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# Qn 266. If the two-digit integers M and N are positive and

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Manager
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Qn 266. If the two-digit integers M and N are positive and [#permalink]

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20 Jan 2004, 07:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Qn 266.

If the two-digit integers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of M and N?

a. 181
b. 165
c. 121
d. 99
e. 44

The answer is 181, and can anyone show a standard working in deriving the answer?
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20 Jan 2004, 08:25
The crack technique is the following:
1) M is A tens plus B units, then M=10*A+B;
2) N is B tens plus A units, then N=10*B+A;
3) M+N=(10*A+B)+(10*B+A)=11*A+11*B=11*(A+B)
4) so the sum of M and N must be divisible by 11.

How to reveal the right answer since now is easy, isn't it?

The all answers except for 181 are divisible by 11.

That's the point.
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20 Jan 2004, 14:17
I checked each and every answer choices, I guess its much easier. 165=87+78, 121=74+47, 99=45+54 and 44=22+22
20 Jan 2004, 14:17
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