{(15!*94)}\) would provide a remainder 15!*94 would provide an even result & \((1)^{(Even No)} = +1\) So, Answer = 1 Answer = A Great explanation!!! I was able to solve till this step:4^{(15!*188)} and then I was struck!



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Re: What is the remainder when ((55)^15!)^188 is divided by 17 [#permalink]
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27 Jun 2015, 01:28
Kris01 wrote: We need to find the remainder when ((17)^10)^123) is divided by 3 As you suggested let us divide, 17/3 and feed in the remainder back into the question. ((2)^10)^123)= (1024)^123/3 You can divide 1024 by 3. The remainder is 1. Hence 1^123=1 This was possible because the remainder of 17/3 was a small number,2 Another way of doing this is ((2)^10)^123)=((31)^10)^123 When you divide the data inside the bracket by 3, 3 would get completely divided by 3 and hence can be ignored. ((1)^10)^123 Any negative number raised to an even integer= positive number 1^123=1 Hope that helps! Richard0715 wrote: Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again! Can we solve your example above like this: (2^10)^123 = 2^1230 ... 2 has cycle of 2, 4 , 8 , 6 .. means 4..1230/4 = 370 + 2...2 remaining means last digit will be 4.. 4/3 leaves a remainder of 1.



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