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 Counting triplets (Posted on 2010-03-31)
For how many ordered triplets (a, b, c) of positive integers, each less than 10 is the product abc divisible by 20?

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 solution | Comment 2 of 7 |

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              62,6,5              62,8,5              6`
`4,4,5              34,x,5              6*6 = 36 where x is neither 2 nor 4 nor 54,5,5              3 `
`8,8,5              38,x,5              6*5 = 30 where x is neither 2 nor 4 nor 8 nor 58,5,5              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

 Posted by Charlie on 2010-03-31 14:06:39

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