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Re: If K is the least positive integer that is divisible by ever
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28 Jan 2017, 23:54
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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.)
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Dmitry Farber | Manhattan GMAT Instructor | New York
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72% (01:18) correct 
