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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
4
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since x and y is not a multiple of 10 but sum of digits of x*y is 1. thus no can be 10,100,1000 etc with 2^1/2/3/4/5 as x and 5^1/2/3/4/5.. as y. since last two digits of 5 powers are always 5 and 2 follows 2,4,8,6 rule, remainders can be 7,9,3,1.
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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
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Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

x and y are positive integers and the sum of the digits of x*y is 1. If x and y are not multiples of 10, which of the following cannot be the remainder when |x - y| is divided by 10?

I. 2
II. 4
III. 5

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


Kudos to Bunuel for sharing such a nice questions with those types of layers lad out beautifully.
Although i took too much time, here's my take. Would love to see an easier approach.
Since each of x and y is not a multiple of 10, there multiple(x*y) can only be a multiple of 10(so that the products digits result in a sum of 1). This is possible only if either of x and y is a multiple of 5 and 2.

Note that 5 and 2 can't be multiplied with other as it would invalidate the condition that one of x and y would be then be a multiple of 10, additionally invalidating another condition that sum of digits of x*y would not be 1.

Now, \(x*y = 10 = 5*2\)
OR
\(x*y = 100 = 5^2*2^2\)
\(x*y = 1000 = 5^3*2^3\)
..
..
so on.

Thus \(x = 5^a\) and \(y = 2^a\) or vice-a-versa, as we are only concerned about |x - y|(modulus being more important.
So we can check for remainders that would result from dividing |x - y| by 10 i.e. the last digit of x - y.
Hence, we have a pattern here
\(5^a - 2^a = ..b\)
\(5^1 - 2^1 = ..3\)
\(5^2 - 2^2 = ..1\)
\(5^3 - 2^3 = ..7\)
\(5^4 - 2^4 = ..9\)
\(5^5 - 2^5 = ..3\)
...

We can be sure of this because of the cyclicity of 2 which results in last digits of 2 as 2, 4, 8, 6.

Hence 3, 1, 7 and 9 are only possible remainders.
Hope this is helpful.

Answer E.
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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
1
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x and y are positive integers and the sum of the digits of x*y is 1. If x and y are not multiples of 10, which of the following cannot be the remainder when |x - y| is divided by 10?

the sum of the digits of x*y is 1 means x*y = 1,10,100,1000

Now, if x and y are not multiple of 10 that means x and y are neither even nor multiple of 5.

Thus, their product can never be equal to 10,100,100

Therefore, none of them can be the remainder.

IMO E
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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
2
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The sum of the digits of x*y will be 1 only if x*y is a power of 10

We know that neither x nor y is a multiple of 10

For x*y=10: (x, y)= (2, 5)
For x*y=100: (x, y) = (4, 25)
For x*y=1000: (x, y) = (8, 125)
And so on...
For x*y=10**n, x=2**n and y=5**n

Since y is always odd, the difference |x-y| will always be odd
The remainder when we divide by 10 will also be odd.
So the remainder cannot be 2 or 4.

y is always a multiple of 5
For the remainder to be 5, x will need to be a multiple of 10 - but that is not possible.

So 2, 4, 5 : none of them can be the remainder
Hence the answer is E

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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
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Re: 12 Days of Christmas GMAT Competition - Day 5: x and y are positive in [#permalink]
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