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Hi,

Given,

“mn” is a two digit number.

Where, “m” is the ten’s digit and “n” is the units digit.

So, “m” can take any integer value from “1” to “9”

And, “n” can take any integer value from “0” to “9”

Question: Is mn is odd ?

i.e.,

“n” has to be either 1,3,5,7,9 ?

Statement I is insufficient:

The least common multiple of m and n is even.

So, either “m” and “n” both are even.

OR,

Anyone of them is even.

For example,

If m = 2 and n = 1, then the L.C.M is 2.

The value of “mn” is 21 and its an odd number. So answer to the question is YES.

But If m = 2 and n = 4, then the L.C.M is 4.

The value of “mn” is 24 and its an even number. So answer to the question is NO.

So insufficient.

Statement II is insufficient:

m is an odd number

Nothing about “n”.

We need to know the value of “n”, to check whether “mn” is odd or not.

So insufficient.

Together it is sufficient.

The least common multiple of m and n is even and “m” is an odd number.

So, “n” has to be even.

Hence answer to the question is always “NO”.

Together it is sufficient.

Answer is C.

Hope this helps.

Regards,
Junaid.
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Bunuel
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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Afc0892
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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Archit3110
Afc0892
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.
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Such high quality question, you just need a bit analysis to solve questions like this ,
Stmt 1 : least common multiple of m and n is even ,

happens only in two cases ,

i. when m and n should be even
ex: lcm of 4 and 6 is 12
ii.when either of m or n can be odd
ex: lcm of 3 and 6 is 6

so, clearly i is insufficient

stmt 2 : alone insufficient to answer for sure
1+2 , now we know m is odd , then obviously n should be even.

Ans :C
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Bunuel
Is a two digit positive integer mn, where 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

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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Bunuel
Is a two digit positive integer mn, where 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

Asked: Is a two digit positive integer mn, where m is the tens digit and n is the units digits, odd?
mn is odd when n is odd

(1) The least common multiple of m and n is even
Either or both of the numbers is even
n may be even or odd
NOT SUFFICIENT

(2) m is an odd number
n is unknown
NOT SUFFICIENT

(1) + (2)
(1) The least common multiple of m and n is even
m or n or both are even
(2) m is an odd number
m is odd & n is even
mn is even
SUFFICIENT

IMO C
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Bunuel
Is a two digit positive integer mn, where 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



M36-60

Official Solution:


Is a two digit positive integer \(mn\), where \(m\) is the tens digit and \(n\) is the units digits, odd?

Notice that the question basically asks whether \(n\) is odd (the units digit of an integer determines eve/odd nature of that integer).

(1) The least common multiple of \(m\) and \(n\) is even

The above means that at least one of \(m\) and \(n\) is even (else how/why the LCM would be even?). Thus, either both \(m\) and \(n\) are even OR one is even and another is odd. So, \(n\) can be even as well as odd. Not sufficient.

(2) \(m\) is an odd number

Clearly insufficient.

(1)+(2) From (2) we know that \(m\) is NOT even and from (1) we know that at least one of \(m\) and \(n\) must be even, thus \(n\) is even. Sufficient.


Answer: C
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