The no. of positive integral solutions of the eqn. a×b×c×d×e=1050

Question

Find the total no. of positive integral solutions to the equation

a×b×c×d×e=1050

I came across this problem while practicing problems in multinomial theorem.....but I have no idea how we can apply it here....also tried out a few other logics but none of them seems to work.....can anyone help me out?

Answer 1

Since $1050 = 2^1 \cdot 3^1 \cdot 5^2 \cdot 7^1$, each factor will be of the form $2^{a_k}3^{b_k}5^{c_k}7^{d_k}$, $k = 1,\ldots, 5$ and \begin{align*} \sum a_k = 1, \quad \sum b_k = 1, \quad \sum c_k = 2, \quad \sum d_k = 1 \end{align*} The number of solutions of the above are $5, 5, 15, 5$ and hence the number of solutions is $5\cdot 5 \cdot 15 \cdot 5 = 1875$

Answer 2

Considering the tag combination attached your question I interpret your question as counting the number of integer sequences $a\ge b\ge c\ge d\ge e\ge1$ such that $abcde=1050$.

One can split this into various cases according as how many of these 5 factors are 1. This leads to far fewer count. Take the prime factorization that yields 5 prime factors (not distinct). $$1050= 5\times5 \times 7\times3\times 2=abcde$$

Case I. $e=1$, $a,b,c,d>1$: By unique factorization, we have to club two of the 5 primes into a single factor. There are 7 ways of achieving this.

Case II. $d=e=1; a,b,c> 1$. Either we can club 3 primes, calling it $a$ and retaining the other two as $b$ and $c$; or club two primes in pairs and call them and retain the other. This also leads to 13 more possibilities

Case III & IV $a>1,c,d,e=1, b=1,b\ne1$. This is number of ways of factoring $1050$ into two factors, which is same as $d(n)$,number of divisors of $1050$, This is 24.

And one more case all the five factors > 1 which has just one possibility. So the answer is $24+7+13+1=45$ I hope I haven't missed anything.