Question
How many three- digit numbers can be formed with at least 2 distinct digits such that the product of the digits is the cube of a positive integer?
Answer: Option A
:
A
The combination of three digits, whose product can give a perfect cube are (1,2,4) (1,3,9) (4,6,9) (3,8,9) and (2,4,8) when all three digits are distinct.With each of these combinations we can have 3! three digit numbers with distinct digits. Therefore, the number of such 3-digit numbers is 5(3!) = 5*6 = 30. There is another case of (8,8,1) and the number of such 3 digit numbers is 3. answer = 30+3= 33.
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:
A
The combination of three digits, whose product can give a perfect cube are (1,2,4) (1,3,9) (4,6,9) (3,8,9) and (2,4,8) when all three digits are distinct.With each of these combinations we can have 3! three digit numbers with distinct digits. Therefore, the number of such 3-digit numbers is 5(3!) = 5*6 = 30. There is another case of (8,8,1) and the number of such 3 digit numbers is 3. answer = 30+3= 33.
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