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

Question

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

chetan2u · Accepted Answer

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..

greenoak · Answer

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.

maratikus · Answer

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

GMAT TIGER · Answer

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.

abdulfmk · Answer

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

gmat6nplus1 · Answer

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

chetan2u · Answer

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..

gmat6nplus1 · Answer

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

ynaikavde · Answer

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.

stonecold · Answer

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

stonecold · Answer

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

JeffTargetTestPrep · 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

rachelapril · Answer

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

benbitz · Answer

can someone help me understand what this formula is? Anyone have a video to reference or something? I really struggle with this specific type of remainder problem

Bunuel · Answer

This is based on algebraic divisibility rules for powers. Here's what the formulas mean:

When n is  even , a^n - b^n is divisible by both (a - b) and (a + b) 
 When n is  odd , a^n - b^n is divisible by (a - b) 
 When n is  odd , a^n + b^n is divisible by (a + b)

In this question, since the exponent 54 is even, we apply the even-power rule: 7^54 - 5^54 is divisible by both 7 - 5 = 2 and 7 + 5 = 12, so it's divisible by 2 * 12 = 24.

That’s why the remainder is 0 when (7^54 - 5^54) is divided by 24. But honestly, I doubt you'll ever need this for the GMAT.

Fisayofalana · Answer

We already know the units digit for 5 will always be 5. 
To find the units digit of 7. You should be familiar with the pattern (7, 9, 3, 1). The closest multiple of 4 for 7^74 is 72, which will end on a 1. Count the next two and you'll have a units digit of 9. I selected a number ending in 9 for 7 and 5 for 5. 7^2-5^2 and found the remainder to be zero.

Alternatively, I just saw the pattern for evens and odds with the first 4 sets of numbers. 
7^1-5^1 = 2, remainder of 2
7^2-5^2 = 24, remainder of 0
7^3-5^3 = 218, remainder of 2
7^4-5^4 = 1776, remainder of 0

Well, with that, you can see 7^74-5^74 will have a remainder of 0

First method: approx 1 min
Second method: approx 2mins

dnb · Answer

Here, power is even.  So let's replace it with 2 instead. 
 7^2 - 5^2  = 49-25 = 24
24/24 = 1 and  remainder = 0

Ans:  A

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?

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