divisibility

2100


First 5 positive multiples of 5 are

5,10,15,20,25,30,35


Break these down in to primes u would need 2^2 5^2 3 and 7 minimum.

140 1400,3500 eliminate not divisible by 3 since sum of digits not divisible by 3

210->3*5*2*7

Missing a 5 and 2 so if u add this u need atleast 2100

anishprabhu wrote:Which of the following is the lowest positive integer that is divisible by the first 7 positive integer multiples of 5?
3500
2100
1400
210
140

So you need the LCM of the first 7 multiples of 5.

= 5 x LCM(1x2x3x4x5x6x7)

Since 1,2 and 3 are factors of 6, LCM(1,2,3,4,5,6,7) = LCM(4,5,6,7)=420

Hence the required number = 5x420 = 2100

The first 7 integer multiples of 5 are 5, 10, 15, 20, 25, 30, and 35. The question is asking for the least common multiple (LCM) of these 7 numbers. Let's construct the prime box of the LCM.

In order for the LCM to be divisible by 5, one 5 must be in the prime box.
In order for the LCM to be divisible by 10, a 5 (already in) and a 2 must be in the prime box.

In order for the LCM to be divisible by 15, a 5 (already in) and a 3 must be in the prime box.

In order for the LCM to be divisible by 20, a 5 (already in), a 2 (already in), and a second 2 must be in the prime box.

In order for the LCM to be divisible by 25, a 5 (already in) and a second 5 must be in the prime box.

In order for the LCM to be divisible by 30, a 5 (already in), a 2 (already in) and a 3 (already in) must be in the prime box.

In order for the LCM to be divisible by 35, a 5 (already in) and a 7 must be in the prime box. Thus, the prime box of the LCM contains a 5, 2, 3, 2, 5, and 7. The value of the LCM is the product of these prime factors, 2100.

The correct answer is D.