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by Night reader » Tue Dec 28, 2010 10:06 pm
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?
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by Anurag@Gurome » Tue Dec 28, 2010 10:48 pm
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
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by shovan85 » Tue Dec 28, 2010 10:55 pm
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
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by Night reader » Tue Dec 28, 2010 11:09 pm
thank you, I see clear logic in the solutions of Anurag and Shovan.
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