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Re: Is a two digit positive integer mn, where x m is the tens digit and n
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03 Dec 2018, 20:53
Question - Whether 10m+n Odd?
From statement 1:
LCM of m and n is even. If m is 2 and n is 3, then LCM = 6. mn = 23 = Odd. If m is 2 and n is 4, then LCM = 4. mn = 24 = even. Insufficient.
From statement 2:
m is odd. If n is even, then mn is even. If n is odd, then mn is odd. Insufficient.
Combining both:
If m is 3 and n is 2, then LCM = 6. mn = 32 = even. If m is 3 and n is 5, then LCM = 15. mn = 35 = odd. Insufficient.
E is the answer.
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Re: Is a two digit positive integer mn, where x m is the tens digit and n
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03 Dec 2018, 22:02
Bunuel wrote:
Is a two digit positive integer mn, where x m is the tens digit and n is the units digits, odd?
(1) The least common multiple of m and n is even (2) m is an odd number
for 10m+n to be odd 'n' has to be odd
from 1: LCM of mn is even so mn m,n: can be ( 3,2) even , (2,3) odd in sufficient
From 2: m is an odd no. it does not say anything about n so insufficeint
from 1 & 2 : m odd and lcm would be even whenever n is even eg : 12,14,16,32,36... so mn is always even not odd , IMO C is sufficient..
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Re: Is a two digit positive integer mn, where x m is the tens digit and n
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03 Dec 2018, 22:04
Afc0892 wrote:
Question - Whether 10m+n Odd?
From statement 1:
LCM of m and n is even. If m is 2 and n is 3, then LCM = 6. mn = 23 = Odd. If m is 2 and n is 4, then LCM = 4. mn = 24 = even. Insufficient.
From statement 2:
m is odd. If n is even, then mn is even. If n is odd, then mn is odd. Insufficient.
Combining both:
If m is 3 and n is 2, then LCM = 6. mn = 32 = even. If m is 3 and n is 5, then LCM = 15. mn = 35 = odd. Insufficient.
E is the answer.
Afc0892 statement 1 : says LCM is to be even ;so whenever m is odd n has to be even .. your eg "If m is 3 and n is 5, then LCM = 15. mn = 35 = odd." isnt correct.. n has to be an even value to get LCM as even with m as odd... IMO C should be correct ..
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Re: Is a two digit positive integer mn, where x m is the tens digit and n
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03 Dec 2018, 22:06
Archit3110 wrote:
Afc0892 wrote:
Question - Whether 10m+n Odd?
From statement 1:
LCM of m and n is even. If m is 2 and n is 3, then LCM = 6. mn = 23 = Odd. If m is 2 and n is 4, then LCM = 4. mn = 24 = even. Insufficient.
From statement 2:
m is odd. If n is even, then mn is even. If n is odd, then mn is odd. Insufficient.
Combining both:
If m is 3 and n is 2, then LCM = 6. mn = 32 = even. If m is 3 and n is 5, then LCM = 15. mn = 35 = odd. Insufficient.
E is the answer.
Afc0892 statement 1 : says LCM is to be even ;so whenever m is odd n has to be even .. your eg "If m is 3 and n is 5, then LCM = 15. mn = 35 = odd." isnt correct.. n has to be an even value to get LCM as even with m as odd... IMO C should be correct ..
Yes, realized it. Thank you again.
_________________
If you are not badly hurt, you don't learn. If you don't learn, you don't grow. If you don't grow, you don't live. If you don't live, you don't know your worth. If you don't know your worth, then what's the point?
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66% (01:09) correct 
