Question: "Find x for 0≤x≤100,000."

My clues are as following. I'm not sure whether there are multiple answers or that only one 'x' remains based on these clues:

* This number isn’t between 20,000 and 40,000 or 60,000 and 80,000.
* This number is divisible by three values between 10 and 50.
* The sum of the prime factors of this number is 53. (20 would be
2+2+5=9).
* Digits multiply up to 0. (So one of the digits is 0)
* This number is odd.

Answer: 
Concerning the 2nd statement, it says that the number is divisible by three values between 10 and 50. But is it exactly 3 values?

If not, then to date I found 3 such numbers exist,namely $5083$, $4807$ and $85085$.

For $5083$, 

**1st statement is valid as** $0<5083<20000$ 

**2nd statement is valid as** $5083=13\cdot17\cdot23$

**3rd statement is valid as** $13+17+23=53$

**4th statement is valid as** $5\cdot0\cdot8\cdot3=0$

**5th statement is valid as** $5083$ is odd.

And similiar for $4807$

Similiar for $85085$,

**2nd statement is valid as** $85085=5\cdot7\cdot11\cdot13\cdot17$ (if more than 3 prime factors are allowed)

**3rd statement is valid as** $5+7+11+13+17=53$

For statement 1,4,5, you can easily see that their validity.

---

**Methodology**

As what I've typed in the comment section, that the sum of prime number is 53 have those combinations mentioned. Among the 3-prime-factor-sum, only 3 of them can constitute 3 factors which is between 10 and 50. They are $(11,13,29)(11,19,23)(13,17,23)$. And their product are $4147,4807$ and $5083$ respectively. And $4147$ is not the answer as it doesn't have a "0".

Among the 4-prime-factor-sum, one can discover all these combinations consist of a "2". But then products of these prime factors,that are, the number to be found, are even, and not odd, so they are excluded from consideration. Fortunately, for the combination $(5,7,11,13,17)$ does not consist of a "2" and can fit all the requirements.
One thing seems not to be particularly clear to me. At the second point, is it allowed that a number can be divisible by more then 3 numbers between 10 and 50? I assume this is the case. I didn't found an easy solution to this problem, so here is a little Python program, which does the job:

```
#!/usr/bin/env python
from itertools import combinations

p = []
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53]

# Find all the prime factorisations, who sum up to 53:
tup = [comb for i in range(1,20) for comb in combinations(primes, i) if sum(comb) == 53]

# Multiply the elements in the primefactorisations, so that we get a list of
# feasable numbers:
for elem in tup:
    x = 1
    for d in elem:
        x *= d
    p.append(x)

# Remove the elements, which are not in the specified range, or are even:
pc = p[:]
for x in pc:
    if (x > 20000 and x < 40000) or (x > 60000 and x < 80000) or x % 2 == 0:
        p.remove(x)

# Remove the elements, which do not contain the digit 0:
pc = p[:]
p = []
for x in pc:
    ch = str(x)
    for c in ch:
        if (c == '0'):
            p.append(x)
            break

# Find the number, which is divisable by at least three values between 10 and 50:
pc = p[:]
p = []
for x in pc:
    c = 0
    for i in range(11,51,2):
        if x % i == 0:
           c += 1

    if c >= 3:
        p.append(x)

# Print the result:
print(p)

```

If you let this program run, it will display the following result:

```
[4807, 5083, 57057, 85085]

```