We note that the prime factorization of $1000000000=2^9\cdot 5^9$ . Thus, $a$ , $b$ , and $c$ must each be in the form $2^m\cdot 5^n$ , so set $a=2^A\cdot 5^D$ , $b=2^B\cdot 5^E$ , and $c=2^C\cdot 5^F$ (where $A$ , $B$ , $C$ , $D$ , $E$ , and $F$ are non-negative integers). Clearly, $abc=2^{A+B+C}\cdot 5^{D+E+F} =2^9\cdot 5^9$ . Equating the exponents, we get that $A+B+C=9$ and $D+E+F=9$ . Additionally, we can note that if we consider any triple, there is only one way to order it if we consider identical numbers indistinguishable. Thus, we only need to find the number of distinct triples which satisfy this equation. We can then use the ball-and-urn method to find that there are $\binom{11}{2}=55$ triples which satisfy $A+B+C=9$ , and also $55$ triples which satisfy $D+E+F=9$ . Since these are independent of one another, we can multiply these to find that there are $3025$ triples. However, we have vastly overcounted. Each triple with no identical elements ( $T_0$ ) is counted $3!=6$ times, each triple with exactly two identical elements ( $T_2$ ) is counted $\frac{3!}{2!}=3$ times, and each triple with three identical elements ( $T_3$ ) is counted $\frac{3!}{3!}=1$ time. It is substantially easier to count the number of triples which satisfy the second and third conditions above than to count the triples that satisfy the first. Hence, an easy way to count each triple once is to count each triple $6$ times, and then divide the number of ways by $6$ . $T_2$ : Let the two equal numbers be $a$ and $b$ . To preserve the generality, we can rearrange the numbers after finding them to maintain the condition $a\le b\le c$ . This implies that $A=B$ and $D=E$ , so $2A+C=9$ and $2D+F=9$ . There are $5$ solutions to both the first and second equations, for a total of $25$ , but one of these is $(A,B,C,D,E,F)=(3,3,3,3,3,3)$ , so we need to subtract one to get $T_2=24$ . Note that we do not consider how these are permuted, as we are only concerned with the number of ordered ( $a\le b\le c$ ) triples. $T_3$ : In this case, $a=b=c$ , which implies $A=B=C$ and $D=E=F$ , giving the equations $3A=9$ and $3D=9$ . Both of these have only one solution each, and thus, $T_3=1$ . Finally, we know that $6T_0+3T_2+T_3=3025$ . Since $3T_2=72$ and $5T_0=5$ , we can add these to get that $6T_0+6T_2+6T_3=3102\implies T_0+T_2+T_3=517$ , so the number of triples is $\boxed{517}$ .

The ordering $a \leq b \leq c$ of the ordered pair makes counting the number of solutions much more difficult. Let's first solve for the number of ordered pairs (a,b,c) and then correct for over-counting.

Step 1: Establishing the Bijection

We first note that $1,000,000,000 = 2^{9}5^{9}$ . Then, Given a set of 3 integers $(a,b,c)$ , we note that $a,$ $b$ , and $c$ can be written as $2^{a_1}5^{a_2}$ , $2^{b_1}5^{b_2}$ , and $2^{b_1}5^{b_2}$ , respectively. we will create a bijection between solutions $(a,b,c,)$ of $abc = 1,000,000,000$ and the equations $a_1 + b_1 + c_1 = 9$ , $a_2 + b_2 + c_2 = 9$ . We will then create one more bijection between the aforementioned equations and 2 mutually exclusive sequences of 11 characters consisting of 2 dividers $(|)$ and 9 "things" $(0)$ .

The construction is as follows. Given a set of 3 integers $(a,b,c,)$ , we note that $a,$ $b$ , and $c$ can be written as $2^{a_1}5^{a_2}$ , $2^{b_1}5^{b_2}$ , and $2^{b_1}5^{b_2}$ , respectively. Then, because $abc = 2^95^9$ , we can equate exponents and get the following:

$2^{a_1 + b_1 + c_1} = 2^9 \rightarrow a_1 + b_1 + c_1 = 9$ $5^{a_2 + b_2 + c_2} = 5^9 \rightarrow a_2 + b_2 + c_2 = 9$

Case 1: powers of 2

we create the sequence corresponding with $(a_1, b_1, c_1)$ as follows. Such a sequence will start with $a_1$ $0$ ‘s, then has a $|$ , then has $b_1$ $0$ ‘s, then has a $|$ ‘s, then has $c_1$ $0$ ‘s,

For example, if $(a_1, b_1, c_1) = (0, 4, 5)$ , then the corresponding sequence is:

$|0000|00000$

The number of such arrangements is $11\choose2$ . We want to be precise in what this number means. $11\choose2$ is the number of possible ways to distribute $9$ factors of $2$ to $a$ , $b$ , and $c$ .

Case 2: powers of five

The result follows identically.

Step 2: (over)Counting the Solutions

Because distributing powers of two and powers of five are independent, the total number of solutions $(a,b,c)$ for $abc = 2^95^9$ is $11\choose2$ $^2 = 3025$ .

Step 3: Fixing Step 2

First off, our answer is much larger than 999, as we've overcounted by a lot.

Case 1

For instance, if $(x,y,z)$ is a solution and $x < y < z$ , we have one good solution,

$(x,y,z)$

and 5 bad solutions:

$(x,z,y)$ , $(y,x,z)$ , $(y,z,x)$ , $(z,x,y)$ , $(z,y,x)$ .

So, if our solution contains 3 distinct numbers, we have overcounted by a factor of 6. Denote the number of solutions satisfying this as $i$ .

Case 2

What if 2 of the integers are the same? WLOG, let (x,x,y) be a solution with $(x<y)$ . (similar analysis on $(x,y,y)$ will produce the same result) We have one good solution:

$(x,x,y)$

and 2 bad solutions:

$(x,y,x)$ $(y,x,x)$ .

So, if our solution contained 2 numbers, we have overcounted by a factor of 3. Denote the number of solutions satisfying this as $j$ .

Case 3

What if all 3 of the integers are the same? The only possible solution is $(100,100,100)$ !

The 3 cases encompass all possible cases.

Step 4: Putting Everything Together

We are interested in $i + j +1$ . With our given information, we have the following equation.

$6i + 3j + 1 = 3025$ .

Let's solve for $j$ . We can establish another bijection relating $j$ to the number of perfect square divisors of $10^9$ .

Let $n^2 | 10^9$ . Then $(n,n, 10^9/n$ ) or $(10^9/n,n,n)$ is a solution.

A perfect square divisor of $10^9$ has the form $2^a5^b$ , where $a$ and $b$ are even. So, $a, b \in \{0,2,4,6,8\}$ , so $j = 5*5 = 25.$

However, we've overcounted again. if $n = 1,000,000$ , then the solution is $(1000,1000,1000)$ , and we have already accounted for this solution with case 3. Therefore, $j = 25 -1 = 24$ .

Back to our equation,

$6i + 3j + 1 = 3025, j = 24 \rightarrow i = 492$

The solution is $492 + 24 + 1 = 517.$

1000000000=2^9 5^9 Hence a,b,c are all expressible in the form 2^a 5^b

There are (9+2) choose 3 = 55 ways of splitting 9 into 3 non negative integers.

Hence there are 55*55=3025 ways of splitting the powers of 2 and 5 However, when all 3 numbers of a set are distinct, it is counted 6 times towards the sum 3025. When exactly 2 numbers are equal, it is counted 3 times towards the sum 3025. When all 3 numbers are equal, it is counted once towards 3025.

When two numbers of equal, the third number, x, is x = 1000000000/square number. There are 25 squares that divide 1000000000.

Among the 25 sets with two numbers equal, there is 1 set with all 3 numbers equal, (1000,1000,1000).

Hence there are (3025-1-24*3)/6 = 492 sets with all 3 numbers distinct.

Thus there are 492+24+1 = 517 sets in total.

If a = 2^x.5^h, b = 2^y.5^j, c = 2^z.5^k, then x + y + z = h + j + k = 9. Hence this is equivalent to sorting 9 factors of 2 into an ordered triplet, and then sorting 9 factors of 5 into another set of triplets, then "multiplying these triplets" together.

There are only 12 ways to choose an unordered triplet (x,y,z) such that x =< y =< z and x + y + z = 9. Of these triplets, one (3,3,3) uses only one distinct number; four ((0,0,9),(1,1,7), (2,2,5), (1,4,4)) use two distinct numbers, and the last seven use three. This becomes important to avoid double-counting when multiplying the triplets of factors of 2 and 5 together.

There is only one way to combine a (3,3,3) with any other triplet. This gives us 12 ordered triplets (a,b,c); where 8 | a, b, c.

However, there are two ways for the 4 two-distinct-number triplet to combine with another 2dn triplet ((x, x, y) x (h, h, j) and(x, y, x) x (h, h, j)) and similarly three ways to combine with a 3dn triplet (it's a matter of which number in the 3dn is combined with y in the 2dn triplet). There is still only one way to combine with a (3,3,3); this gives us a total of (4x2 + 3x7 + 1x1) x (4 2dn triplets) ways; or 120 ways where two of a, b, c share the same highest power of 2.

Lastly, 3dn triplets can combine with another 3dn triplet in a total of 6 ways; it's the number of permutations of 3 ordered objects. This gives us a total of (7x6 + 4x3 + 1x1) * 7 ways, or 385.

(Throughout this process, it doesn't matter whether the triplets given by the multiplications have their numbers in ascending order; all that matters is that no two triplets share the same numbers.)

Total triplets = 12 + 120 + 385 = 517.

$1,000,000,000 = 2 ^ 9 \cdot 5 ^ 9$ . Each $2$ and each $5$ is a factor of one of three positive integers, so we want to partition nine items into three sets twice. There are twelve ways to partition the $2$ s: $(9, 0, 0), (8, 1, 0), (7, 2, 0), (7, 1, 1), (6, 3, 0), (6, 2, 1), (5, 4, 0), (5, 3, 1), (5, 2, 2), (4, 4, 1), (4, 3, 2), (3,3,3)$ . Likewise, there are the same twelve ways to partition the $5$ s.

For each partition of $2$ s into three different values, there are $6$ ways to match it with a partition of $5$ s into three different values, $3$ ways to match it up with a partition containing exactly two equal values, and $1$ way to match it up with $(3, 3, 3)$ . For every partition of $2$ s containing exactly two equal values, there are $3$ ways to match it up with a partition of $5$ s into three different values, $2$ ways to match it up with a partition containing exactly two equal values, and $1$ way to match it up with $(3, 3, 3)$ . The partition $(3, 3, 3)$ has one match up with any partition of $5$ s.

We have $7$ partitions into three different values, $4$ containing exactly two equal values, and $(3, 3, 3)$ . So the answer is the sum of all the combinations of partitions, or $7 \cdot (7 \cdot 6 + 4 \cdot 3 + 1) + 4 \cdot (7 \cdot 3 + 4 \cdot 2 + 1) + 1 \cdot 12 = 517$ .

For $n=1,2,3$ let $T_n=\{ (a,b,c) \in \mathbb{N}^3 :abc=10^9, \left|\{a,b,c\} \right|=n\}$ and $S_n=\{ (a,b,c) \in T_n : a \le b \le c \}$ . Also let $T=T_1 \cup T_2 \cup T_3$ and $S=S_1 \cup S_2 \cup S_3$ . Our goal is to compute $|S|$ . (It is to be understood that $| \cdot |$ here means set cardinality.)

Then $S_1 = T_1= \{(10^3, 10^3, 10^3 )\}$ so $|S_1| = |T_1| = 1$ . And $|S_2 | = |\{(x,x,y)\in T_2 \}|$ $=|\{ x \in \mathbb{N} :x^2 | 10^9\}| - |T_1|$ $=|\{ x \in \mathbb{N} :x| 10^4\}| -1$ $=5^2 -1=24$

On the other hand, $|T |= |\{ (a,b,c) \in \mathbb{N}^3 :abc=2^9\}| \cdot |\{ (a,b,c) \in \mathbb{N}^3 :abc=5^9\}|$ $= {11 \choose 2} \cdot {11 \choose 2} =55^2$

Now we can use \( |T|= |T_1| + |T_2| +

|T_3|

|S_1| + 3|S_2| + 6|S_3| \), which is to say $6|S|= 6|S_1| + 6|S_2| + 6|S_3| = |T| +5|S_1| +3|S_2|$ $=55^2+5(1)+3(24) =3102$ , so $|S|=3102/6=517$ .

Assume a,b,c are all equal. This gives one solution a=b=c=1000. Assume two of them are equal. Since there are only two distinct values, the condition $a \leq b \leq c$ doesn't matter. WLOG, let $a=b=2^x5^y$ . Then $x,y \leq 4$ . So a can take $5 \times 5=25$ values. But the case a=b=1000 is repeated, so we have 25 sets until now. Now assume all are distinct. We ignore the condition $a \leq b \leq c$ first, because if we get a solution (a,b,c) where all are distinct, in its 6 permutations only one satisfies condition $a \leq b \leq c$ . WLOG, we choose a first. Let $a=2^m5^n$ . Then $b=2^p5^q, p\leq 9-m, q\leq 9-n$ . There is then a unique solution for c. If both m and n are odd, then bc is a perfect square, and in one of the solution b=c which contradicts our assumption. There are 25 cases when m and n are odd(there are 5 odd numbers from 0 to 9). If $m,n\leq 4$ , some b will be equal to a. There are again 25 such cases(there are 5 integers from 0 to 4). If $m,n\leq 4$ , some c will be equal to a. There are again 25 such cases(there are 5 integers from 0 to 4). In these three special cases, a=b=c is repeated twice, so we minus 73 from our answer. The answer before minus is (1+2..+10)^2, because m is 0 to 9, so 9-m has 1 to 10 choices, same for n. So the final answer is $\frac{55^2-73}{6}+25=517$ .

First, we solve the problem for $abc = 2^ 9 5^9$ . By considering the prime powers, this corresponds to solving $a_2+b_2+c_2 = 9$ and $a_5 + b_5 + c_5 = 9$ , hence by bars and stripes has ${ 11 \choose 2} ^2 = 55^2 =3025$ solutions.

Now, we count the number of solutions in which $a = b =c$ . There is clearly 1 solution. We count the number of solutions where $a = b$ . This corresponds to solving $2A + C = 9$ twice, hence has $5^2=25$ solutions. We count the number of solutions where $a, b, c$ are pairwise distinct. Let $X, Y, Z$ denote the sets where $a=b, b=c, c=a$ respectively. By the Principle of Inclusion and Exclusion,
$\begin{aligned} |X \cup Y \cup Z| &= |X| + |Y| + |Z| - |X \cap Y| - |Y \cap Z| - |Z \cap X| + |X \cap Y \cap Z| \\ &= 25 + 25 + 25 - 1 - 1 - 1 + 1 \\ &= 73 \\ \end{aligned}$

Hence, there are $3025 - 73 = 2952$ solutions with $a \neq b \neq c$ .

Now, we deal with the condition that $a \leq b \leq c$ . If $a, b, c$ are distinct, they will generate 6 sets of $abc= 10^9$ , hence there are $\frac {2952}{6} = 492$ solutions. If only 2 of them are the same, then they generate 1 set of solutions, hence there are $25-1=24$ solutions. If all 3 of them are the same, there is 1 solution. Hence, there is $492 + 24 + 1 = 517$ solutions in total.