Hey all,
Great question - I love these divisibility problems and I've found that the more of these you do the more they just "click" and they become less about testing numbers and more about following the logic that allows you to predict exactly what the product here will look like. A few thoughts here:
Amit2k9 - your only real mistake there is that you used 1; remember, the question says positive EVEN number, so 1 doesn't apply.
Overall:
If you look at what we'll get, we're taking an even number n, adding three different even numbers to it, and multiplying them all together. So we know that we'll have:
Even * Even * Even --> So at a minimum we'll have at least 2*2*2 represented in the prime factorization.
But wait...there's more! We're multiplying together three CONSECUTIVE even numbers. So, actually, we can guarantee that one of them will be a multiple of 4 and bring an extra multiple of 2 to the table. Try it - any set of three consecutive integers will provide at least one that's divisible by 4, because every SECOND even number is divisible by 4: (6, 8*, 10); (8*, 10, 12*); (10, 12*, 14). So now we know that we have at least 2*2*2*2 present in the factorization of this product.
But wait...there's more! Note that if we start with n as a number NOT divisible by 4, then that's an even integer that's off the divisible-by-4 cycle:
Even, Even AND divisible by 4, Even, Even AND divisible by 4, Even...
2, 4, 6, 8, 10...
So we know, then, that n + 6 and n + 10 will get us back ON that cycle, so the bookends of this (n+6)(n+8)(n+10) are both divisible by 4. So our product is going to look like:
(Even AND divisible by 4)(Even)(Even AND divisible by 4)
So...we know that, as prime factors, we'll have at minimum:
(2*2)(2)(2*2) ---> our product will be divisible by 2^5.
By similar logic, and knowing that we're testing for divisibility by 24 (which is 3*2^3) and 96 (3*2^5), we can use the consecutive-even-integers setup to prove that we'll definitely have a 3 in there, too. If you take any three integers in a row, one of them is divisible by 3 because every THIRD integer is divisible by 3: 1, 2, 3*, 4, 5, 6*, 7, 8, 9*.....
And for consecutive even integers, we're just taking a set of three straight integers and multiplying each by 2. So we know for sure that one - and exactly one - of the series (n+6), (n+8), (n+10) will be divisible by 3, so we'll have exactly one factor of 3 to go with our factors of 2^5. So we can prove that the product will be divisible by at least 2*2*2*2*2*3, and since those factors satisfy 24, 32, and 96, the answer is E.
Now, as a teaching tool - if you don't 100% follow this logic in the abstract form without plugging in some numbers to make it more concrete, that's perfectly fine. It takes some time to think this way. But you know that the GMAT tests divisibility and number properties, so when you're plugging in numbers in practice, try to think behind them to see if you can use that trial-and-error to become more aware of properties and rules like we just discussed. The more you learn to think like the testmaker, the easier this test becomes...
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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