xy+1 divisibility by 3

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jsl: Master | Next Rank: 500 Posts; Posts: 203; Joined: Wed Jun 04, 2008 1:01 am; Location: Windsor; Thanked: 5 times; GMAT Score:650

by jsl » Thu Nov 06, 2008 5:30 am

If x and y are integers, is xy+1 divisible by 3?

1) When x is divided by 3, the remainder is 1
1) When y is divided by 9, the remainder is 8

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Source: Beat The GMAT — Data Sufficiency | Thu Nov 06, 2008 5:30 am

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Thu Nov 06, 2008 6:10 am

I would go with C)

Please post OA.

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

Re: xy+1 divisibility by 3

by sudhir3127 » Thu Nov 06, 2008 6:14 am

jsl wrote:If x and y are integers, is xy+1 divisible by 3?

1) When x is divided by 3, the remainder is 1
1) When y is divided by 9, the remainder is 8

clearly 1 and 2 are insufficient..
therefore

X = 3a+1
Y= 9b+8

XY would be (3a+1)(9b+8)

u dont have to solve the equation ..its enuf if u realise that whole equation except the constant part will be will divisible by 3.

the constant part will be 8.
thus XY+ 1 will make the constant as 9 .hence divisble by 3.

hence C .

do let me know the OA

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jsl: Master | Next Rank: 500 Posts; Posts: 203; Joined: Wed Jun 04, 2008 1:01 am; Location: Windsor; Thanked: 5 times; GMAT Score:650

Re: xy+1 divisibility by 3

by jsl » Thu Nov 06, 2008 6:21 am

sudhir3127 wrote:XY would be (3a+1)(9b+8)

u dont have to solve the equation ..its enuf if u realise that whole equation except the constant part will be will divisible by 3.

the constant part will be 8.
thus XY+ 1 will make the constant as 9 .hence divisble by 3.

Thanks for the explanation. OA is C. Howevever, can I just question your comment here?

I think that you do need to know the non-constant part of the equation because if this part is not a multiple of 3, then the whole numerator will not be a multiple of 3. The answer is correct because a is a multiple of 3 and b is a multiple of 9.

Is my logic correct here?

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Thu Nov 06, 2008 6:25 am

Yes

Or Take (I think u can take k for both quotients)

x= 3k+1
y = 9k+8

xy + 1
= (3k+1)(9k+8) + 1
= 27 k^2+33 k + 8+1
= 27 k^2+33 k + 9

The above term will be divisible by 3

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