Multiplying three consecutive even numbers gives 87*****8, where each "*" stands for a digit. What are the missing numbers?

(n-2)*n*(n+2) = (n^2 – 2n)(n+2) = n^3 + 2n^2 –2n^2 – 4n = n^3 – 4n

Well, I guess we could find what even cubes put us in the range of 87*****8 since n^3 is so much bigger than 4n. Also, since we are subtracting 4n, we should only look at numbers that are greater than 87000000.

The cubes that are in reasonable range are 444^3 = 87528384, and 445^3 = 88121125 but 445 isn’t even. So let’s look at n^3 – 4n for n = 444.

444^3 – 4*444 = 87526608

So the answer is 52660

I think the "trick" to this problem is to look at (n-2)*n*(n+2) and not n*(n+2)*(n+4). I think many people might say "ok, three consecutive even numbers. That’s n*(n+2)*(n+4) = n^3 + 6n^2 + 8n." Using this equation, you can still look at the n^3 number that put you in range, you’ll just have to be willing to look a little farther away from a number that starts with 87 because 6n^2+8n is still pretty big, compared with 4n. I think that looking at (n-2)*n*(n+2) just focuses your solution set better.

Posted by nikki on 2004-09-07 14:27:04