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# If the two-digit integers M and N are positive and have the

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Eternal Intern
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If the two-digit integers M and N are positive and have the [#permalink]

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23 Jul 2003, 17:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
1) 181
2) 163
3) 121
4) 99
5) 44
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23 Jul 2003, 17:38
Curly05 wrote:
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
1) 181
2) 163
3) 121
4) 99
5) 44

1 & 2 are not possible.My logic goes like this. Say two digits are x & y. So
M=10x+y and
N=10y+x
----------
M+N = 11(x+y) . so the sum has to divisible by 11. But 1 & 2 are not divisible by 11.
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23 Jul 2003, 22:47
It should be 165 for option 2. This question had a mistake or typo in the book.
Eternal Intern
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24 Jul 2003, 08:30
Hey boarder,

How does 181 go into M and N, I'm not sure.
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25 Jul 2003, 12:06
Actually, 181 cannot be the sum of two numbers where digits interchange their place. That's why (A) 181 is the answer.

Because,
(B) 165 = 96 + 69
(C) 121 = 83 + 38
(D) 99 = 54 + 45
(E) 44 = 31 + 13
Thanks
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27 Jan 2007, 16:12
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) 163
(C) 121
(D) 99
(E) 44
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27 Jan 2007, 18:33
M = 10x + y
N= 10y + x

M + N = 11 ( x + y ) ... sum should be divisible by 11

but here A & B both are not divisible by 11.
A/B ..... not sure
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28 Jan 2007, 02:33
AK , text book style answer...........great job
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29 Jan 2007, 00:11
AK wrote:
M = 10x + y
N= 10y + x

M + N = 11 ( x + y ) ... sum should be divisible by 11

but here A & B both are not divisible by 11.
A/B ..... not sure

Great working)
Maybe some typo in Answer choices?
Stuck between A and B
_________________

IE IMBA 2010

29 Jan 2007, 00:11
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