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

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

Thanks, guys. Appreciate your quick response.

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.

7^54 /24 leaves a remainder of 1. similary 5^54 leaves a remainder of 1 => 1-1=0

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..

You're right, I was just trying to offer an alternative point of view to this question. Your approach is definitely the fastest

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.

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

peace out
Stone cold

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 ..
Peace out

This is by far the cleanest fastest answer.

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

Hi All,

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

Thanks

Hello!

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

Kind regards!

Solution with only 11s

We know that any number under the format n^2-1 is divisible by 12 or 24 (r=0)

7^74-5^74 = (7^74-1)-(5^74-1)

Then 7^74-1 is divisible by 24
5^74-1 is divisible by 24
=> remainder =0

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