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Interesting GMATFix Problem-46

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Interesting GMATFix Problem-46

by arora007 » Mon Oct 04, 2010 4:21 am
x and y are positive odd integer. What is the remainder when the product of xy is divided by 18?
1)When x is divided by 9, the remainder is 3
2) y-1 is a multiple of 6
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by goyalsau » Mon Oct 04, 2010 4:32 am
I think it should be E,
Am i correct?
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by selango » Mon Oct 04, 2010 5:16 am
Clearly stmt1 and stmt2 alone not sufficient.

combining 1 and 2,

x=9a+3 (21,39,57,75...)

y-1=6b

y=6b+1 (7,13,19,25)

watever the combination we try the remainder is 3.

Pick C
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by okletsdothis » Mon Oct 04, 2010 9:52 am
how did u get 3 ? and what is the answer here...pls always put the OA with your question.
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by blaster » Tue Oct 05, 2010 1:30 am
i don't think that answer should be 3 (it must be whether 2 or 4)

but my pick is C
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by mj78ind » Tue Oct 05, 2010 4:54 am
selango wrote:Clearly stmt1 and stmt2 alone not sufficient.

combining 1 and 2,

x=9a+3 (21,39,57,75...)

y-1=6b

y=6b+1 (7,13,19,25)

watever the combination we try the remainder is 3.

Pick C
It is ironic that your approach gives me an idea which proves the answer you suggest is correct but the reasoning presented by you is slightly off (I had solved it using brute force and come to C as well!)

x*y = (9a+3)*(6b+1) = 54a + 18b+9a + 3

When we divide this by 18, we get 0 remainders from the first two terms. As we know a is odd, let a = 2p+1 then the last two terms become 9*(2p+1) + 3 when divided by 18 this will always leave a remainder 12

Hence C
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by reply2spg » Tue Oct 05, 2010 6:05 am
C is correct.

x = {21, 39, 57}
y = {7, 13, 19}

you will always get remainder 1 whenever you divide xy by 18.
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by arora007 » Tue Oct 05, 2010 6:30 am
OA is C
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by anantbhatia » Tue Oct 05, 2010 8:30 am
As we know a is odd, let a = 2p+1
@MJ

why have you assumed a as odd?
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by selango » Tue Oct 05, 2010 8:45 am
anantbhatia wrote:
As we know a is odd, let a = 2p+1
@MJ

why have you assumed a as odd?
x=9a+3

Given X is odd.So a must be even in order for X to be odd.

So a=2p

Last 2 terms=18p+3[p=1,2...]

Sub any values of p u ll get remainder as 3
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by anantbhatia » Tue Oct 05, 2010 9:02 am
:(
missed out reading tht x, y are odd.
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by pharmxanthan » Tue Oct 05, 2010 9:55 am
mj78ind wrote:
selango wrote:Clearly stmt1 and stmt2 alone not sufficient.

combining 1 and 2,

x=9a+3 (21,39,57,75...)

y-1=6b

y=6b+1 (7,13,19,25)

watever the combination we try the remainder is 3.

Pick C
It is ironic that your approach gives me an idea which proves the answer you suggest is correct but the reasoning presented by you is slightly off (I had solved it using brute force and come to C as well!)

x*y = (9a+3)*(6b+1) = 54a + 18b+9a + 3

When we divide this by 18, we get 0 remainders from the first two terms. As we know a is odd, let a = 2p+1 then the last two terms become 9*(2p+1) + 3 when divided by 18 this will always leave a remainder 12

Hence C
if we pick numbers, we ger remainder 1 when xy is divided by 18.?
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by [email protected] » Sat Oct 16, 2010 1:52 am
Hi..
Why cant the values of X be 12,21,30,39 and so on...

So considering this..
X=12,21,30,39....
Y=7,13,19,25 and so on...
Case 1:
xy ie. when x=12 and y =7
SO XY =84 and wen divided by 18 the remninder is 12

Case 2: x=21 and y =7
xy= 147 and wen divided by 18 the reminder is 3.
Pls help.
hz do i answer this.
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by [email protected] » Sat Oct 16, 2010 1:54 am
oops sorry sorry.. mistake.. i got it.. oits oly odd positive integers.. ahh..
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