If the least common multiple of m and n is 24,

Question

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

Source:   Nova GMAT  
Difficulty Level:   600

Abhishek009 · Accepted Answer

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)

stonecold · Answer

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.

EMPOWERgmatRichC · Answer

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:  A

GMAT assassins aren't born, they're made,
Rich

Kinshook · Answer

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

Papai21 · Answer

Given, least common multiple of m and n is 24 = 2^3 * 3
Thus either m or n has 2^3 term and 3 term in them. No other prime factor is present. And the powers denote the highest powers of 2 and 3.
So for a number to be devided by m and n, it must be divided by 2^3 = 8 and 3.
Considering the given options only option A is both divisible by 8 and 3.
Thus ANSWER: A

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BrushMyQuant · Answer

Given that LCM of m and n is 24 and we need to find what is the first integer larger than 3070 that is divisible by both m and n

LCM(m,n) = 24 => Both m and n are multiples of 24

=> For a number to be divisible by m and n it should be divisible by 24 for sure.

Number greater than 3070 which is divisible by 24 is 3072 (as 072 is divisible by 8 and 3072 is divisible by 3)

Watch this video  to learn about  Divisibility Rules

So ,  Answer will be A 
Hope it helps!

To learn more about LCM and GCD watch the following videos

https://www.youtube.com/watch?v=7jPV6JD0908

https://www.youtube.com/watch?v=vtSsAjcF-0Y

bumpbot · Answer

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If the least common multiple of m and n is 24,

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