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by AIM TO CRACK GMAT » Sun Oct 20, 2013 6:31 am
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]
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by theCodeToGMAT » Sun Oct 20, 2013 6:38 am
(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]
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by theCodeToGMAT » Sun Oct 20, 2013 6:42 am
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
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by mevicks » Sun Oct 20, 2013 6:43 am
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]
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by Uva@90 » Sun Oct 20, 2013 6:46 am
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.
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by ganeshrkamath » Sun Oct 20, 2013 6:48 am
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

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by Brent@GMATPrepNow » Sun Oct 20, 2013 6:55 am
Just want to mention two things.

1. ganeshrkamath and theCodeToGMAT are using something called "modular arithmetic" to solve this question. While this works, the GMAT does not require you to know this for the test.

2. Uva@90 and mevicks are using a different approach that does not require modular arithmetic. This strategy is covered in this free video: https://www.gmatprepnow.com/module/gmat- ... ts?id=1031

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by theCodeToGMAT » Sun Oct 20, 2013 6:56 am
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!
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