How to find an $n$-digit number with $m$ number of factors?

Hey Vinayak, here is an explanation of the formula for the number of factors: Let $n = p_1^{m_1} \cdot p_2^{m_2} \cdots$. Every factor of $n$ is a product of the prime numbers risen to some power: $f = p_1^{n_1} \cdot p_2^{n_2} \cdots$ with $0 \leq n_1 \leq m_1, 0 \leq n_2 \leq m_2 ...$. For example let $n = 12 = 2^2 \cdot 3^1$. All the factors of 12 are:

$2^0 \cdot 3^0 = 1$

$2^1 \cdot 3^0 = 2$

$2^2 \cdot 3^0 = 4$

$2^0 \cdot 3^1 = 3$

$2^1 \cdot 3^1 = 6$

$2^2 \cdot 3^1 = 12$

As you can see, for each power of the prime numer $p_n$ there are $m_n + 1$ choices $0, 1, ... m_n$. Therefore, the number of factors is $(m_1 + 1) \cdot (m_2+1) \cdots$. I hope this helps!

Interestingly, although there are 3,013 six-digit numbers with 28 factors, there are some numbers of factors for which there is only one six-digit number with that many factors. These are, I believe, 7, 13, 19, 38, each of which has only one six-digit number with that many factors. (edit: there are probably a lot more solo numbers of factors >50 but I aggregated everything with 50 or more factors to save on processor time.)

A factor $f$ of a number $n = p_{1}^{m_{1}} \cdot p_{2}^{m_{2}} \cdot p_{3}^{m_{3}} \ldots$ must be a number with prime exponents which are less than or equal to the exponents of the number. To make things simple, here is an example: $60 = 2^2 \cdot 3^1 \cdot 5^1$. Without Brute-forcing, we can think logically here. If a factor $f$ can be represented here, say 15 which is equal to $3^1 \cdot 5^1$, the factor's prime exponents will be less than or equal to the prime exponents of $n$. In this case, they are equal

Taking this a step further, for each prime exponent say $m_{1}$, there will be $m_{1} + 1$ options to choose for a power. For the number 60 and specifically prime power of 2, we have $2^{m_{1}}, m_{1} \in \{0, 1, 2\}$ so there are 3 options.

Similarly each of the $m_{i^{th}}$ exponent will have $m_{i} + 1$ options.

The total options will thus equal the product of all $= (m_{1} + 1) \cdot (m_{2} + 1) \cdot (m_{2} + 1) \ldots$

I think it's possible that I don't understand the question. (edit: Narrator: "He didn't understand the question.")

The lowest composite number I can generate with $2$ distinct factors is $2 \times 3 = 6 ( = 3! )$

The lowest composite number I can generate with $3$ distinct factors is $2 \times 3 \times 4 = 24 ( = 4! )$

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The lowest composite number I can generate with $n$ distinct factors is $(n+1)!$

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The lowest composite number I can generate with $28$ distinct factors is $29! = 8841761993739701954543616000000$. This has a lot more than 6 digits. What is it that I don't understand here?

(Additional, just for laughs, if I insist that the $28$ factors are prime factors, I get: $2566376117594999414479597815340071648394470$.)

125000?

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@Vinayak Srivastava

Great Question, But yes has way too many solutions.

@Vinayak Srivastava,if product of factors are given,then a unique solution can be determined.

Task: Find the least number n that can we represented as a product n = a ∙ b in k (1 ≤ k ≤ 50) ways. Products a ∙ b and b ∙ a are the same, all numbers are positive integers.

This isn't the same problem. For example 4 has 3 divisors(1,2,4), but we can write this in only two ways : 1x4 and 2x2.

The result is:

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1   1
2   4
3   12
4   24
5   36
6   60
7   192
8   120
9   180
10  240
11  576
12  360
13  1296
14  900
15  720
16  840
17  9216
18  1260
19  786432
20  1680
21  2880
22  15360
23  3600
24  2520
25  6480
26  61440
27  6300
28  6720
29  2359296
30  5040
31  3221225472
32  7560
33  46080
34  983040
35  25920
36  10080
37  206158430208
38  32400
39  184320
40  15120
41  44100
42  20160
43  5308416
44  107520
45  25200
46  2985984
47  9663676416
48  30240
49  233280
50  45360