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.
50 = 2*5*5
Let P={5,10,15,20}, q = set of remaining less than 21
to be divisible by 50, at least 2 must be from p.
Case 1: 4 chosen from p
No = 1
Case 2: 3 chosen from p and 1 from q
4C3 * 16 = 4*16 = 64
Case 3: 2 from p and 2 from q
if chosen = 5,15
one from remaining 2 must be even.
No = 8*15 = 120
(8 ways even can be chosen and remaining in 15 ways)
if chosen not equal to 5,15.
Total no. of ways 2 chosen p = 4C2 = 6
excluding previous case its 5
product of these 2 will be divisible by 50 itself
No = 5*16C2 = 5*8*15 = 600
Total = 1+64+120+600 = 785
|
|
Posted by Praneeth
on 2010-08-20 17:29:13 |
Solution
