canbtg wrote:If K is the least positive integer that is divisible by every integer from 1 to 8 inclusive, then K =
A) 840
B) 2,520
C) 6,720
D) 20,160
E) 40,320
Here's another approach:
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) = [spoiler]840 = A[/spoiler]
Cheers,
Brent
