Digit Product of 10000

10000 has a few factors that are less than 10: 1, 2, 4, 5, and 8. Our 7-digit integers will be comprised of these five different digits only.

$10000 = 2^4 \times 5^4.$ Our integer must contain four 5s within it because the factorization contains four 5s.

This leaves three possible cases. The product of the remaining three digits must be $2^4 or 16.$

The three digits could be (1, 2, 8), (1, 4, 4), or (2, 2, 4). Thus we have three possible permutations that we could form.

Any number that satisfies the conditions of the problem must be a permutation of 1285555, 1445555, or 2245555.

Calculating the possible permutations, we get $\frac{7!}{4!} + \frac{7!}{4!2!} + \frac{7!}{4!2!} = 420$ permutations.

Thus there are 420 different 7-digit positive integers such that the product of its individual digits is 10000.

First, prime factorization of 10000 is 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5.

Beside 5, there are no other divisible by five, single digit factor for 10000. This implies that we have to take four 5's. And we are left with 3 digit for 16 = 1 \times 2 \times 8 = 1 \times 4 \times 4 = 2 \times 2 \times 4.

thus all we need to do is to add the permutation of 1285555, 1445555, 2245555.

number of permutation for 1285555 = \frac {7!} {1! \times 1! \times 1! \times 4!} = \frac {7!} {4!} = 7 \times 6 \times 5 = 210

number of permutation for 1445555 = \frac {7!} {1! \times 2! \times 4!} = \frac { 7 \times 6 \times 5 } {2} = 105

number of permutation for 2245555 = \frac {7!} {1! \times 2! \times 4!} = \frac { 7 \times 6 \times 5 } {2} = 105

total would be 420 .

10000 = 10^4 = (5^4)*(2^4) We see that the single digit factors of 10000 are 1,2,4,5 and 8 Here, 4=2^2 and 8=2^3 i.e. 2^2 and 2^3 are both single digits, but 5^2 and above are all more than single digits, hence, it is then compulsory to use all the 4 fives in our 7 digit number. 10000/5^4 = 16

Hence, we need to choose remaining 3 digits from 1,2,4,and 8 (each can be used more than once) such that their product is 16. Possibilities are 2,2,4 ; 1,4,4 and 1,8,2.

Hence we have 3 cases for the 7 digit number. Case 1 : 5,5,5,5,2,2,4 Total number of permutations = 7! /(4! 2!)

Case2 : 5,5,5,5,1,4,4 Total number of permutations = 7!/(4! 2!)

Case 3 : 5,5,5,5,1,2,8 Total number of permutations = 7!/4!

Hence, our final answer will be their sum , on simplification, we will get 420

10000 is equal to 2^4 times 5^4. We already know that 4 of the digits should be the 5's, since we cannot pair them to another prime factor because we will have a two-digit number then. Thus, we have to have 3 one-digit numbers which multiply to 16. These are A{1,4,4}, B{1,2,8} and C{2,2,4}. Using set A and the 4 5's, we have (7C4)(3P2) possibilities = 105. This is the same for using set C = 105. Lastly, using set B, we have (7C4)(3!) possibilities = 210 Adding all up, we get 420.

Note that $10000=10^4=2\cdot 2\cdot 2\cdot 2\cdot 5\cdot 5\cdot 5\cdot 5.$ The number $22225555$ and its permutations would work if the original problem did not specify that the integers had to be $7$ digits in length. Therefore, we must combine two or more digits to produce a $7$ digit integer while still keeping the product of the digits at $10000$ . We have three cases to tackle.

CASE 1: We can combine two of the 2's to get a number of the form $2245555$ , which satisfies the conditions. There are $\dfrac{7!}{2!1!4!}=105$ ways to rearrange the digits of this number, and each number has the property that the product of the digits is $10000$ .

CASE 2: We can combine two pairs of 2's and attach a 1 to get a number of the form $1445555$ , which also satisfies the conditions. There are $\dfrac{7!}{1!2!4!}=105$ ways to rearrange the digits of this number.

CASE 3: We can combine three of the 2's and attach a 1 to get a number of the form $1255558$ , which furthermore satisfies the conditions. There are $\dfrac{7!}{1!1!4!1!}=210$ ways to rearrange the digits of this number.

The next several sentences will explain why these three cases are the only ones that we need to count. Combining four $2$ s gives us $16$ , which introduces a factor of $6$ into our product (making it impossible to equal $10000$ ). Combining two or three fives gives $25$ and $125$ , and these combinations of digits do not give rise to any new cases. Combining four $5$ s gives $625$ , which also introduces an unwanted factor of $6$ into our product. Finally, combining any number of $2$ s and $5$ s gives a number that is a multiple of $10$ , guaranteeing that the product of the digits will be $0$ . Hence we have exhausted all cases, and the number of possible integers satisfying the conditions is $105+105+210=\boxed{420}$ .

Since $10000=2^4 \cdot 5^4$ , we seek 7 digits that multiply to $2^4 \cdot 5^4$ . Clearly, four of these 7 digits must be 5, for the next highest multiple of 5, which is 10, has two digits. We have 3 cases to consider:

(1) 5555422 in some order. Then we have ${7 \choose 4} \cdot 3 = 105$ ways to arrange these digits, because there are 4 slots of 7 to select to place the 5's, and 3 ways to order 4, 2, 2 (namely 422, 242, 224) in the 3 remaining slots. (2) 5555441 in some order. For the same reasons as (1), there are 105 ways to order these numbers. (3) 8555521 in some order. There are ${7 \choose 4} \cdot 3! = 210$ ways to order these numbers, for there are $3!$ ways to order 8, 2, 1.

We have $105+105+210=420$ solutions in all.

$10000 =10^4 = (2 \cdot 5)^4 = 2^4 \cdot 5^4$ and so due the the uniqueness of prime factorisations we know that the digits of our candidate numbers will involve these primes. Since $5 \cdot 2 = 10$ which cannot be a digit in base $10$ we must have that four of our digits are $5$ . So the product of our final three digits must be $2^4 = 16$ , which gives three possibilities: $(1, 2, 8), (1, 4, 4), (2, 2, 4)$ where again, we disregard any combinations that use $16$ because it cannot be a digit in base $10$ . summing the permutations of these possibilities gives $\frac {7!} {4!} + \frac {7!} {4! \cdot 2!} + \frac {7!} {4! \cdot 2!} = 420$

We know that $10000 = 2^4 \cdot 5^4$ , so 4 of the 7 digits need to be a $5$ . Then, the three remaining digits must have a product of $2^4=16$ , with the restrictions that the largest digits is $8=2^3$ and the smallest digit is $1 = 2^0$ . Also, the largest of the three is at least $4$ , because otherwise, their product is smaller than $16$ . By simply checking, we see that the three remaining digits are

• $8,2,1$ in some order, or
• $4,4,1$ in some order, or
• $4,2,2$ in some order.

Therefore, the total number of 7-digits positive integers that satisfy the conditions is equal to the sum of the number of ways we can arrange the sequences

$5,5,5,5,8,2,1 \quad \text{and} \quad 5,5,5,5,4,4,1 \quad \text{and} \quad 5,5,5,5,4,2,2,$

which is

$\frac{7!}{4!} + \frac{7!}{4!2!} + \frac{7!}{4!2!} = \frac{7!}{4!} \cdot \left( 1 + \frac{1}{2!} +\frac{1}{2!} \right) = 5 \cdot 6 \cdot 7 \cdot 2 = \boxed{420}.$

We factor this as $5^4 \cdot 2^4$ . Since the only digit divisible by 5 is 5, there must be 5 5's. The other digits could be 441, 422, or 821 (in some order). These cases give 105, 105, and 210 possible arrangements, so the total number of positive integers is 420.

Since $10000=2^2 5^2$ , we must have all of the $5$ 's appear in the $7$ digit number, since no power of $5$ , other than $5^1$ , is a one digit number. The rest of the digits must multiply to $2^4=16$ . There are $3$ ways to do this with $1$ digit numbers: $1*2*8$ , $1*4*4$ , and $2*2*4$ . There are $\frac{7!}{4!}=210$ numbers for the first possibility, $\frac{7!}{4!2!}=105$ numbers for the second possibility, and $\frac{7!}{4!2!}=105$ numbers for the third possibility, so our answer is $210+105+105=\fbox{420}$ .

On factorizing $10000$ we get, $2*2*2*2*5*5*5*5$ , so it is obvious that, four $5\,'s$ should be present in the number, among the rest three nos. there are the following possibilities: $1)\:2*8*1*5*5*5*5\:which\:can\:be\:rearranged\:in\:\frac{7!}{4!}\:=\:210\:ways.$ $2)\:4*4*1*5*5*5*5\:which\:can\:be\:rearranged\:in\:\frac{7!}{4!2!}\:=\:105\:ways.$ $3)\:2*2*4*5*5*5*5\:which\:can\:be\:rearranged\:in\:\frac{7!}{4!2!}\:=\:105\:ways.$ $Hence,\,the\,total\,number\,of\,positive\,integers,\,is\,210+105+105=\fbox{420}.$

The prime factorization of $10000$ is $2^4\times5^4$ , as should be clear from the example given.

However, the example gives an eight digit integer; we are concerned with seven digit ones. Any power of five exceed one digit, so we focus on the powers of two that are only one digit, specifically: $1, 2, 4, 8$ .

Any seven digit number for which the product of its digits equals 10000 will be one of the following permutations: $5555422$ $5555441$ $5555821$ The factorial function gives the number of permutations of $n$ distinguishable items. But in the strings above, there are occasions of indistinguishable items. To count the number of unique strings, we can divide the number of permutations by the factorial of the number of objects that are identical. Formally we have $\frac{7!}{4!2!} + \frac{7!}{4!2!} + \frac{7!}{4!} = 7\times6\times5\times2 = \boxed{420}$

Use prime factorization to find which primes multiplied with each other equal 10000. They are:

2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

We need a 7-digit number only, so we can take two 2s and multiply them with each other.

2 x 2 x 4 x 5 x 5 x 5 x 5

Now, we have seven digits. To solve for the number of combinations you can make with these, use this formula. Let n be the number of digits in the entire combination. Let a (b, and c) be the redundancies (repeated numbers; ex: 4 because there's four 5s)

n! / (a!b!c!)

In the first case, it's 7! / (2!4!) = 105

Find a second case, which would be:

1 x 4 x 4 x 5 x 5 x 5 x 5

And repeat the same step. It'll also equal to 105 combinations.

Lastly, find the remaining case. There is only one because you can put two 5s together, nor two 4s. Only 2 x 4 or 2 x 2.

1 x 2 x 8 x 5 x 5 x 5 x 5

This one equals to 210 because there is only one redundancy (4!).

Add up 210 + 105 + 105 and it'll equal the answer. Yay.

Permutations of $2245555$ , $1255558$ , $1445555$ : $7!\left(\frac{1}{4!2!}+\frac{1}{4!}+\frac{1}{4!2!}\right)=3\frac{7!}{4!}=420$

There must be 4 5's and the product of the last 3 digits must be 16, which leaves (8,2,1), (4,4,1), (4,2,2). There are 210 possible combinations of the first, and 105 combinations of each of the last.

10000=2.2.2.2.5.5.5.5 7 digit integers => there are three way 1. 10000=1.2.8.5.5.5.5 there are 7C4.3C2 numbers 2. 10000=2.2.4.5.5.5.5 there are 7C4.3C2 numbers 3. 10000=1.4.4.5.5.5.5 there are 7C4.3.2 numbers so total numbers are 7C4.3C2+7C4.3C2+7C4.3.2=420