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Salmon
Group Title
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
 10 months ago
 10 months ago
Salmon Group Title
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
 10 months ago
 10 months ago

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zpupster Group TitleBest ResponseYou've already chosen the best response.2
Let 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
 10 months ago

Salmon Group TitleBest ResponseYou've already chosen the best response.0
thanks @zpupster
 10 months ago

Loser66 Group TitleBest 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
 10 months ago
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