.

So if I'm right, one solution for all equations is ```x = !128/!8-5``` as the right side is always 5 smaller then the divider on the right side and the equations cold be rewritten as ```x+5 mod d = 0``` so 9 to 128 have to divide x+5. When multiplying all the dividers together, the result can be divided by all of them and is exactly 5 bigger than x.
But this obviously is not the smallest number, because many numbers from 9 to 128 are multiples (e.g. 9 and 18) so we only need to multiply with the biggest multiple below 129.
My program simply tests if x is a multiple of the divider. If it's not, it multiplies x to the divider making it a divider. This procedure is repeated with dividers from 128 to 9 downwards.
```
x = 1
while True:
    for i in range(128, 8, -1):
        if x % i == 0:
            pass
        else:
            x *= i
            break
    else:
        x -= 5
        break
```
So ```x = 3692046484038964353473548580043636599281147520791807426559995```


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To support your work, I also upvoted your post!


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I am a bit late but here is my solution. The equations can written like x mod 9=-5, x mod 10=-5,
x mod 11=-5...
So we just need to look for the least common multiple of 9 10 ... 128 and subtract 5. That can be done for example by iteratively computing the lcm of 2 numbers I think . My answer is 13353756090997411579403749204440236542538872688049071995
The python code is below
```python
def gcd(x, y): 
  
   while(y): 
       x, y = y, x % y 
  
   return x 
a = []  
for x in range(9,129):
    a.append(x)
lcm=1
for x in a:
    lcm=int(abs(lcm*x))//int(gcd(lcm,x))
    print(lcm%x,lcm)

for x in range(9,129):
    if((lcm-5)%x==x-5):
        pass
    else:
        print("Nope")
print(lcm-5)