# What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c$?

What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?

My try:

The prime factorization of $1000$ is $2^3\cdot 5^3$

$a\cdot b \cdot c = 2^3\cdot 5^3$

$a=2^{a_1}\cdot 5^{a_2}$

$b=2^{b_1}\cdot 5^{b_2}$

$c=2^{c_1}\cdot 5^{c_2}$

$abc=2^{a_1+b_1+c_1}\cdot 5^{a_2+b_2+c_2}=2^3\cdot 5^3$ $a_1+b_1+c_1=3$

How many ways are there such that $a_1+b_1+c_1=3$

Star's and Bar's method: -

Number of ways to chose $2$ separators($0s$) in a string of $5$$= {5\choose 2 }=10$ $N(a_1+b_1+c_1=3)=10$ Similarly, $N(a_2+b_2+c_2=3)=10$ $N(abc=1000)=10\cdot 10=\boxed{100}$

Is that okay ? Please write down any notes

• Have you remembered to take into account the condition that $a\leq b\leq c$? I think not. – JMoravitz Jul 12 '17 at 20:44
• @JMoravitz I can't understand what to do – user373141 Jul 12 '17 at 20:49

For all $a,b,c$ you got 100 but that didn't take into account $a \le b \le c$.

So count $a=b=c$ that is $a=b=c=10$ you counted that once. The remaining $99$ were overcounted.

Consider $a=b, c\ne a$. Then $a=b= 2^j5^k; c=2^{3-2j}3^{3-2j}$ so there are $3$ such cases ($j = 0,1; k = 0,1$ but not $j=k=3-2j=3-2k= 1$). You counted each of those $3$. So that accounts for $9$ when you should have only counted $3$. You have $90$ more to account for.

These are $a,b,c$ distinct. You counted each of these $6$ times when you should have counted them once. So you counted $90$ when you should have counted $15$.

So the number should be $15 + 3 + 1 = 19$.

• You meant $c = 2^{3 - 2j}3^{3 - 2\color{red}{k}}$. – N. F. Taussig Jul 12 '17 at 21:45

Your computation of $N=10$ is correct and $100$ is the number of ordered triples that have product $1000$. You have failed to account for the condition that $a \le b \le c$. All of the unordered triples that have three distinct elements have shown up six times, so you have overcounted. Those that have two or three equal elements have been counted differently. Keep going.

• okey , help me finish it , i got stuck here – user373141 Jul 12 '17 at 20:47
• Well count the the $a=b=c$. That was counted once. count the $a=b, c \ne b$ those were counted three times. Divide them by 3. That leaves $a,b,c$ distinct. Those were counted 6 times. Divide those by 6. – fleablood Jul 12 '17 at 20:51
• As a check: In Mathematica Solve[a b c == 1000 && 0 < a <= b <= c, Integers] shows 19 solutions. – David G. Stork Jul 12 '17 at 20:54
• Can we write a general solution, for arbitrary $n$, without having to do the analysis "by hand"? For instance, for $n=10^6$ the number of ways is $78$. For $n = 10^9$ it is $517$. – David G. Stork Jul 12 '17 at 21:26
• ...and for $n=10^{12}$ it is $1405$. – David G. Stork Jul 12 '17 at 21:35