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Counting triplets (Posted on 2010-03-31) Difficulty: 3 of 5
For how many ordered triplets (a, b, c) of positive integers, each less than 10 is the product abc divisible by 20?

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Solution 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              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


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