Which of the following is the lowest positive integer that is

Option (E) can be ignored as it is not divisible by 9.

Next lowest number is 1890, which is not divisible by 4 or 8.

Option (C) is divisible by 3,6,7,8,9. Hence Ans - C.

It must be devisible by both 2 and 5 so by 10.
The last two digit must be devided by 4
Finally the sum of all digits must be devided by 9
OA is C

Sent from my Nexus 5 using GMAT Club Forum mobile app

LCM of these numbers: 2,520 answer
IMO C

To be divisible by all of those numbers, it must have all of the prime factors of each of those numbers.

Break down of prime factors
2 - 2
3 - 3
4 - 2 x 2
5 - 5
6 - 2 x 3
7 - 7
8 - 2 x 2 x 2
9 - 3 x 3

The number must have all the qualifying factors for each of those numbers so:
2^3 x 3^2 x 5 x 7 = 2520

Since we have to check the numbers for the divisibility by 2, 3, 4, 5, 6, 7, 8, 9 ... We just need to check the divisibility by 5,7,8,9

So we will start checking the options one by one for the divisibility by 5,7,8, 9.

To solve the problem in a faster way, we need to go through options very smartly and hence recheck the question. (I solved the question in 2:40 seconds which is on the higher side because I checked the divisibility of each number by 5,7,8,9 and there were 2 numbers divisible by all then I noticed that the question is asking for lowest positive integer.)

So how to solve this problem in smartly and in timely manner.

what to keep in mind.
1. As the question says we need to find the lowest positive integer divisible. So we should start from the lowest integer in the options.
2. We need to start from the number for which divisibility check is easier e.g. 5

Shortest way : Since this question ask for the lowest positive integer divisible by numbers .
This is the LCM of numbers 5x7x8x9 = 2520 Answer C.

Shorter way : If the question would have asked for lowest positive integer from the options.
Then,
1st check 1680 : Divisible by 5, Divisible by 8 , But not divisible by 9. No need to check divisibility by 7.
2nd 1890 : Divisible by 5, Not divisible by 8. No need to check divisibility by 9 and 7
3rd 2520 : Divisible by 5, divisible by 8, divisible by 9 , divisible by 7

So C is the correct option.

Finally don't forget to give kudos..

We need to determine the least common multiple (LCM) of 2, 3, 4, 5, 6, 7, 8, and 9. Breaking each of these into primes, we have:

2^1, 3^1, 2^2, 5^1, 2^1 x 3^1, 7^1, 2^3, 3^2

To find the LCM, we need to use each prime number from the above list and we need that prime number as many times as its highest exponent. Let’s first consider 2. Look at all of the occurrences of the prime number 2 from the previous list: 2^1, 2^2, 2^1, 2^3. Since the highest power of 2 is 3, we will use 2^3 as one of the factors in the LCM. Similarly, the prime number 3 occurred in the list as 3^1, 3^1, and 3^2. Thus, we will use 3^2 as one of the factors in the LCM. Note that 5 occurred only as 5^1 and that 7 occurred only as 7^1; thus, the numbers 5^1 and 7^1 will be the final two factors of the LCM.

Thus, the LCM is 2^3 x 3^2 x 5 x 7 = 8 x 9 x 5 x 7 = 2,520.

Answer: C

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.