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

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If x and y are integers between 10 and 99, inclusive, is [#permalink] New post 26 Apr 2005, 07:49
24. 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.

Last edited by Rupstar on 26 Apr 2005, 08:30, edited 2 times in total.
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 [#permalink] New post 26 Apr 2005, 08:26
Ques seems incomplete.
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 [#permalink] New post 26 Apr 2005, 09:43
A)...

1) x=10z+w and y=10w+z => insert in eq => 10z+w-(10w+z)/9 => 10z+w-10w-z/9 => 9z-9w/9 => z-w => sufficient => both are multiples of 9

2) insufficient x and y can be the same values as above as well as different values. try 53 and 35 YES or 20 and 35 NO
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 [#permalink] New post 26 Apr 2005, 10:49
christoph wrote:
A)...

1) x=10z+w and y=10w+z => insert in eq => 10z+w-(10w+z)/9 => 10z+w-10w-z/9 => 9z-9w/9 => z-w => sufficient => both are multiples of 9

2) insufficient x and y can be the same values as above as well as different values. try 53 and 35 YES or 20 and 35 NO


ok agreed that A is sufficient in this case IF you assume both the digits are different
but it is possible that we could have 44 or 55 right; thus immplying that x and y are the same integers.
Why is this possibility being ruled out?
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 [#permalink] New post 26 Apr 2005, 11:04
"A"....even when x=y....x-y = 0....which is divisible by all numbers....suff
  [#permalink] 26 Apr 2005, 11:04
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