We can use only 1, 2, 4, 5, 7 as the presence of other digits will not give us the product as 140.
positive integers less than 5,000 so the number can be of 4 digit or 3 digit as 2 and 1 digit number's product of digit will not give us 140.
Consider 4 digit numbers (1000 < X < 5000)
ABCD be the structure of the number.
A cannot be 5 or 7 and 5 and 7 are the minimum requirement to get 140, we can manipulate the other two digits as (1,4) or (2,2)
When A = 1, BCD can be arranged by 4, 5, and 7. Thus 3! = 6 ways.
When A = 2, BCD can be arranged by 2, 5, and 7. Thus 3! = 6 ways.
When A = 4, BCD can be arranged by 1, 5, and 7. Thus 3! = 6 ways.
Total 4 digits = 18 numbers
When 3 digit number.
Only possible digits are 4,5, and 7. Thus 3 digits in 3 place can be arranged in 3! ways = 6 ways.
Total = 18+6 = 24
If the problem is Easy Respect it, if the problem is tough Attack it
