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

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Manager
Joined: 28 Apr 2003
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Qn 266. If the two-digit integers M and N are positive and [#permalink]  20 Jan 2004, 06:35
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?
Intern
Joined: 20 Jan 2004
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[#permalink]  20 Jan 2004, 07:25
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.
Manager
Joined: 26 Dec 2003
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[#permalink]  20 Jan 2004, 13:17
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
[#permalink] 20 Jan 2004, 13:17
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# Qn 266. If the two-digit integers M and N are positive and

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