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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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What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m  [#permalink]

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29 Sep 2018, 10:42
2
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Difficulty:

65% (hard)

Question Stats:

55% (02:17) correct 45% (02:30) wrong based on 78 sessions

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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  [#permalink]

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29 Sep 2018, 10:52
1
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.

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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m  [#permalink]

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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
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Re: What is the remainder when X^Y is divided by Z , Where X = 32, Y = 3[m  [#permalink]

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29 Sep 2018, 11:02
4
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  [#permalink]

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29 Sep 2018, 11:11
Ans C-----6 (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  [#permalink]

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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  [#permalink]

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30 Sep 2018, 01:33
3
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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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   [#permalink] 05 Mar 2019, 09:17
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