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If K is the least positive integer that is divisible by

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by GMATGuruNY » Sat Apr 05, 2014 5:30 am
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

OA A
Divisibility by 2: the units digit is even.
Divisibility by 3: the sum of the digits is a multiple of 3.
Divisibility by 4: the tens digit and the units digit form a multiple of 4.
Divisibility by 5: the units digit is 0 or 5.
Divisibility by 6: the units digit is even and the sum of the digits is a multiple of 3.
Divisibility by 8: the last 3 digits form a multiple of 8.

Any integer divisible by 6 will also be divisible by 2 and 3.
Any integer divisible by 8 will also be divisible by 2 and 4.
Check whether the smallest answer choice (840) is divisible by 5, 6, and 8.

Answer choice A: 840
Divisibility by 5: the units digit (0) is 0 or 5.
Divisibility by 6: the units digit (0) is even and the sum of the digits is a multiple of 3 (8+4+0=12).
Divisibility by 8: the last 3 digits form a multiple of 8 (840/8 = 105).
Check whether 840 is also divisible by 7:
840/7 = 120.

Since 840 is divisible by 5, 6, 7 and 8, 840 is divisible by every integer between 1 and 8, inclusive.

The correct answer is A.
Last edited by GMATGuruNY on Sun Apr 06, 2014 2:58 am, edited 1 time in total.
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by Brent@GMATPrepNow » Sat Apr 05, 2014 7:45 am
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
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by [email protected] » Sat Apr 05, 2014 10:48 am
Hi canbtg,

While this question appears to require LOTS of division, there are some Number Property shortcuts that you can use to avoid some of the math:

1) Any number that's divisible by 6 is ALSO divisible by 2 and 3, so you don't have to bother testing for 2 or 3 - just test for 6

2) Any number that's divisible by 8 is ALSO divisible by 2 and 4, so you don't have to bother testing for 2 or 4 - just test for 8

So in this question, we're looking for the SMALLEST number that's divisible by 5, 6, 7 and 8

All the numbers end in 0, so they're all divisible by 5; this ultimately means that you need to find the smallest number that's divisible by just 3 numbers: 6, 7 and 8.

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by Jeff@TargetTestPrep » Tue Apr 10, 2018 10:18 am
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

OA 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

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