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If the least common multiple of m and n is 24, then what is
[#permalink]
30 May 2018, 01:22

1

Question Stats:

If the least common multiple of m and n is 24, then what is the first integer larger than 3070 that is divisible by both m and n?

(A) 3072

(B) 3078

(C) 3084

(D) 3088

(E) 3094

(A) 3072

(B) 3078

(C) 3084

(D) 3088

(E) 3094

Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4805**

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Re: If the least common multiple of m and n is 24, then what is
[#permalink]
05 Jun 2018, 13:10

1

Expert Reply

Least common multiple of \(m\) and \(n\) is 24.

So for a number larger than 3070 to be divisible by both \(m\) and \(n\), it must be divisible by the least common multiple of m and n as well.

Using the gre calculator \(\frac{3070}{24}=127.9166\).

Clearly multiple of 24 larger than 3070 is \(128 \times 24=3072\).

_________________

So for a number larger than 3070 to be divisible by both \(m\) and \(n\), it must be divisible by the least common multiple of m and n as well.

Using the gre calculator \(\frac{3070}{24}=127.9166\).

Clearly multiple of 24 larger than 3070 is \(128 \times 24=3072\).

_________________

Sandy

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Re: If the least common multiple of m and n is 24, then what is
[#permalink]
08 Jun 2018, 09:26

1

24=23∗324=23∗3

As per divisibility rule -

1. For a number to be divisible by 8, the last 3 digits must be divisible by 8.

2. For a number to be divisible by 3, the sum of the digits must be divisible by 3.

Check the options only (A) follows

1. 072 - Divisible by 8

2. 3072 - Sum of the digits is 12 , which is divisible by 3

As per divisibility rule -

1. For a number to be divisible by 8, the last 3 digits must be divisible by 8.

2. For a number to be divisible by 3, the sum of the digits must be divisible by 3.

Check the options only (A) follows

1. 072 - Divisible by 8

2. 3072 - Sum of the digits is 12 , which is divisible by 3

gmatclubot

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