Each of A, B, C and D is a positive integer with the proviso that A ≤ B ≤ C ≤ D ≤ 20.

Determine the total number of quadruplets (A, B, C, D) such that A*B*C*D is divisible by 50.

In order to be divisible by 50, two of the integers must be some combination of 5,10,15, & 20.  There are 6 combinations of this [4!/(2!2!)].  Once one of these pairs is picked, there are 153 combinations of the remaining integers [18!/(2!16!)].  This yields 918 total.

Of the 6 original pairs, all satisfy being divisible by 2 except for 5,15.  So this pair will require at least one even number included among the remaining two integers.  To do this, the 28 odd-odd combinations [8!/(2!6!)] can be subtracted from the total.  This leaves 890.

Answer:  890

Edited on August 20, 2010, 8:06 pm

Posted by hoodat on 2010-08-20 19:58:08