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# 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?

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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?

Jan 19, 2020
edited by mathmathj28  Jan 21, 2020
edited by Melody  Jan 22, 2020
edited by Melody  Jan 22, 2020
edited by mathmathj28  Jan 23, 2020

#1
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The prime factorization of 10,000  =  5 x 5 x 5 x 2 x 2 x 2 x 1      (using 7 digits).

So the 7 digit possibilities are  5 x 5 x 5 x 2 x 2 x 2 x 1

a)  5 x 5 x 5 x 2 x 2 x 2 x 1

b)  5 x 5 x 5 x 4 x 2 x 1 x 1

c)  5 x  5 x 5 x 8 x 1 x 1 x 1

There are  7! / (3! x 3!)  =  140 ways to do part a.

There are  7! / (3! x 2!)  =  420 ways to do part b.

There are  7! / (3! x 3!)  =  140 ways to do part c.

Jan 19, 2020
#2
+1

Wouldn't 5 x 5 x 5 x 2 x 2 x 2 x 1 be the prime factorization for 1,000?

The combinations of digits that I found were 5 x 5 x 5 x 5 x 2 x 2 x 4, 5 x 5 x 5 x 5 x 4 x 4 x 1, and 5 x 5 x 5 x 5 x 8 x 2 x 1

mathmathj28  Jan 19, 2020
edited by mathmathj28  Jan 19, 2020
#3
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How many ways are there to form a 7-digit code where each digit can be from 0 to 9 and the product of all the digits in the code is 10,000?

Thanks Geno and Mathmathj28,

Since there is disagreement I will add my 10cents worth.

factor(10000) = 2^4*5^4

So I can only use 2,4,8,5,1  and 5 must be used 4 times

10000/5^4 = 16

5 5 5 5

I need 3 digits that multiply to 16       8,2,1       4,4,1     4,2,2,  and that is it.

So the digits can be

5 5 5 5 8 2 1         7!/4!         =210

5 5 5 5 4 4 1          7!/(4!2!)  = 105

5 5 5 5 4 2 2          7!/(4!2!)  =  105

210+105+105 = 420 ways

Jan 19, 2020
#5
+1

Thank you!!

mathmathj28  Jan 19, 2020
#4
+1

My not-so-short computer code agrees with Melody's number:

a=0;b=0;c=0;d=0;e=0;f=0;g=0;p=0; cycle:n=a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g; if(a*b*c*d*e*f*g==10000, goto loop, goto next);loop:printn,", ",;p=p+1; next:g++;if(g<10, goto cycle, 0);g=1;f++;if(f<10, goto cycle, 0);g=1;f=1;e++;if(e<10, goto cycle,0);g=1;f=1;e=1;d++;if(d<10, goto cycle,0);g=1;f=1;e=1;d=1;c++;if(c<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b++;if(b<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b=1;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT:

1255558 , 1255585 , 1255855 , 1258555 , 1285555 , 1445555 , 1454555 , 1455455 , 1455545 , 1455554 , 1525558,.............................  8525551 , 8551255 , 8551525 , 8551552 , 8552155 , 8552515 , 8552551 , 8555125 , 8555152 , 8555215 , 8555251 , 8555512 , 8555521 , Total =  420

Jan 19, 2020
#6
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What happened to the original question?

Jan 22, 2020