In [2]:

```
def egyptian_divmod(dividend, divisor):
assert divisor != 0
pwrs, dbls = [1], [divisor]
while dbls[-1] <= dividend:
pwrs.append(pwrs[-1] * 2)
dbls.append(pwrs[-1] * divisor)
ans, accum = 0, 0
for pwr, dbl in zip(pwrs[-2::-1], dbls[-2::-1]):
if accum + dbl <= dividend:
accum += dbl
ans += pwr
return ans, abs(accum - dividend)
# Test it gives the same results as the divmod built-in
from itertools import product
for i, j in product(range(13), range(1, 13)):
assert egyptian_divmod(i, j) == divmod(i, j)
# Mandated result
i, j = 580, 34
print(f'{i} divided by {j} using the Egyption method is %i remainder %i'
% egyptian_divmod(i, j))
```

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