Number property.Pg 180.#17

[manigraja100]

b,c and d are consecutive even integers such that 2<b<c<d.What is the largest positive integer that must be a divisor of bcd ?
i understood everything in the explanation except how they got the fourth 2 in the prime factorisation.
and also
let b=4,c=6,d=8 then bcd will be 192. isn't 192 the largest divisor of 192 ?

[tommywallach]

Hey Mani,

You get an extra 2 because every other even number is a multiple of 4, and every multiple of 4 has two 2s in its prime factorization.

2 4 6 --> four twos in there
4 6 8 --> six twos in there (because 8 has three 2s in it)
6 8 10 --> five twos in there
8 10 12 --> five twos in there
10 12 14 --> four twos in there

You can never get fewer than four 2s.

As for your second question, you're forgetting that just because a number is bigger than another number doesn't mean it has all the same factors. Yes, the smallest thing that can be a product of three consecutive integers greater than two is 192. But that doesn't mean 192 will go into the product of any three consecutive even integers. For example:

6 * 8 * 10 = 480

That's not divisible by 192. This is why we need our primes rules!

Hope that helps!

-t