Exponent Issues

(1027)^199/5

(2)^199/5 [as 1027/5 leaves remainder "2"]

199/4 = Remainder = 3

So, rewrite (2)^3/5

8/5 = 3

Answer [spoiler]{C}[/spoiler]

Another Approach

(2)^199/5 [as 1027/5 leaves remainder "2"]

{[(2)^4]49 * 2^3}/5 [break 2 in such a power which yields remainder "1" when divided by 5]

As, (2^4)/5 leaves remainder 1

So,

(1)^49 * 2^3/5
= 8/5 = 3

AIM TO CRACK GMAT wrote:What is the remainder when 1027^199 is divided by 5?

a) 1
b) 2
c) 3
d) 4

[spoiler]OA = C[/spoiler]

Cyclicity of 7 is 4 (7,9,3,1). Thus considering only the units digit of 1027 we can find the remainder.
units digit of 7^200 is 1
thus units digit of 7^199 would be 3
Any number having 3 as the units digit when divided by 5 would yield a remainder of 3.
eg: 13/5 , 113/5 ..

[spoiler]Answer C : 3[/spoiler]

AIM TO CRACK GMAT wrote:What is the remainder when 1027^199 is divided by 5?

a) 1
b) 2
c) 3
d) 4

[spoiler]OA = C[/spoiler]

Hi,
Last Digit of any number is sufficient to find whether the number is divisible by 5 or not(or to find the reminder)

So 1027^199 can be rephrased as 7^199

Now note the patter of positive powers of 7 below,

7^1 ends in 7
7^2 ends in 9(7*7=49)
7^3 ends in 3(9*7)
7^4 ends in 1(3*7)

So series goes like this 7,9,3,1,7,9,3,1...........

We are asked to find 7^199.
You can note that series repeats after every 4 number.

Divide 199/4 gives the remainder 3 so 7^199 = 7^3
7^3 End with 3

Hence Answer is C

Regards,
Uva.

AIM TO CRACK GMAT wrote:What is the remainder when 1027199 is divided by 5?What is the remainder when 1027^199 is divided by 5?

a) 1
b) 2
c) 3
d) 4

[spoiler]OA = C[/spoiler]

1027 mod 5 = 2
2^199 mod 5 = ((2^2)^99 * 2) mod 5
= (4^99 * 2) mod 5
= ((-1)^99 * 2) mod 5
= (-1 * 2) mod 5
= (-2) mod 5
= 3

Choose c

Cheers

AIM TO CRACK GMAT wrote:What is the remainder when 1027199 is divided by 5?What is the remainder when 1027^199 is divided by 5?

a) 1
b) 2
c) 3
d) 4

[spoiler]OA = C[/spoiler]

AIM TO CRACK GMAT, although you have received solutions for the problem, I would advise you to correct the question; this will help to maintain the quality of the post.

Thanks!