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Intern  Joined: 27 Oct 2011
Posts: 12
Concentration: Finance, Technology
GMAT Date: 04-10-2012
What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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2
30 00:00

Difficulty:   85% (hard)

Question Stats: 47% (02:01) correct 53% (02:18) wrong based on 324 sessions

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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 600-700 difficulty level.
Senior Manager  B
Joined: 24 Aug 2009
Posts: 445
Schools: Harvard, Columbia, Stern, Booth, LSB,
Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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9
5
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
={[17-1]^(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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Intern  Joined: 08 Nov 2012
Posts: 19
Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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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!
Manager  Joined: 24 Sep 2012
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GMAT 1: 730 Q50 V39 GPA: 3.2
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Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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2
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!
Intern  Joined: 08 Nov 2012
Posts: 19
Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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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!
Manager  Joined: 24 Sep 2012
Posts: 81
Location: United States
Concentration: Entrepreneurship, International Business
GMAT 1: 730 Q50 V39 GPA: 3.2
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Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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1
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)=((3-1)^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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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 (18-1) where 18 is divisible by 3 and the rest about -1 is already explained by Kris01.

Hope this helps
Intern  Joined: 21 Jul 2014
Posts: 15
Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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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
={[17-1]^(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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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 600-700 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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Last edited by SravnaTestPrep on 24 Aug 2014, 21:04, edited 2 times in total.
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What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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6
6
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
={[17-1]^(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 $$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{(n-1)} b^n + ........... & so on$$, we can conclude that all terms except the last is divisible by 17; so focus only on the last term

$$4^{(15!*188)} = 4^{(2*15!*94)} = (4^2)^{(15!*94)} = 16^{15!*94}$$

$$16^{15!*94} = (17-1)^{15!*94}$$

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

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What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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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
={[17-1]^(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 $$(a+b)^n = n_{C_0} a^n b^0 + n_{C_1} a^{(n-1)} b^n + ........... & so on$$, we can conclude that all terms except the last is divisible by 17; so focus only on the last term

$$4^{(15!*188)} = 4^{(2*15!*94)} = (4^2)^{(15!*94)} = 16^{15!*94}$$

$$16^{15!*94} = (17-1)^{15!*94}$$

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

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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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)=((3-1)^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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Re: What is the remainder when ((55)^15!)^188 is divided by 17  [#permalink]

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_________________ Re: What is the remainder when ((55)^15!)^188 is divided by 17   [#permalink] 19 Dec 2018, 10:30
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