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ds - divisible

by ccassel » Sun Mar 27, 2011 12:08 pm
Hi,

Does anyone have a quick way to answer this question?

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

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

Answer is C

Thanks,
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by ccassel » Sun Mar 27, 2011 12:10 pm
Please include your reasoning in the explanation.

thanks,
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by clock60 » Sun Mar 27, 2011 12:58 pm
hi ccassel
it will be great if next time you put the answer in spoiler,
If x and y are integers, is xy+1 divisible by 3?
(1) When x is divided by 3, the remainder is 1. or x=3k+1
insuff as we know nothing about y
(2) When y is divided by 9, the remainder is 8. or y=9m+8
insuff no info about x
both:simply multiply x*y
(3k+1)(9m+8)+1=27km+24k+9m+8+1 clearly it is divisible by 3 suff
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by MAAJ » Sun Mar 27, 2011 2:25 pm
Agree with Clock60, I arrived to the exact same solution.
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by force5 » Mon Mar 28, 2011 12:05 am
agreed its the best and standard way.Clock got it correct. Answer is C
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by ccassel » Mon Mar 28, 2011 7:38 am
The question requires us to determine if it is sufficient but to take it a step further, how can we determine that the equation "xy + 1" IS certainly divisible by 3 with both 1 and 2?

Does anyone have material on remainder questions that you can direct me to? I need some work in this area.

Thanks in advance.

Chris
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by MAAJ » Tue Mar 29, 2011 9:41 am
Sent you a PM. Check it.

x MUST be 4, 7, 10, 13... etc or any other number that when divided by 3 the remainder is 1 (x = 3k+1)

y MUST be 17, 26, 35, 44... etc or any other number that when divided by 9 the remainder is 8 (y = 9m+8)

xy+1 is divisible by 3 because:

(3k+1)(9m+8) +1
27km+24k+9m+8+1
27km+24k+9m+9 is Divisible by 3
Note that (9km + 8k + 3m + 3) * 3 = 27km+24k+9m+9

If you want to see the inputs:
(4*17)+1 = 69 is Divisible by 3
(7*26)+1 = 183 is Divisible by 3
(4*26)+1 = 105 is Divisible by 3
(13*17)+1 = 222 is Divisible by 3
ccassel wrote:The question requires us to determine if it is sufficient but to take it a step further, how can we determine that the equation "xy + 1" IS certainly divisible by 3 with both 1 and 2?

Does anyone have material on remainder questions that you can direct me to? I need some work in this area.

Thanks in advance.

Chris
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by GMATGuruNY » Tue Mar 29, 2011 12:08 pm
ccassel wrote:Hi,

Does anyone have a quick way to answer this question?

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

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

Answer is C

Thanks,
Another approach is to list values that satisfy the conditions given and to plug in different combinations.

Statement 1: When x is divided by 3, the remainder is 1.
x = 1, 4, 7, 10, 13...

Statement 2: When y is divided by 9, the remainder is 8.
y = 8, 17, 26, 35, 44...

Now list possible values of xy + 1 and determine the remainder when each value is divided by 3:
1*8 + 1 = 9. 9/3 = 3 R0.
4*8 + 1 = 33. 33/3 = 11 R0.
10*17 + 1 = 171. 171/3 = 57 R0.
4*26 + 1 = 105. 105/3 = 35 R0.

Since in every case R=0, sufficient.

The correct answer is C.
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