Answer to https://fivethirtyeight.com/features/can-you-find-a-number-worth-its-weight-in-letters/

riddler.py

#!/usr/bin/env python3

singles = {
    0: "", # This is wrong for "zero", but is right for everything else.
    1: "one",
    2: "two",
    3: "three",
    4: "four",
    5: "five",
    6: "six",
    7: "seven",
    8: "eight",
    9: "nine",
    10: "ten",
    11: "eleven",
    12: "twelve",
    13: "thirteen",
    14: "fourteen",
    15: "fifteen",
    16: "sixteen",
    17: "seventeen",
    18: "eighteen",
    19: "nineteen",
}

tens = {
    2: "twenty",
    3: "thirty",
    4: "forty",
    5: "fifty",
    6: "sixty",
    7: "seventy",
    8: "eighty",
    9: "ninety",
}

large = [
    (10**3,  10**2, "hundred"),
    (10**6,  10**3, "thousand"),
    (10**9,  10**6, "million"),
    (10**12, 10**9, "billion"),
]

def word(n):
    if n in singles:
        return singles[n]

    if n < 100:
        d = n // 10
        r = n % 10
        return tens[d] + word(r)

    for (limit, step, name) in large:
        if n < limit:
            d = n // step
            r = n % step
            return word(d) + name + word(r)
    
    raise RuntimeError("Number %d out of range" % n)

def weight(n):
    return sum(ord(c) - ord('a') + 1 for c in word(n))

## Test cases
assert(weight(1) == 34)
assert(weight(2) == 58)
assert(weight(1417) == 379)
assert(weight(3140275) == 718)

# We need the overall delta to be positive
delta = lambda n : weight(n) - n

# Adding anything "two hundred" or greater is a net loss
for i in range(200, 10000, 100):
    assert(delta(i) < 0)

# The loss from adding a "one thousand" (or anything higher) offsets any gain
# from lower numbers < 200.
min_loss = min([-delta(i) for i in range(1000,10000,1000)])
max_gain = max([delta(i) for i in range(0, 200)])
assert(min_loss > max_gain)

# Get the answer
i_max = 0
for i in range(0,1000):
    d = delta(i)
    if d > 0:
        i_max = i
print(i_max)