To enter her secret lab, SuperMathHeroine must punch in a 7-digit code, where each digit can be from 0 to 9. Unfortunately, SuperMathHeroine has forgotten her code. She only remembers that the product of the digits in her code is 10000. How many different codes could there be?
Well, none of the digits can be 0 because the product is than going to be 0. You can't have more than 2 1's either.
To get started, we turn to the factorization of 10,000. It factors out as 2^4 * 5^4. This means that the only possible combinations of numbers are:
1,2,8,5,5,5,5 -> 7!/4! = 210 ways to arrange this
2,2,4,5,5,5,5 -> 7!/(4!*2!) = 105 ways to arrange this
1,4,4,5,5,5,5 -> 7!/(4!*2!) = 105 ways to arrange this
1,1,8,5,5,5,5 -> 7!/(4!*2!) = 105 wasy to arrange this
So the number of different possible codes is 120+105+105+105 = 435 possible codes.
I don't understand how "1,1,8,5,5,5,5 -> 7!/(4!*2!) = 105" is incorperated into this question because it makes 5000. Also, you add 120 and not 210.
We need to find how many 7 digit codes are possible where each digit is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 and the product of all the digits is 10000.
Do the prime factorization for 10000 to get = 2^4*5^4
We now know that 2,4,8,5,1 and 5 must be used 4 times
10000/5^4 = 16
I need 3 digits that multiply to 16
8*2*1
4*4*1
4*2*2
Therefore....
7!/4! =210
7!/(4!2!) = 105
7!/(4!2!) = 105
And we add these together to get 210+105+105 = 420 ways
Hope this helped!
Melody Jan 19, 2020