**Part 1:** *P(n)* is defined such that for each number *i*, add *10i* to any number divisible by *i*. Find the first value *n* such that *P(n)* is at least a given target number.

Let’s throw some memory (and numpy) at it:

```
target = int(sys.argv[1])
presents = numpy.zeros(target)
for i in range(1, target):
presents[i::i] += 10 * i
for i in range(len(presents)):
if presents[i] >= target:
print(i)
sys.exit(0)
```

It barely makes it in under a minute, but it does. You can speed it up even more if you guess on where the answer will be an initialize to only the first `numpy.zeros(target / 10)`

. In only shaves off about ^{1}⁄_{6} of the time on my run though, so I’m not sure it’s worth it.

**Part 2:** Do the same thing, only use *11i* instead of *10i* but only to the first 50 multiples.

Nothing much changes:

```
target = int(sys.argv[1])
presents = numpy.zeros(target / 10)
for i in range(1, target):
presents[i:i*50:i] += 11 * i
for i in range(len(presents)):
if presents[i] >= target:
print(i)
sys.exit(0)
```