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PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]
 01 Aug 2008, 03:14
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
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Re: incomplete positive integers [#permalink]
01 Aug 2008, 04:48
expln pls nikhil... so tat others can learn...
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Re: incomplete positive integers [#permalink]
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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Re: incomplete positive integers [#permalink]
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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Re: incomplete positive integers [#permalink]
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?
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Re: incomplete positive integers [#permalink]
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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Re: incomplete positive integers [#permalink]
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
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Re: incomplete positive integers
[#permalink]
03 Aug 2008, 07:00
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