Number Properties Guide : Pg no. 171 Hard Practice : Q17

[xerocoool]

Hi,

Although the answer is right, but was unable to get the explanation given at the back of the book. Please let me know if this is the right way of going about.

Q) b,c,d are consecutive even integers such that 2<b<c<d. What is the largest positive integer that must be a divisor of bcd.

Soln. I believe, the question is asking for the largest positive factor of bcd
Using LCM and the condition 2<4<6<8
4: 2 x 2
6: 2 x 3
8: 2 x 2 x 2
LCM : 2x3x2x2x2 (basic building blocks for the numbers after prime factorization of 4,6,8)
And hence 48

Thanks,
-X

[tommywallach]

Hey Xero,

The Lowest Common Multiple you found there has an extra 2. Remember, you wouldn't use the 2 that's inside the 6. You already have three 2s (from the 8), so that covers the 2 you get from the 6. So:

The LCM of and 4 6 and 8 is 24: 2 * 2 * 2 * 3

HOWEVER, this question isn't asking for the Lowest common multiple. NOR is it asking for the Greatest Common Factor. It's asking what the biggest factor that MUST go into AN UNKNOWN is (because b, c, and d could be 4, 6, and 8 OR 6, 8, and 10 OR 120, 122, and 124 OR any three consecutive even integers).

So we'll look at the smallest thing that bcd could be, which is:

4 * 6 * 8 = 192

Let's look at the next thing that bcd could be:

6 * 8 * 10 = 480

The biggest thing that goes into both of these is 48.

-t

P.S. The theoretical way to approach this is like this. Three even numbers will always have at LEAST 4 twos in their primes. This is because every other even number is divisible by 4, meaning it has two 2s in its primes. So if you take three even numbers in a row, at least one of them will be divisible by 4 (and the other two will be divisible by just 2).

That's four 2s altogether, or 2 * 2 * 2 * 2 = 16

And if you have three even numbers in a row, one of them will be a multiple of 3, so you'll also have a 3.

Putting it together, we have 2 * 2 * 2 * 2 * 3 = 48

Hope that helps!

-t