Which of the following is a prime factor of 5^8 - 2^8 ?

Question

I. 2
II. 3
III. 7

(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III

pushpitkc · Answer

Rule:    a^2 - b^2 = (a+b)(a-b)

5^8 - 2^8  can be re-written as  (5^4)^2 - (2^4)^2

Further simplifying the expression, we get
 (5^4)^2 - (2^4)^2 = (5^4 + 2^4)(5^4 - 2^4) = (625+16)(5^2 + 2^2)(5^2 - 2^2) = (641)(29)(21)

5^8 - 2^8  is divisible by prime number(s) 3 and 7(but not 2) and  Option D(II and III)  is our answer!

Gladiator59 · Answer

5^8 - 2^8 = (5^4 + 2^4)(5^4 - 2^4)  ....  a^2 - b^2 = (a + b)(a - b)

= (5^4 + 2^4)*(5^2 + 2^2)*(5^2 - 2^2)

= (5^4 + 2^4)*(5^2 + 2^2) * (5 + 2) ( 5 - 2)

hence  7 and 3 are prime factors  of the given number.

Also note that  for all other factors... we are adding an odd number  ( power of 5)  to an even number  ( power of 2) ... hence it is always odd.

Hence 2 is not a prime factor of given number.

Hence  Option (D)  is correct.

Best,
Gladi

aggvipul · Answer

5^8 - 2^8 is similar to a^n - b^n
Pls note that (a^n - b^n) is always divisible by (a-b); if n is even (a^n - b^n) is always divisible by (a+b)

Here:  a+b =  7  and  a-b =  3  are the factors(both prime) of 5^8 - 2^8.
Also note that, 5^8 is an odd no. and 2^8 is an even no. and therefore their subtraction will be an odd no. hence 2 cant be the factor

Ans. D

ScottTargetTestPrep · Answer

Using the difference of squares, we have:

(5^4 - 2^4)(5^4 + 2^4) = (5^2 - 2^2)(5^2 + 2^2)(5^4 + 2^4) = 21 x 29 x 641

We can quickly see that 21 x 29 x 641 does not contain a prime factor of 2.

We can also see that 21 is divisible by both 3 and 7, so 3 and 7 are both prime factors of 5^8 - 2^8. 7.

Answer: D

PVOG · Answer

If anyone is interested, you don't need to actually figure out what the sum is to determine whether or not it is a multiple of 2, this can be ruled out instantly.

Whenever we have a multiple of an integer 'n' added to or subtracted from a NON-multiple of that same integer 'n', the result of the sum will never be a multiple of n.

5^n will never be a multiple of 2, 5^n will always be odd, its units digit will always be 5... Whereas 2^n will always be a multiple of 2, this completely rules out Statement 1 from being a possibility, bin it

This leaves you only with B and D

JaiPavuluri · Answer

Option 1:

Odd - Even = Odd, So 2 not possible -- A, B and E eliminated

(5^2+2^2) * (5^2-2^2) * (5^4+2^4)

5^2 - 2^2 = 25 - 4 = 21 = 3*7

Option D is Answer

GmatPoint · Answer

Since a^2 - b^2 = (a + b)*(a - b)
 5^8 - 2^8 = ( 5^4 - 2^4 )*( 5^4 + 2^4 )
Now, 5^4 - 2^4 = ( 5^2 - 2^2 )*( 5^2 + 2^2 )
Also, 5^2 - 2^2 = ( 5 + 2 )*( 5 - 2 ) = 3*7

Thus, 5^8 - 2^8 = 3*7*( 5^2 + 2^2 )*( 5^4 + 2^4 )

Thus, 3 and 7 are prime multiples of 5^8 - 2^8.
Also, since 5^4 and 5^2 are odd numbers, and 2^2 and 2^4 are even numbers, their summation will always be an odd number.
Thus, 5^8 - 2^8 is not divisible by 2.

The correct answer is D.

raghav1708 · Answer

5^8 - 2^8

Here, upon my quick calculations, I am able to understand that the power of both of the numbers can tell a lot about the prime factors.

Firstly, any square of 5 will give a 5 in the end, and any square of 2 will give an even number in the end, so  ODD - EVEN = ODD   (Hence option 2 is out)

Now with the powers, if there is an odd power on the numbers, then the number will be divisible by 3, only.   Example (5^1 - 2^1 = 3)

If there is an even power on the numbers, then the number will be divisible by 3 & 7.   Example = (5^2 - 2^2 = 21) Divisible by both 3 & 7

More examples  (5^3 - 2^3 = 117)  Divisible only by  3 ;  (5^4 - 2^4 = 609)  Divisible by both  7 & 3 .

So, the question has an even power, so it must be divisible by 3 & 7.

Bunuel , can you please correct me if I am anywhere wrong here?

Thank you

Which of the following is a prime factor of 5^8 - 2^8 ?

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Which of the following is a prime factor of 5^8 - 2^8 ?

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