How many positive integers less than 5,000 are such that the product of their digits is 140? I have checked again and came to an answer 33 intgrs; it's interesting what your answers will be?

Night reader wrote:How many positive integers less than 5,000 are such that the product of their digits is 140? Let's break 140 in its factor in such a way that each of the factors is of single digit. Then we could use the factors to make a number in which the product of the digits will be 140. 140 = 2*2*5*7 = 4*5*7 Thus the numbers are either made of (2, 2, 5, 7) or (1, 4, 5, 7) or (4, 5, 7). For (2, 2, 5, 7) Only numbers starting with 2 are valid. Number of such numbers = 3! = 6 For (1, 4, 5, 7) Only numbers starting with 4 and 1 are valid. Number of such numbers = 3! + 3! = 12 For (4, 5, 7) All numbers are valid. Number of such numbers = 3! = 6 Therefore number of positive integers less than 5,000 such that the product of their digits is 140 = 6 + 12 + 6 = 24

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

thank you, I see clear logic in the solutions of Anurag and Shovan.

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