For n=2, minimum value of x=1

number having unit digit in 0 or 5 are divisible by 5

No power raise to 4 will give 0 as unit digit, so option A is ruled out

Ans. B) 1

deepaks04 wrote:For n=2, minimum value of x=1

number having unit digit in 0 or 5 are divisible by 5

No power raise to 4 will give 0 as unit digit, so option A is ruled out

Ans. B) 1

B is right, but a better explanation is still awaited.

sanju09 wrote:For what least value of x, both 4^(n + 1) + x and 4^(2 n) - x are divisible by 5, n is an even positive integer?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


[spoiler]Made up[/spoiler]

4^(n + 1) will always end with a 4 in its unit digits provided n is an even positive integer. So we need X to be 1 so that

4^(n + 1) + x is divisible by 5

4^(2 n) always end with 6 in its unit's digits provided n is an even positive integer. So we need X to be 1 so that

4^(2 n) - x is divisible by 5.

But Sanju, only 1 can satisfy the given condition.. why all other options?? Not sure whether i was rite! feeling sleepy

I think u can modify the options like 1,6,11,16...

sanju09 wrote:For what least value of x, both 4^(n + 1) + x and 4^(2 n) - x are divisible by 5, n is an even positive integer?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Hi sanju09,

Nice idea for a problem, and I guess gmatmachoman "got the picture" and realized that the "key" was controlling the unit´s digit of the integer powers of 4...

Just a small detail: I guess you should have put that x is (for instance) a non-negative integer in the question stem, otherwise it is not hard to see that x=-4 is also possible, and also x=-4-5=-9 , x=-14, etc...

fskilnik wrote:

sanju09 wrote:For what least value of x, both 4^(n + 1) + x and 4^(2 n) - x are divisible by 5, n is an even positive integer?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Hi sanju09,

Nice idea for a problem, and I guess gmatmachoman "got the picture" and realized that the "key" was controlling the unit´s digit of the integer powers of 4...

Just a small detail: I guess you should have put that x is (for instance) a non-negative integer in the question stem, otherwise it is not hard to see that x=-4 is also possible, and also x=-4-5=-9 , x=-14, etc...

Thank you very much fskilnik for liking the idea, yeah gmatmachoman did justice with that and he came up with a nice suggestion too. Although, we can ignore his this statement that "4^(2 n) always end with 6 in its unit's digits provided n is an even positive integer", but we cannot ignore his suggestion to modify answers, really; let's not forget that he was feeling sleepy too.

I must confess in the end that the question should have been worded as under:

For what least non-negative integer value of x, both 4^(n + 1) + x and 4^(2 n) - x are divisible by 5, n is an even positive integer?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Thanks

sanju09 wrote:Thanks

My pleasure! (Sorry for this delay; I did not see you reply till now!!)

fskilnik wrote:
Just a small detail: I guess you should have put that x is (for instance) a non-negative integer in the question stem, otherwise it is not hard to see that x=-4 is also possible, and also x=-4-5=-9 , x=-14, etc...

I must say you have a excellent approach, Its been pleasure reading all your comments and suggestions.....

Every now and then you prove the point that are very critical..
Thanks a lot

goyalsau wrote:

fskilnik wrote:
Just a small detail: I guess you should have put that x is (for instance) a non-negative integer in the question stem, otherwise it is not hard to see that x=-4 is also possible, and also x=-4-5=-9 , x=-14, etc...

I must say you have a excellent approach, Its been pleasure reading all your comments and suggestions.....

Every now and then you prove the point that are very critical..
Thanks a lot

Thank YOU for your kind words, goyalsau!

All the best,
Fabio.