Is it possible to find a multiple knowing only: the count of its divisors, the upper limit and some of its divisors (more details)?

Question

In other words, say I am looking for multiple X

let:

X < 1000005

let the fist 18 divisors of X be: 1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 25 | 32 | 40 | 50 | 64 | 80 | 100 | 125 | 160 | 200

finally, I also know: X has exactly 49 divisors.

I will tell you what the answer is...frankly if you google it it will probably showed up...but again: is this possible knowing only the count/number of divisors and that x cannot be bigger than some number? thanks

Answer 1

If $X = \prod p_i^{k_i}$ then $X$ will have $\prod (k_i + 1)$ factors so having $\prod (k_i + 1) = 49$ factors means .....

Either $\prod (k_i + 1)= (k_1 + 1)(k_2 + 1)$ where $k_1=6$ or $k_2 = 6$ or $\prod(k_i + 1) =k_1 + 1= 49$ and $k_1 = 48$.

And that means

Either $X = p^6q^6$ for prime $p, q$ or $X=p^{48}$ for prime $p$.

So...

What can some of the prime factors be.

They must be

$5$ is a factor and $2$ is a factor so there must be at least two prime factors.

So that means $X = ....$

$X = 2^65^6$

So $X = .....$

$X = 1,000,000$

Answer 2

Easy answer using no number theory.

1) If those are the first factors then what are the prime factors?

Not $3$ or $7$ or $11$ etc. because those aren't listed. We have $2$ and $5$ and none others. If there are any more the must be prime higher than $200$. If $p > 200$ then $64*125*p$ is a factor. But $8000p > 8000*200 = 16000000 > 1000005$. So that's impossible.

So $2$ and $5$ are the only prime factors.

So $X = 2^a*5^b$ for some powers $a,b$.

2) Note $128 = 2^7$ is not a factor but $64 = 2^6$ is.

So $2^7$ is not a factor so $X = 2^6*5^b$.

3) If $m$ and $n$ are factors than so is $\gcd(m,n)$.

So can you name some more of the factors.

$1, 2,4, 8, 16, 32, 64$ and $5,10, 20, 40, 80, 160, 320$ and $25, 50, 100, 200, 400,800, 1600$ and $125, 250, 500, 1000, 2000, 4000, 8000$ are all factors

4) Those are $28$ factors and we need more. We can't get any more by higher powers of $2$. We can only get more by higher powers of $5$.

If $X = 2^6*5^b$ and $b \ge 4$ we will have $625, 1250, 2500, 5000, 10000,20000$ as factors but that is only $35$ factor.

If $x = 2^6*5^b$ and $b \ge 5$ how many more factors will we get?

And so on.