Reply To: Programming

X = 29031

Here’s a python solution; I challenged myself to write it all in one line. (I don’t usually write unreadable code like this.)
print(next(filter(lambda x: all(x % (4, 5, 7, 11, 17)[i] == (3, 1, 2, 2, 12)[i] for i in range(5)), range(5000, 50000))))

I’ll break it down step by step.
First, I generate a list of numbers from 5000-50000 with range(5000, 50000).
Next, we check each number in turn. I check the remainder division of the number with each divisor we care about, and check it against the expected result.
That’s this section: all(x % (4, 5, 7, 11, 17)[i] == (3, 1, 2, 2, 12)[i] for i in range(5)). It uses list comprehension for efficiency.
We filter out each number from the list by using the in-built filter function: filter(lambda x: ...). Lambda simply creates an anonymous function for this purpose.
Finally, we print out the first (and only) number using print(next(...)). It is important to use the next keyword rather than indexing (e.g. list[0]), as we have created a filter object that isn’t subscriptable.

Manually checking the result:
29031 / 4 = 7257r3
29031 / 5 = 5806r1
29031 / 7 = 4147r2
29031 / 11 = 2639r2
29031 / 17 = 1707r12

Naturally, I also decided to write a one-liner for this:
print(list(divmod(29031, v) for v in (4, 5, 7, 11, 17)))