searching smallest number that has $40$ distinct positive divisors

What is the smallest natural number such that it has $40$ distinct positive (integer) divisors (inclusive of $1$ and itself?
At first I was stunned of seeing the problem.It's not possible to find all the divisors of all the numbers. I think it's a huge calculation. How can I solve this type of problem easily? help me.How can I proceed?

• Hint: consider to use the factorial. – Anatoly Aug 6 '14 at 10:41
• Knowing that every positive integer has a unique prime factorisation, you should be able to think of how to get 40 distinct divisors, whilst not making the number any larger than it has to be. – FireGarden Aug 6 '14 at 10:42
• Start by solving a simpler version of the problem. What positive integers $n$ have (say) exactly two divisors, inclusive of 1 and $n$? Which is the smallest of these? 40 distinct divisors does sound like a lot, but once you see the pattern you will be able to answer with confidence. – hardmath Aug 6 '14 at 10:48

Letting $$N={p_1}^{q_1}\cdot {p_2}^{q_2}\cdots \cdot {p_k}^{q_k}\ \ (p_i\ \text{are primes}, q_i\ge 1\in\mathbb N,k\in\mathbb N)$$ be your number, the following has to be satisfied (see here for details): $$(q_1+1)(q_2+1)\cdots(q_k+1)=40=2^3\cdot 5.$$ (Here, LHS represents the number of the positive divisors of $N$.)

So, separate it into cases as the followings :

(1) Since $40=40$, $N=p^{39}$. Hence, we have $2^{39}.$ (Note this is the smallest number in this case)

(2) Since $40=2\times 20$, $N={p_1}^{1}\cdot {p_2}^{19}$. Hence, we have $3^1\cdot 2^{19}.$

(3) Since $40=4\times 10$, $N={p_1}^{3}\cdot {p_2}^{9}$. Hence we have $3^3\cdot 2^9$.

Can you take it from here? Note that there are still several cases.

• @fazlaRabbiMashrur: Is my answer not enough? I hope you can find what you want in the way I wrote... – mathlove Aug 6 '14 at 12:44
• at last my answer is $2^{4}*3*5*7$ isn't it correct? – Fazla Rabbi Mashrur Aug 7 '14 at 1:53
• @fazlaRabbiMashrur: I think the answer is $2^4\cdot 3\cdot 5\cdot 7=1680$. – mathlove Aug 7 '14 at 7:33

Hint: Consider first how to determine the number of distinct divisors of a number from its prime factorisation.

So any prime has two factors, namely $1$ and itself.

If you can do this, you will be able to find numbers with $40$ factors quite easily - but there will be a little work to do to find the smallest.

Stated a slightly different way, consider the function d(n), which is the gives the number of divisors for integer n. d(n) is multiplicative, in that if the prime decomposition of n is:

$$n = {p_1}^{q_1} * {p_1}^{q_2} \cdots \cdot {p_k}^{q_k}\ \ p_i\ \text{are primes}$$

The number of divisors in n is the product of the number of divisors in each factor of n.

$$d(n) = d({p_1}^{q_1}) * d({p_1}^{q_2}) \cdots \cdot d({p_k}^{q_k})\ \ p_i\ \text{are primes}$$

Remembering that the number of divisors in ${p_1}^{q_i}$ is $q_i + 1$ (since it includes 1 and itself) this should be a good start, as mathlove stated. You're looking for the product to be 40, so you have several combinations to try out (such as 4*10, 8*5, 5*4*2, etc).

Good luck!