+0  
 
+1
112
3
avatar

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?

 Mar 21, 2020
 #1
avatar+108 
+1

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.

 Mar 21, 2020
 #2
avatar
0

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.

 Mar 21, 2020
 #3
avatar+2277 
+1

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

 Mar 21, 2020

12 Online Users

avatar
avatar
avatar