If K is the least positive integer that is divisible by ever

LCM of 1 to 8 numbers should be a three digit number (= 840)

All 4 digit OA can be discarded

Answer = A

I am confused!! Can someone please explain it to me in a more basic language?

Integer that is divisible by every integer from 1 to 8 inclusive is : 8!

\(8! = 8*7*6*5*4*3*2\)

Check the options to find the least value which is divisible by 2,3,5 & 7

Only option (A) 840 , matches as it is divisible by 2,3,5 & 7

So, Correct answer must be (A)

We need to determine the LCM of 2, 3, 4, 5, 6, 7, and 8. Factoring each number into primes, we have:

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

So the LCM is 2^3 x 3 x 5 x 7 = 840.

Answer: A

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)

K is divisible by every integer from 1 to 8 inclusive
This means that there's a 2 "hiding" within the prime factorization of K, a 3 "hiding" within the prime factorization of K, a 4 "hiding" within the prime factorization of K, etc.

So, let's begin with a 2 "hiding" within the prime factorization of K.
This means that K = (2)(other numbers)

Also, if there's a 3 "hiding" within the prime factorization of K, then we need to add a 3 like so: K = (2)(3)

There's a 4 "hiding" within the prime factorization of K.
Since 4 = (2)(2), then we need to add a SECOND 2 to get: K = (2)(2)(3)

There's a 5 "hiding" within the prime factorization of K, so we'll add a 5 to get: K = (2)(2)(3)(5)

There's a 6 "hiding" within the prime factorization of K.
Since 6 = (2)(3), we can see that we ALREADY have a 6 "hiding" in the prime factorization: K = (2)(2)(3)(5)

There's a 7 "hiding" within the prime factorization of K, so we'll add a 7 to get: K = (2)(2)(3)(5)(7)

There's an 8 "hiding" within the prime factorization of K.
Since 8 = (2)(2)(2), we need to add a THIRD 2 to get: K = (2)(2)(2)(3)(5)(7)

We have now ensured that K is divisible by every integer from 1 to 8 inclusive. This means that we have found the LEAST possible value of K that satisfies the given conditions.
So, K = (2)(2)(2)(3)(5)(7) = 840

Answer: A

Cheers,
Brent

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