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# PR and MN are 2-digit positive integers, where P, R, M, and

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Current Student
Joined: 11 May 2008
Posts: 556
PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]

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01 Aug 2008, 03:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

PR and MN are 2-digit positive integers, where P, R, M, and N are digits of the two numbers, and are different from each other. If the tens' digit of sum of PR and MN is M, which of the following must be true?
I. M<9
II. R+N>9
III. P>8
A. I only
B. II only
C. III only
D. I and II
E. I, II and III
Current Student
Joined: 11 May 2008
Posts: 556

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01 Aug 2008, 04:48
expln pls nikhil... so tat others can learn...
Senior Manager
Joined: 16 Jul 2008
Posts: 289

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01 Aug 2008, 05:03
I'm going for E.

This forum is fun

EDIT: so, we know that "something plus M equals 10 + M". From there we get:

R + N > 9
1 + P + M = 10 + M --> P = 9 --> P > 8

But P, R, M and N are different, therefore M cannot be 9. So: M < 9.
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VP
Joined: 17 Jun 2008
Posts: 1381

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03 Aug 2008, 01:10
arjtryarjtry wrote:
PR and MN are 2-digit positive integers, where P, R, M, and N are digits of the two numbers, and are different from each other. If the tens' digit of sum of PR and MN is M, which of the following must be true?
I. M<9
II. R+N>9
III. P>8
A. I only
B. II only
C. III only
D. I and II
E. I, II and III

say x=10p+r and y=10m+n => x+y= 10(p+m)+r+n

p+m=m when r+n > 9 hence II is must
now carry from r+n is 1 hence p+m+1 =10a+m a is any number
is possible when p=9 and m <9 => III and I are must

hence I,II and III are must
IMO E
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Its Now Or Never

Intern
Joined: 02 Aug 2008
Posts: 29

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03 Aug 2008, 02:10
a number like 41 and 32 proves that (2) need not be right. Apart from m <9, i don't find anything "must"
did i do anything wrong?
Senior Manager
Joined: 16 Jul 2008
Posts: 289

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03 Aug 2008, 04:22
nishchals wrote:
a number like 41 and 32 proves that (2) need not be right. Apart from m <9, i don't find anything "must"
did i do anything wrong?

41 + 32 = 73. The tens digit is 7, I think you are looking at the units digit.
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Manager
Joined: 23 Apr 2008
Posts: 86

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03 Aug 2008, 07:00
IMO E

PR
+ MN
_____

CLEARLY IF P IS NOT EQUAL TO M,THEN P+M CANT BE EQUAL TO M
UNLESS R+N>9

NOW THE CARRY CARRY FWD DIGIT WILL BE 1 ONLY (CANT B GREATER THAN 1)

SO P+1+M IS EQUAL TO M(ACC TO QUES)
WHICH IS NOT POSSIBLE IF P<8

NOW COS P CANT B LESS THAN 8,IT HAS TO BE 9
THEREFORE M HAS TO BE LES THAN 9

HENCE ALL TRUE
Re: incomplete positive integers   [#permalink] 03 Aug 2008, 07:00
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