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Re: PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]
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: PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]
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: PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]
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: PR and MN are 2-digit positive integers, where P, R, M, and [#permalink]
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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