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Salmon
 one year ago
How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer?
a)21 b)24 c)36 d)30
Salmon
 one year ago
How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer? a)21 b)24 c)36 d)30

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zpupster
 one year ago
Best ResponseYou've already chosen the best response.2Let one digit is fixed to be 1. Then product of two distinct digits must be a perfect cube i.e. (2,4) such that 1*2*4 = 8 = 23 total 3! = 6 combinations (3,9) such that 1*3*9 = 27 = 33 total 3! = 6 combinations One digit is fixed to be 2. (4,8) such that 2*4*8 = 64 = 43 total 3! = 6 combinations One digit is fixed to be 3. (8,9) such that 3*8*9 = 216 = 63 total 3! = 6 combinations One digit is fixed to be 4 (6,9) such that 4*6*9 = 216 = 63 total 3! = 6 combinations

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@zpupster I got you except 23 total, 33 total, 43 total.... what do they mean, please explain me. btw , your explanation is perfect
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