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17^23 is divided

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17^23 is divided

by sanju09 » Mon Jun 28, 2010 3:07 am
What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
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by kvcpk » Mon Jun 28, 2010 3:17 am
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
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by sanju09 » Mon Jun 28, 2010 3:32 am
kvcpk wrote:
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
B-)
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by gmatmachoman » Mon Jun 28, 2010 3:39 am
kvcpk wrote:
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
I would prefer to follow Rule of Cyclicity over here !!

17 ^1 ends with 7 in unit's digit
17 ^2 ends with 9
17^3 ends with 3
17 ^4 ends with 1

17^5 ends with 7....

So from here on it follows a cyclicty of 4. after that it follows a pattern.

Now 23 can be written as 4*5 +3

So if u find the remainder of 17 ^3 , it will be same as 17^23.

remainder of 17^ 3 when divided by 16 is 1 .

So for 17^23 also it will be 1.
Pick B
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by gmatmachoman » Mon Jun 28, 2010 3:41 am
sanju09 wrote:
kvcpk wrote:
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
B-)
Sanju Bhai,
even i also want a smiley frm u ....!! After all I am also one of ur Quant Commando's..Isnt bhai??
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by nathanalgren » Mon Jun 28, 2010 4:07 am
gmatmachoman wrote:
kvcpk wrote:
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
I would prefer to follow Rule of Cyclicity over here !!

17 ^1 ends with 7 in unit's digit
17 ^2 ends with 9
17^3 ends with 3
17 ^4 ends with 1

17^5 ends with 7....

So from here on it follows a cyclicty of 4. after that it follows a pattern.

Now 23 can be written as 4*5 +3

So if u find the remainder of 17 ^3 , it will be same as 17^23.

remainder of 17^ 3 when divided by 16 is 1 .

So for 17^23 also it will be 1.
Pick B
Although it is the same logic, this method is much easier (and probably more reliable) like that:

17^1 = 1 (mod 16)

So any power of 17 = 1 (mod 16)

This could be implied to any kind of remainder questions.
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by sanju09 » Mon Jun 28, 2010 4:26 am
gmatmachoman wrote:
kvcpk wrote:
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
(16+1)^23 = 16^23 +1^23 +16*(.......)

When divided by 16 it leaves 1 as reminder

B
I would prefer to follow Rule of Cyclicity over here !!

17 ^1 ends with 7 in unit's digit
17 ^2 ends with 9
17^3 ends with 3
17 ^4 ends with 1

17^5 ends with 7....

So from here on it follows a cyclicty of 4. after that it follows a pattern.

Now 23 can be written as 4*5 +3

So if u find the remainder of 17 ^3 , it will be same as 17^23.

remainder of 17^ 3 when divided by 16 is 1 .

So for 17^23 also it will be 1.
Pick B
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by amising6 » Mon Jun 28, 2010 4:30 am
sanju09 wrote:What is the remainder when 17^23 is divided by 16?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
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by sumanr84 » Mon Jun 28, 2010 4:52 am
amising6 wrote: c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
17^23 / 16 does not mean 17*17*17*17*...23times / 16*16*16*16..23 times...why are you dividing each 17 with 16. Ideally, only one time such a division should be made.

Its not, 17+17+17.....23 times / 16 , which you can split into several parts.

I see that a lot of people are trying various innovative approach here. Good for practice..

Don't try to innovate this simple problem. Any wrong strategy that does not hold for general will prove FATAL in exam.

I liked gmatmachoman conventional approach. Stick to it and you won't land up in trouble.
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by amising6 » Mon Jun 28, 2010 5:11 am
sumanr84 wrote:
amising6 wrote: c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
17^23 / 16 does not mean 17*17*17*17*...23times / 16*16*16*16..23 times...why are you dividing each 17 with 16. Ideally, only one time such a division should be made.

Its not, 17+17+17.....23 times / 16 , which you can split into several parts.

I see that a lot of people are trying various innovative approach here. Good for practice..

Don't try to innovate this simple problem. Any wrong strategy that does not hold for general will prove FATAL in exam.

I liked gmatmachoman conventional approach. Stick to it and you won't land up in trouble.
okie dude
lets take 17*17 i.e 17^2=289/16=remainder 1
so isnt 17*17 516=1
check and then say wrong concept dude
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by gmatmachoman » Mon Jun 28, 2010 5:16 am
sumanr84 wrote:
amising6 wrote: c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
17^23 / 16 does not mean 17*17*17*17*...23times / 16*16*16*16..23 times...why are you dividing each 17 with 16. Ideally, only one time such a division should be made.

Its not, 17+17+17.....23 times / 16 , which you can split into several parts.

I see that a lot of people are trying various innovative approach here. Good for practice..

Don't try to innovate this simple problem. Any wrong strategy that does not hold for general will prove FATAL in exam.

I liked gmatmachoman conventional approach. Stick to it and you won't land up in trouble.
Thx suman bhai!! Is there any software that can type as we say..if so let me know...
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by nathanalgren » Mon Jun 28, 2010 5:18 am
Guys, if you know what modular arithmetic is, this is the easiest and fastest way to solve these kind of questions.


17^1 = 1 (mod 16)

So any power of 17 = 1 (mod 16)

This could be implied to any kind of remainder questions. The other approaches written here are okay, but this is far best.

Another example: Suppose it asked the remainder of 18^34 divided by 13.

18^1=5 (mod 13)
18^2=12 (mod 13)
18^3=8 (mod 13)
18^4=1 (mod 13)

The remainder of 34/4 is 2. So we look at 18^2=12 (mod 13).

Hence the remainder of 18^34 divided by 13 is equal to 12.
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by sumanr84 » Mon Jun 28, 2010 5:26 am
gmatmachoman wrote:
sumanr84 wrote:
Thx suman bhai!! Is there any software that can type as we say..if so let me know...
There is one I used long time back when it was free. Just see if you can get any crack version for installation.

https://www.downloadatoz.com/dragon-naturallyspeaking/
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by sumanr84 » Mon Jun 28, 2010 5:33 am
amising6 wrote: c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
All I meant was that you could have phrased it little differently to arrive at the conclusion as to why are we multiplying 1*1*1*1*.... though there is only one 16 for division.

Even when you use this approach it may result in chaos if the problem was 17^23 / 15

You will get 2*2*2*2*..23 times(if I have correctly understood your point) ...then how will you proceed from here...??

nathanalgren has put up a nice approach.
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by amising6 » Mon Jun 28, 2010 5:37 am
sumanr84 wrote:
amising6 wrote: c 17^23 means 17*17*17*17*...23timesif you divide all of this by 16 every 17 will give 1 remainder i.e 1*1*1*1...23 times=1 remainder
All I meant was that you could have phrased it little differently to arrive at the conclusion as to why are we multiplying 1*1*1*1*.... though there is only one 16 for division.

Even when you use this approach it may result in chaos if the problem was 17^23 / 15

You will get 2*2*2*2*..23 times(if I have correctly understood your point) ...then how will you proceed from here...??
now make a group of four 2's i.e 16 *16*16*16*16*8 divide by 15
so every 16 will give 1so u will be left with 1*8 mod 15=8

nathanalgren has put up a nice approach.
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