For how many ordered triplets (a, b, c) of positive integers, each less than 10 is the product
abc divisible by 20?
20 factors into 2*2*5, so at least one of a,b,c must be a 5.
2,2,5 accounts for 3 permutations.
2,4,5 6
2,6,5 6
2,8,5 6
4,4,5 3
4,x,5 6*6 = 36
where x is neither 2 nor 4 nor 5
4,5,5 3
8,8,5 3
8,x,5 6*5 = 30
where x is neither 2 nor 4 nor 8 nor 5
8,5,5 3
6,6,5 3
-----
102
Computer verification:
CLS
FOR a = 1 TO 9
FOR b = 1 TO 9
FOR c = 1 TO 9
s = a * b * c
IF s MOD 20 = 0 THEN
ct = ct + 1
END IF
NEXT
NEXT
NEXT
PRINT ct
102
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Posted by Charlie
on 2010-03-31 14:06:39 |
solution
