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DS - Integer

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karthikpandian19: Legendary Member; Posts: 1665; Joined: Thu Nov 03, 2011 7:04 pmFollowed by: 70 members; Thanked: 165 times

by karthikpandian19 » Mon Dec 05, 2011 10:11 pm

If j andk are positive integers, what is the remainder when (8 x 10power k + j)is divided by 9?

(1) k = 13
(2) j = 1

(sorry, i could not able write better the "10 power k")

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Source: Beat The GMAT — Data Sufficiency | Mon Dec 05, 2011 10:11 pm

amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members; Location: pune; Thanked: 36 times

by amit2k9 » Mon Dec 05, 2011 10:27 pm

sum of the digits = divisible by 9 such as 9,18,81 and so.

8*10^k + j means for k=1,2,3 the values are 80,800,8000 and so

so it depends on the value of j,such as j=1,2 giving 81,82.

a for j=1,2 it is both divisible and not divisible. not sufficient.

b for j=1,since k > 0 (positive integer) meaning the 8*10^k + j = 81,801 and so on.
divisible by 9 always.

hence B it is.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Dec 05, 2011 11:11 pm

karthikpandian19 wrote:If j and k are positive integers, what is the remainder when (8 x 10power k + j)is divided by 9?

(1) k = 13
(2) j = 1

(sorry, i could not able write better the "10 power k")

Any number is divisible by 9 if the sum of digits of the number is divisible by 9.
Now 8 * 10^k + j, where k is any positive integer.
For any positive integer k, the sum of the digits of the expression 8 * 10^k will always be 8.
If k = 1, 8 * 10^1 = 80; sum of digits = 8
If k = 3, 8 * 10^3 = 8000; sum of digits = 8

(1) k = 13 implies 8 * 10^13 + j or the sum of the digits of 8 * 10^13 will be 8, but we do not know the value of j. Hence NOT sufficient.

(2) j = 1 clearly implies that the sum of the digits of 8 * 10^k + 1 will always be 9; SUFFICIENT.

Th correct answer is B.

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