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# counting/probability problemo

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Find the number of 7-digit positive integers, where the product of the digits is equal to 10000.

Nov 15, 2019

#1
+397
+2

Each digit must be a factor of 10000, so the only possible digits are 1, 2, 4, 5, and 8. Furthermore, since the prime factorization of 10000 is there must be exactly four digits of 5. Among the remaining three digits, we also need four factors of 2. We have the following cases.

Case 1: The remaining three digits are 1, 2, and 8.

We want to arrange the digits 5, 5, 5, 5, 1, 2, and 8. There are $$\frac{7!}{4!} = 210$$ ways to arrange these digits.

Case 2: The remaining three digits are 1, 4, and 4.

We want to arrange the digits 5, 5, 5, 5, 1, 4, and 4. There are $$\frac{7!}{4! 2!} = 105$$ ways to arrange these digits.

Case 3: The remaining three digits are 2, 2, and 4.

We want to arrange the digits 5, 5, 5, 5, 2, 2, and 4. There are $$\frac{7!}{4! 2!} = 105$$ ways to arrange these digits.

Thus, there are a total of $$210 + 105 + 105 = \boxed{420}$$ 7-digit numbers.

Nov 15, 2019
edited by sinclairdragon428  Nov 15, 2019
#2
+1

This computer code confirms the above theoretical calculations and gives exactly same answer of 420 integers.

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

Nov 15, 2019