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a and b are two positive integers that are not divisible by

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a and b are two positive integers that are not divisible by [#permalink] New post 06 Mar 2007, 00:31
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a and b are two positive integers that are not divisible by 10, and the sum of the digits of the product of a and b is 1. Which of the following cannot be the remainder when |a-b| is divided by 10?

(I) 5
(II) 9
(III) 0

(A) I only (B) II only (C) III only (D) II and III only (E) I, II and III

Last edited by kevincan on 06 Mar 2007, 10:06, edited 1 time in total.
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Re: PS: Remainder of Special Product [#permalink] New post 06 Mar 2007, 02:55
According to me the answer should be (B).

a*b = 10^n as the sum of the digits is 1.
=> a*b = (5^n) * (2^n)
=> |a-b| is odd but not divisible by 5.
=> the answer is (B).

is the answer and explanation correct?

~newbie here...
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 [#permalink] New post 06 Mar 2007, 08:21
There are three numbers 1, 10, 100 can be product of a and b, and the sum of digits would be 1. Beyond 100 one of the numbers has to be divisible by 10.

1 = 1*1
10 = 2*5
100= 4*25

|1-1| = 0, remainder=0, when divided by 10
|2-5| = 3, remainder=3, when divided by 10
|4-25| = 21, remainder=1, when divided by 10

so the answer is (I) and (II). But do not see the choice with (I) and (II). Am I doing something wrong?
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 [#permalink] New post 07 Mar 2007, 17:29
I go with B.

rdg,
You can try 100 = 5* 20 and then rule out 1.
I solved this similar to how you did it.
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 [#permalink] New post 07 Mar 2007, 19:01
the product of a and b would be :

1 = 1*1 remainder is zero
10= 5*2 remainder is 3
100 = 25*4 remainder is 1
1000 = 125*8 remainder is 7
10000 = 625*16 remainder is 9
100000 = 3125*32 remainder is 3
and so on......

'A' seems to be the answer.
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 [#permalink] New post 07 Mar 2007, 20:29
Thanks. I should have considered with more numbers.
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 [#permalink] New post 07 Mar 2007, 21:13
Kevincan question seems to be wrong then...as all three can be remainder.
And there is no option that says 'none'
  [#permalink] 07 Mar 2007, 21:13
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