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What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 10:42
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Weekly Quant Quiz Question 9 What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13? a) 4 b) 5 c) 6 d) 7 e) 8 ONLY TEXT SOLUTIONS ARE ALLOWED.
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 10:52
What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13? a) 4 b) 5 c) 6 d) 7 e) 8 Ans: x^y => 32^(32^32) Write 32 as 26+6 The exponent is a very large integer call it M. x^y => (26+6)^M In the binomial expansion of the above binomial entity every term will have a factor of 26 except the last term. Hence 13 will divide all the terms except 6. The remainder will be 6. Hence IMO Option (C) is correct. Best, Gladi
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 10:56
What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13? a) 4 b) 5 c) 6 d) 7 e) 8 By calculation the cyclicity of remainders when 13 divides 32 and powers of 32= 6,10,8,4 therefore as 32 is a factor of therefore the remainder is 4 Answer=A
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 11:02
Should be 6, since 32÷13 gives 6 As remainder, and then x^y would be in the form of 32*32*32.... hence comes out to be 6
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 11:11
Ans C6 (13*2 +6) =32
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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29 Sep 2018, 12:53
Ans is C, 6
X* Y = (32) raised to the power of 33
This can be written as (26+6) to the power of 33.
Using binomial expression, we will get (6) power of 33
We now have to check it’s divisibity by 13. Thus remainder is 6
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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30 Sep 2018, 01:33
Hey all, hoping someone can clarify one aspect of this for me as I'm hitting my head against a wall here!
My thought process is: When expanding the binomial (26+6)^M the last term will be 6^M.
6^M/13 doesn't always leave remainder 6. e.g. 6^2/13 = remainder 10... So how can we confidently say it does.
What simple fact am i missing here!



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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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12 Jan 2019, 06:59
This question is incorrect.
\(32^{{32}^{32}}\) mod \(13\) is \(9\) (checked with wolfram alpha)
So how do we get to \(R=9\)?



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What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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Updated on: 12 Jan 2019, 15:40
Hi @gladiator56, thanks for your explanation but I will appreciate if you can explain a bit on this part "" The exponent is a very large integer call it M. x^y => (26+6)^M In the binomial expansion of the above binomial entity every term will have a factor of 26 except the last term. Hence 13 will divide all the terms except 6."" ""From what I know 13 will be divisible by all terms except 6^M and not just 6. Did you just assume or check for cyclicity of remainder of 6^M as well before arriving at OA. I am simply asking the same question MattyE is asking ?. Thanks. Posted from my mobile device
Originally posted by Kem12 on 12 Jan 2019, 14:20.
Last edited by Kem12 on 12 Jan 2019, 15:40, edited 1 time in total.



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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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12 Jan 2019, 14:30
Kem12 wrote: Hi @gladiator56, thanks for your explanation but I will appreciate if you can explain a bit on this part "" The exponent is a very large integer call it M. x^y => (26+6)^M In the binomial expansion of the above binomial entity every term will have a factor of 26 except the last term. Hence 13 will divide all the terms except 6."" From what I know 13 will be divisible by all terms except 6^M and not just 6. Did you just assume or check for cyclicity of remainder of 6^M as well before arriving at OA. I am simply asking the same question MattyE is asking ?. Thanks. Posted from my mobile device Kem12 Question is incorrect IMO, as are all answers. The remainder is 9 which you can easily check if you plug in the numbers as x mod y into wolfram alpha



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What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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05 Mar 2019, 09:17
gmatbusters wrote: Weekly Quant Quiz Question 9 What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32)^(32) and Z = 13? a) 4 b) 5 c) 6 d) 7 e) 8 ONLY TEXT SOLUTIONS ARE ALLOWED. The answer is 9 as per my deduction. gmatbusters please confirm?[
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What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m
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