How many 7 digit positive integers are there such that the product of the individual digits of each number is equal to 10000?
4455551 = 4x4x5x5x5x5x1=10000
the proof is not valid.
To find the number of 7-digit positive integers whose digits multiply to 10,000, we need to consider the prime factorization of 10,000.
Prime factorization of 10,000:
10,000 = 2^4 × 5^4
So, the number 10,000 can be represented as a product of its prime factors, with each factor corresponding to one of the digits in the 7-digit number. We have two prime factors, 2 and 5, with exponents from 0 to 4.
Let's consider the digits:
1. The digit 2:
- It can be one of the digits in the 7-digit number.
- There are 7 positions to choose for the digit 2.
2. The digit 5:
- It can be one of the digits in the 7-digit number.
- There are 7 positions to choose for the digit 5.
3. The digit 1:
- We have 4 options: using 2^0 × 5^0, 2^1 × 5^0, 2^2 × 5^0, or 2^3 × 5^0.
- There are 2 positions to choose for the digit 1.
Now, we need to calculate the number of ways to arrange the digits in the 7-digit number.
The number of ways to arrange the digits (including repeated digits) in a 7-digit number is 7 factorial (7!). However, we need to account for the repeated digits, so we divide by the factorial of the number of times each digit repeats.
In this case, the digit 2 repeats 4 times (2^4), the digit 5 repeats 4 times (5^4), and the digit 1 repeats 2 times (2^0 × 5^0).
The total number of 7-digit positive integers whose digits multiply to 10,000 is:
Number of ways to arrange the digits = 7! / (4! × 4! × 2!)
= (7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1) × (4 × 3 × 2 × 1)
= 35,280.
Therefore, there are 35,280 7-digit positive integers such that the product of the individual digits is equal to 10,000.
10000 = 10^4 = 2^4 * 5^4
so, the digits must be all 1's, 2's and 5's
Since you need all 8 2's and 5's to get 10,000, no 7-digit number will do it.