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

Question

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

Bunuel · Accepted Answer

The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), 7 and 8=2^3. The least common multiple of these integers is 2^3*3*5*7 = 840. Answer: A. Similar question to practice from OG: what-is-the-lowest-positive-integer-that-is-divisible-by-138733.html Hope it helps.

DmitryFarberMPrep · Answer

leonidbasin1 , the question is asking for the smallest number that can be divided by all the integers from 1 to 8. At first glance, it would seem that you could just multiply all those numbers together. 8*7*6*5*4*3*2*1 = 40,320. However, that number is far too big. It's not the  least  number that's divisible by 1 through 8; it's just the simplest to find.

Why is it too big? Well, if a number is divisible by 8 (which is 2*2*2), then it will automatically be divisible by 4 (2*2) and 2, so we don't need to multiply those numbers in. Similarly, if it is divisible by all those 2's, and also by 3, then it will automatically be divisible by 6. So what do we really need to multiply together? 8*7*5*3*1 = 840.

You could also start at the bottom. Some people might find this more intuitive, but others may find it confusing. The idea is to work up from the lowest numbers, multiplying as you go, and see what else is needed. The first three integers give us 1*2*3 = 6. Next on our list is 4. We already have a 2, so we just need one more 2 to make a multiple of 4. That makes 6*2 = 12. We don't have a 5 yet, so put that in: 12*5 = 60. We get to 6 and see that our number (60) is already a multiple of 6. Skip it. We don't have 7 yet: 60*7 = 420. Our last number is 8. We already have a multiple of 4, so we just need one more 2 to make a multiple of 8: 420 * 2 = 840.

(Of course, in both methods you can skip multiplying by 1. I just included it since it's in the list.)

PareshGmat · Answer

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

All 4 digit OA can be discarded

Answer = A

leonidbasin1 · Answer

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

Abhishek009 · Answer

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)

JeffTargetTestPrep · Answer

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

BrentGMATPrepNow · Answer

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

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

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