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If x and y are integers between 10 and 99, inclusive, is (x

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If x and y are integers between 10 and 99, inclusive, is (x [#permalink] New post 10 Aug 2009, 10:03
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If x and y are integers between 10 and 99, inclusive, is
(x - y)/ 9 an integer?

(1) x and y have the same two digits, but in reverse order.
(2) The tens digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.
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Re: place value [#permalink] New post 10 Aug 2009, 10:55
D for me.

1. Picking numbers, 23-32, 21-12, 95-59, 79-97 all satisfy, the (x - y)/ 9 condition. Suff.

2. Picking number, 35-53, 79-97, 86-68, all will have a difference of 18 which when divided by 9 yields 2. Suff.
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Re: place value [#permalink] New post 10 Aug 2009, 11:02
A. for me.

1) e.g. 11-11/9 = 0 integer. This also holds for any other combinations such as 21-12, 23-32 etc etc

2) x= 20 and y = 24. Insuff.
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Re: place value [#permalink] New post 10 Aug 2009, 23:13
A for me ...

i is suffiecient

but ii) you can have 35 and 97
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Re: place value [#permalink] New post 10 Aug 2009, 23:26
yezz wrote:
If x and y are integers between 10 and 99, inclusive, is
(x - y)/ 9 an integer?

(1) x and y have the same two digits, but in reverse order.
(2) The tens digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.


from 1

10a+b = x, 10b+a

10a+b - 10b-a = 9a-9b= 9(a-b) divisible by 9......suff

from 2

10a+(a-2) = x = 9a-2, 10b+(b+2) = y = 11b+2

x-y = 9a-11b...it depends on the values of b....insuff

A
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Re: place value [#permalink] New post 02 Sep 2009, 08:02
absolutely A, because in ii) there are tons of variants. 20 and 46 (31 and 13 as well). In first case x-y is not divisible by 9, but in second it IS.
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Re: place value   [#permalink] 02 Sep 2009, 08:02
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If x and y are integers between 10 and 99, inclusive, is (x

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