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What is the remainder when 7^74 - 5^74 is divided by 24?

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What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 03 Jul 2008, 13:19
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What is the remainder when 7^74 - 5^74 is divided by 24?

A. 0
B. 1
C. 2
D. 3
E. None of these
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 06:22
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abdulfmk wrote:
7^54 /24 leaves a remainder of 1. similary 5^54 leaves a remainder of 1 => 1-1=0

hi abdul..
there is a straight formula for such kind of questions..
1) a^n-b^n is divisible by both a-b and a+b if n is even..
2)a^n-b^n is divisible by a-b if n is odd.
3)a^n+b^n is divisible by a+b if n is odd...
we will use 2 and 3 here..

so 7^54-5^54 = (7^37-5^37)(7^37+5^37)..
now we will use 2 and 3 here..
so (7^37-5^37) is div by 7-5 =2 and (7^37+5^37) by 7+5 or 12..
combined by 2*12 or 24
so remainder 0..
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 03 Jul 2008, 14:45
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In addition to maraticus's solution:

It may be useful to remember that a^n-b^n is always divisible by (a-b).

So, when we write 49^37-25^37, we can note that 49-25=24, and thus, the expression can be evenly divided by 24.
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 03 Jul 2008, 14:20
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wizardofwashington wrote:
What is the remainder when 7^74-5^74 is divided by 24?

Is there an easy way to solve this?


easiest way for me: 7^74 - 5^74 = (49)^37-25^37 = (24*2+1)^37 - (24+1)^37 -> remainder is 1^37 - 1^37 = 0
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 03 Jul 2008, 14:57
greenoak wrote:
In addition to maraticus's solution:

It may be useful to remember that a^n-b^n is always divisible by (a-b).

So, when we write 49^37-25^37, we can note that 49-25=24, and thus, the expression can be evenly divided by 24.


Thanks, guys. Appreciate your quick response.
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Jul 2008, 23:20
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maratikus wrote:
wizardofwashington wrote:
What is the remainder when 7^74-5^74 is divided by 24?

Is there an easy way to solve this?


easiest way for me: 7^74 - 5^74 = (49)^37-25^37 = (24*2+1)^37 - (24+1)^37 -> remainder is 1^37 - 1^37 = 0


I like your approach. however i figured out as under:

7^1 - 5^1 = 2 so reminder = 2.
7^2 - 5^2 = 24 ...... so reminder 0.
7^3 - 5^3 = 48 ...... so reminder 0.
7^4 - 5^4 = 218 ....... so reminder = 2

similarly;
7^73 - 5^73 should have a reminder of 2.
7^74 - 5^74 should have a reminder of 0.
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 05:43
7^54 /24 leaves a remainder of 1. similary 5^54 leaves a remainder of 1 => 1-1=0
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What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 06:25
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wizardofwashington wrote:
What is the remainder when 7^74 - 5^74 is divided by 24?

A. 0
B. 1
C. 2
D. 3
E. None of these


If you test a couple of divisions a pattern emerges

7/24 reminder 7; \(7^2\)/24 reminder 1; \(7^3\)/24 reminder 7. We can conclude that when 7^even non-negative integer, reminder is going to be 1; when 7^odd positive integer, reminder is going to be 7.

Thus \(7^7^4\) will have a reminder of 1

except for 5^1/24, which yields reminder 5. 5^2/24 yields reminder 1, 5^3/24 yields reminder 5, 5^4 yields reminder 1. Thus we can assume that 5^74 will yield reminder 1.

Now R1-R1=R0=multiple of 24.

Answer A
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 07:16
gmat6nplus1 wrote:
wizardofwashington wrote:
What is the remainder when 7^74 - 5^74 is divided by 24?

A. 0
B. 1
C. 2
D. 3
E. None of these


If you test a couple of divisions a pattern emerges

7/24 reminder 7; \(7^2\)/24 reminder 1; \(7^3\)/24 reminder 7. We can conclude that when 7^even non-negative integer, reminder is going to be 1; when 7^odd positive integer, reminder is going to be 7.

Thus \(7^7^4\) will have a reminder of 1

except for 5^1/24, which yields reminder 5. 5^2/24 yields reminder 1, 5^3/24 yields reminder 5, 5^4 yields reminder 1. Thus we can assume that 5^74 will yield reminder 1.

Now R1-R1=R0=multiple of 24.

Answer A


hi gmat6nplus1,
there are various ways to do these type of questions ..
but remember, its all about time, so very important to the easiest way .
ill just tell u three ways ..
1) just explained above by me. if u know these rules, the ans will take exactly 10 seconds..
2) as you have written by finding a pattern. may be slightly time consuming.
3) remainder theorem... for example this very Q..
mod for 24or 2^3*3 here will be=2^3*3*(1/2)(2/3)=8. It means for 24, whatever it has to divide say 'a', a^8x will be divisible by 24..
now back to the Q.. 7^74 =7^(8*9+2)= 7^(8*9)+7^2=0 +49..
similarily 5^74=5^(8*9+2)= 5^(8*9)+7^2=0 +25..
combining remainder of 7^74 - 5^74 = 49-25=24, which itself is div by 24...
here this method takes a bit longer but required where the eq is not of this form..
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 07:53
Quote:
hi gmat6nplus1,
there are various ways to do these type of questions ..
but remember, its all about time, so very important to the easiest way .
ill just tell u three ways ..
1) just explained above by me. if u know these rules, the ans will take exactly 10 seconds..
2) as you have written by finding a pattern. may be slightly time consuming.
3) remainder theorem... for example this very Q..
mod for 24or 2^3*3 here will be=2^3*3*(1/2)(2/3)=8. It means for 24, whatever it has to divide say 'a', a^8x will be divisible by 24..
now back to the Q.. 7^74 =7^(8*9+2)= 7^(8*9)+7^2=0 +49..
similarily 5^74=5^(8*9+2)= 5^(8*9)+7^2=0 +25..
combining remainder of 7^74 - 5^74 = 49-25=24, which itself is div by 24...
here this method takes a bit longer but required where the eq is not of this form..


You're right, I was just trying to offer an alternative point of view to this question. Your approach is definitely the fastest :)
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 04 Feb 2015, 08:17
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a^n – b^n = (a – b)(a^n – 1 + a^(n – 2)b + a^(n – 3)b^2 + ··· + + ab^(n – 2) + b^(n – 1))

This might be useful.
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 17 Mar 2016, 06:03
I actually didi it this way=>
7-5=> remainder =2
7^2-5^2=> remainder =0
7^3-5^3 => remainder =2
hence the pattern => 2,0,2,0.....
hence remainder =>0 as 74 is even
if the terms were both 73 => the remainder would be 2
hence A

Alternatively dont get into this mess => use the identity
hence A^2-B^2=> hence answer => zero
i will try and do it this way next time

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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 21 Mar 2016, 08:58
Okay I don't know many rules as most you do But i surely know binomial.
Here is what i did
7^74 => 49^37 => (48+1)^37 => 12P +1 for some p
and 5^74 => (24+1)^37 => 12Q +1
now subtracting them => 12P+1 - 12Q -1 => 12 (P-Q) => remainder => ZERO
i hope it helps ..
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 21 May 2017, 15:37
greenoak wrote:
In addition to maraticus's solution:

It may be useful to remember that a^n-b^n is always divisible by (a-b).

So, when we write 49^37-25^37, we can note that 49-25=24, and thus, the expression can be evenly divided by 24.



This is by far the cleanest fastest answer.
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 16 Jun 2018, 09:56
wizardofwashington wrote:
What is the remainder when 7^74 - 5^74 is divided by 24?

A. 0
B. 1
C. 2
D. 3
E. None of these


Note that 7^2 = 49 produces a remainder of 1 when divided by 24. Since 7^3 = 7^2 x 7, the remainder from the division of 7^3 by 24 is 7 and since 7^4 = 7^2 x 7^2, the remainder from the division of 7^4 by 24 is 1. We can generalize this as follows: 7^n produces a remainder of 1 when n is even and produces a remainder of 7 when n is odd.

Similarly, 5^2 = 25 produces a remainder of 1 when divided by 24. Since 5^3 = 5^2 x 5, the remainder from the division of 5^3 by 24 is 5 and since 5^4 = 5^2 x 5^2, the remainder from the division of 5^4 by 24 is 1. We can generalize this as follows: 5^n produces a remainder of 1 when n is even and produces a remainder of 5 when n is odd.

Thus, the remainder from division of 7^74 and 5^74 by 24 are both 1 and the remainder from the division of 7^74 - 5^74 is 1 - 1 = 0.

Answer: A
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 05 Aug 2018, 20:45
Hi All,

Will it be possible to find the answer using the cyclicity of digits?

Thanks
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?  [#permalink]

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New post 05 Jan 2019, 13:01
chetan2u wrote:
abdulfmk wrote:
7^54 /24 leaves a remainder of 1. similary 5^54 leaves a remainder of 1 => 1-1=0

hi abdul..
there is a straight formula for such kind of questions..
1) a^n-b^n is divisible by both a-b and a+b if n is even..
2)a^n-b^n is divisible by a-b if n is odd.
3)a^n+b^n is divisible by a+b if n is odd...
we will use 2 and 3 here..

so 7^54-5^54 = (7^37-5^37)(7^37+5^37)..
now we will use 2 and 3 here..
so (7^37-5^37) is div by 7-5 =2 and (7^37+5^37) by 7+5 or 12..
combined by 2*12 or 24
so remainder 0..


Hello!

Could you please explain why are we using rules 2 and 3 if in this particular problem n=74=enven?

Kind regards!
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Re: What is the remainder when 7^74 - 5^74 is divided by 24?   [#permalink] 05 Jan 2019, 13:01
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