Isnochys Math Problem #5.1

in STEMGeeks3 months ago

Congratulations, @servelle and @enforcer48!
You send an answer, that was OK.
;)

3099044504245996706400

That is the calculated number I got.
How was this done?

i = 1
for k in (range(1, 51)):
    if i % k > 0:
        for j in range(1, 51):
            if (i*j) % k == 0:
                i *= j
                break
print(i)

We take k as a number going from 1 to 50.
I is a number, that we just need as check and printout later.
Now, k is slowly going up, one step each time the first for loop goes around.
The inner for loop is only reached, if k and that number i have a rest bigger 0 when you divide i by k.

Let's start with i=1.
k=1
i%k=0
k=2
i%k=1
j=1
(ij)=1
%k =1
j=2
(i
j)=2
%k=0
i=12
break
k=3
j=1,2 (i
j% k!=0)
i = 2*3
break
...
etc.

OOPH.
That one is a headscratcher, isn't it?
You need all prime numbers and mulitple of it, "other numbers" but not those, that could be already build out of the existing ones inside:))))

Don't worry, it will only get harder from now on!
:D

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Great explaination
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