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
If K is the least positive integer that is divisible by
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Divisibility by 2: the units digit is even.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 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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Here's another approach: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
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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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich
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We need to determine the LCM of 2, 3, 4, 5, 6, 7, and 8. Factoring each number into primes, we have: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
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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