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Is a two digit positive integer mn, where x m is the tens digit and n

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Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 03 Dec 2018, 21:19
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  55% (hard)

Question Stats:

64% (01:36) correct 36% (01:34) wrong based on 161 sessions

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Re: Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 03 Dec 2018, 21:53
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  [#permalink]

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New post 03 Dec 2018, 22:30
1
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


mn is a two digit positive integer.
m is 10's digit, n is unit's digit.

Question : mn is odd or not.

Statement 1: LCM of m and n is even i.e. either m or n or both is even.
Not sufficient

Statement 2: m is an odd number.
So we can't say anything about n. n can be odd or even.
Not sufficient

Combined :
From statement 1 , either m or n or both is even.
From statement 2, m is odd.
So, n has to be an even number.

Answer C
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Re: Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 03 Dec 2018, 22:57
1
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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Re: Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 03 Dec 2018, 23:02
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  [#permalink]

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New post 03 Dec 2018, 23:04
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  [#permalink]

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New post 03 Dec 2018, 23:06
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.
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Re: Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 10 Dec 2018, 10:32
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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Re: Is a two digit positive integer mn, where x m is the tens digit and n  [#permalink]

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New post 24 Dec 2018, 01:17
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Re: Is a two digit positive integer mn, where x m is the tens digit and n   [#permalink] 24 Dec 2018, 01:17
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