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divisibility and sums

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divisibility and sums

by acecoolan » Sun Nov 09, 2008 8:18 pm
x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Which of the following is NOT a possible value for y + b?
24
21
20
17
15
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Re: divisibility and sums

by logitech » Sun Nov 09, 2008 8:40 pm
acecoolan wrote:x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Which of the following is NOT a possible value for y + b?
24
21
20
17
15
well, y>6 and b>9 x+y>15 Hence, E

Why y is greater than 6 is your homework. :wink:
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by acecoolan » Sun Nov 09, 2008 9:02 pm
well ..logitech. I already know that from the solution to this problem - this is a MGMAT question ... :)

They have followed the same logic that you have but I was hoping to see an alternate soln if at all.
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by stop@800 » Sun Nov 09, 2008 9:21 pm
acecoolan wrote:well ..logitech. I already know that from the solution to this problem - this is a MGMAT question ... :)

They have followed the same logic that you have but I was hoping to see an alternate soln if at all.
There is no alternate solution

Let me try and justify the same

x/y gives remainder as 6
since division by y is giving a reminder 6 than y has to be greater than 6

try for y less than or equal 6, the remainder can never be 6

so y is 7 8 9 ............


similarly

b will be > 9
b = 9 10 11 12 13.......


min of y+b will be 9+7 = 16
hence
value < 16 is not possible

Hope this helps
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by acecoolan » Sun Nov 09, 2008 9:44 pm
Cool. Thanks stop and logitech
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by raunekk » Sun Nov 09, 2008 10:44 pm
when X/Y the remainder is 6

thus Y has to be greater than 6

Also When A/B the reminder is 9

thus B has to be greater than 9

Thus A+B has to be greater than 9+6=15

Hence E
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by cramya » Mon Nov 10, 2008 6:13 am
Every 6 th number is divisible by 6 (the maximun reminder u can get dividing by 6 is 5) so to get a remainder of 6 y has to be atleast 7. Every 9th number(8 is the maximum remainder u can get dividiing by 9) is divisible by 9 so b has to be alteast 10 which makes the sum 17 (15 cannot be the sum in any case)

E)

Nice problem though...
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