2 ways to do this

METHOD - I

3070/24 = 127.xx

So, the next multiple 128*24 = 3072 must be divisible by both m & n.

METHOD - II

\(24 = 2^3*3\)

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

Thus, answer must be (A)
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Nice Question.
For a number to be disable by both m and n => it must be divisible by its LCM.

Hence the number would be of the form => 24k

3070=24k'+22
Adding 2 to both sides => 3072 is divisible by 24.

So the smallest number would be 3072.


SMASH THAT A.
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Hi vikasp99,

This question can be solved by TESTing VALUES and taking advantage of certain Number Properties.

To start, we're told that the LCM of M and N is 24. Since we're not told anything else about those two variables, we can TEST any two values that fit that information....

Let's TEST M = 3 and N = 8

We're asked to find the SMALLEST integer GREATER than 3070 that is divisible by both M and N. Let's start with Answer A....

Using the "rule of 3", (if the digits of a number sum to a multiple of 3, then that number is divisible by 3), we know that 3072 is divisible by 3 (since 3+0+7+2 = 12 and 12 is a multiple of 3).

The number 8 divides evenly into 1,000 (125 times), so it will divide evenly into 3,000 (375 times). Since all of the answer choices are 3,000+, we really just have to determine whether 8 divides into the 'rest' of the number or not. Thus, we really just have to think about the "72" in 3072. Does 8 divide into 72? Yes it does (9 times). Answer A fits everything that we're told, so it MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Asked: 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?

LCM(m,n) = 24

Any integer that is divisible by both m and n = 24k

24k>3070
k > 3070/24 = 127 22/24 = 127 11/12
Smallest k = 128

The first integer larger than 3070 that is divisible by both m and n = 24*128 = 3072

IMO A
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