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

(In reply to Answer by K Sengupta)

Let the three consecutive even numbers be denoted by (2n-2), 2n and (2n+2)

Denoting  T = 87*****8, we obtain:
8(n^3-n) = 87*****8, so that:

87000000< 8(n^3-n)< = 87000008
or, 10875000< n^3 - n < = 10875001

Now, n^3 - n< n^3  for all n>0, while:
n^3 - n > (n-1)^3, for all n>0

Accordingly, n^3> 10875000 and (n-1)^3 < = 10875001

This yields n> 221.5524 and (n-1) < = 221.55239
Accordingly,

221.5524< n < = 222.55239

The only integer satisfying the above relationship occurs at n = 222.

This gives: (2n-2, 2n, 2n+2) = (442, 444, 446), and:
442*444*446 = 87526608

Consequently, the missing digits are 5,2,6,6,0 (in this order) and the  three consecutive even numbers are 442, 444 and 446

 

Edited on March 18, 2008, 3:33 pm

Posted by K Sengupta on 2007-06-25 15:47:35