## Problem 32 – Pandigital Products

October 30, 2009 by

We say that $n$ is a pandigital product if $n=a\cdot b$ and the string “abn” obtained by concatenating the base-10 representations of $a$, $b$, and $n$ contains all of the digits 1-9 exactly once (with no 0s). In problem 32 we are asked to find the sum of all pandigital products.

I ended up being not particularly fancy about this problem. Loop through enough values for $n$, and, for each $n$, loop through enough values for $a$, and see if you get a pandigital product with $n$, $a$, and $n/a$. We know that it is enough for $a$ to venture up to $\sqrt{n}$, but what about bounds on $n$? If $n$ has 5 digits, then to be a pandigital product, with factors $a$ and $b$, the factors must be either 1 and 3 digits, or 2 and 2 digits. But 9*999 and 99*99 aren’t 5 digit numbers, so we know we can stop looking for $n$ once we get to 10000. Similar reasoning shows that $n$ must have more than 3 digits, and so, in fact, to be a pandigital product, $n$ must have 4 digits (and since they must not be 0, and must be distinct, 1234 is the smallest possibility).

The following code seems to get the job done:

def ispandigprod(a,b,n):
""" assumes a*b==n """
# consider the digits in the string "abn"
# remove 0s and duplicates (with filter, set)
# if you've still got 9 elements, you're pandigital
return len(set(filter(lambda t:t, map(int, "%s%s%s" % (a,b,n))))) == 9

def solve():
""" return the sum of the pandigital products """
ret = 0
for n in xrange(1234,9876+1):
a = 1
included = False
while(a < n**.5 + 1 and not included):
if(n % a == 0):
b = n//a
if(ispandigprod(a,b,n)):
ret += n
included = True
a += 1
return ret


I’d honestly like a nicer solution for this problem. The loop I have is pretty stupid, and loops through far more choices than is worthwhile. In particular, I’d like to not think about any $n$ or $a$ that have a digit that is 0. To ignore $n$ with a 0 digit, one could replace the “included = False” line (apparently line 10) with

  included = str(n).count("0") > 0


This seems to cut the run-time down by about 20% (or more), so is maybe worthwhile. I still expect this code could be improved by a bit, but I think I’ll stop here for now.

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## Current Problems

October 25, 2009 by

To date, there has been a widget in the right-hand column with the “current problems”. The original intention here was to have two per week. Of course, the original intention also involved some other authors.

I’m going to remove the “current problems” widget. I’ll still hope to do at least one problem per week.

## Problem 31 – Coins

October 19, 2009 by

In problem 31, we are given a list of coin values and a target value, and asked how many ways we can make that target value with coins of the given denominations. I think this is probably a pretty standard programming exercise, and I’m embarassed it took me as long as it did. Anyway, here’s what I came up with, nothing too special:

import sys, time

def worker(target, coins):
""" assume coins are sorted decreasing """
if target == 0:
return 1 # there's only one way to not have money?

# get rid of coins bigger than the target
while(len(coins) and target < coins[0]):
coins = coins[1:]

# some easy cases
if not len(coins): # all coins are > target
return 0
if len(coins) == 1:
return (target % coins[0] == 0) and 1 or 0

# there are >=2 types of coins, all <= target
return sum([worker(target - v, coins[1:])
for v in xrange(0, target + 1, coins[0])])

def solve(target, coins):
t = time.time()
coins.sort()
coins.reverse()
ret = worker(target, coins)
t = time.time() - t
return [ret, t]

if __name__ == "__main__":
print solve(int(sys.argv[1]), map(int,sys.argv[2:]))


I also had fun trying this another way. Instead of breaking down the target amount and doing things recursively, I thought about having a huge amount of coins around. The example problem had a target of 200, with denominations 1, 2, 5, 10, 20, 50, 100, 200. Suppose I had 200/1=200 of the 1s, 200/2 = 100 of the 2s, 200/5 of the 5s, etc. I could then take any number of my 1s, along with any number of my 2s, etc, see how much I have, and see if the total is 200. So I’ve got a product space $[0,200/1]\times [0,200/2]\times [0,200/5]\times\cdots\times [0,200/200]$ of possible combinations, and I want to pick out all those that have a value of 200. This presents a fun application of the map/reduce functions, which I can’t seem to get enough of.

Suppose I start with the list of coins available, [1,2,5,10,20,50,100,200], and my target value in mind. I want to replace each item in this list, say the value $c$, with the list of values I could make using only $c$-valued coins (and only enough to get to 200). I’m going to do something a little silly, and store each value as it’s own list (this’ll make the next step easier, I think). So I first do

target = 200
coins = [1, 2, 5, 10, 20, 50, 100, 200]
var = map(lambda c: [ [v] for v in xrange(0,target+1,c)], coins)


Next, I want to make the cartesian product of all of these list of lists. I’ll build up my product, working my way through var. That is, I think about $L_1\times L_2\times \cdots \times L_k$ as $(((L_1\times L_2)\times L_3)\times \cdots )\times L_k$, where each $L_i$ here is a list. This is the sort of thing reduce is for. Keeping in mind that all of my lists are lists of one-element lists (so “+” below is really “concatenate”), I can build the cartesian product via

var = reduce(lambda l,r: [ x+y for x in l for y in r ], var)


Now each element of var is a list with as many values as there are coins, e.g. [7, 6, 0, 40] corresponds to 7 in 1s, 6 in 2s, 0 in 5s, 40 in 10s (of course, in the running example, I need more values in the list, but I hope the point comes across). To find the combined value of the coins in each list, I just need the sum, so I do

var = map(sum, var)


Finally, I only want the values where the sum is my target value, so I filter. We’re asked for the number of ways to make the target value, so I take the length:

var = filter(lambda t: t == target, var)
print len(var)


This is not particularly efficient. In the stated problem, the product of my lists has 1.28 billion elements (significantly larger than the actual answer, which is less than 100,000). That’s rather a lot of space to be taking up. I tried running this code on a target value of 100, and gave up on it after about 20 minutes. For contrast, the first solution above, the recursive solution, takes about half a second on my computer. Still, I like this exercise of getting the cartesian product of a bunch of lists.

## Problem 30 – Sums of Digit Powers

October 1, 2009 by

In problem 30 we are asked to find all the numbers that are the sum of the $n$-th powers of their digits. Well, the sum of those numbers. And $n=5$. You can just about do it in one line…

Most of this problem is fairly easy: Given a number, compute the sum of the powers of its digits, and see if you get the number you started with. If you wanna find all such numbers, just loop for a while. The trouble is how long? If you’ve got some range in mind you want to look at, you can do the following:

def solve(pow, low, hi):
""" Find the sum of all numbers that are the sum of their pow-th powers of digits in the interval [low,hi) """
return sum(filter(lambda n:n==sum(map(lambda d:int(d)**pow,str(n))) xrange(low, hi)))


Ok, sure, it’s a longish line, but it’s all there. Python’s syntactic sugar for handling all the little looping is awesome.

The remaining trick is to figure out those bounds. The problem mentions not using 1 as one of the numbers, so we could start at 2. However, that means we could start at 2**pow, because that’ll be the lowest digit sum we’d make. On the high end, the biggest digit-power-sum will come from digits being 9. So I figure that to find an upper bound, you can think about digit sums for longer and longer strings of 9. If you think about $k$ 9s in a row, that’ll give a digit sum of $k\cdot 9^p$ (if $p$ is the specified power). Find the largest $k$ where this value is still a $k$-digit number (or bigger), and you’ve got yourself an upper bound to loop to. Here’s some code to find the bound:

def bound(pow):
""" Find bound on nums that can be sum of their pow-th powers of digits """
if pow <= 1:
return 9
nine = 9**pow
coeff = 1
while(len(str(coeff*nine)) >= coeff):
coeff += 1
return (coeff - 1)*nine


Now, these bounds I’ve set up are pretty poor. In the case where the power is 4, as in the example problem, we’d loop from 2**4=16 to 32805, when the actual numbers that have the property we’re hoping for are between 1000 and 10000. We’ve overshot by something like 3 times too many values.

So, we could try to be more clever and come up with a smaller set to search in. But we could still use that one-line solver to pick from that set the appropriate values. And that makes me happy.

## Problem 29 – One Liner

September 29, 2009 by

In problem 29 we are supposed to determine how many unique values there are for the value $a^b$ where $a_0\leq a\leq a_1$, $b_0\leq b\leq b_1$ (all integers). I know there is clever counting one can do to determine this, but I thought for comparison I’d also just do up the brute force approach. Find all the values, get rid of duplicates, and count how many are left.

Python has a built-in type “set”, which will eliminate duplicates as I go. And you can ask for the length of the set. And list comprehension is awesome. So you can solve this in one line:

def solve(lowa, hia, lowb, hib):
return len(set([a**b for a in xrange(lowa, hia+1) for b in xrange(lowb, hib+1)]))


Having wrapped up solution code into a function, I also like to use command line arguments to pass in… well, arguments. In this case, I’d like to allow two different options for command line options. The stated problem has the bounds 2 to 100 for both $a$ and $b$. So if I take one number in, I want to assume it is the upper bound, and that 2 is the lower bound, for both variables. Alternatively, I could take all four bounds in as arguments. I know the following is crude, and doesn’t fail gracefully, but it’s got a lot of nice things going for it:

if __name__ == "__main__":
if len(sys.argv) == 2:
print solve(*(2*[2,int(sys.argv[1])]))
else:
print solve(*map(int, sys.argv[1:]))


I love this. Using * to change a list into the form my solve function expects, with 4 arguments… that’s awesome. Using [1:] to grab the arguments besides the called command… nice. 2*list to do a list concatenated with itself… delightful. And, of course, map. Who could forget map?

Anyway, I may return to the more mathematical, counting based solution for this. At the same time, though, this code is quick (to write and run (for the stated bounds)) and easily checked. Maybe I’ll leave the more mathematical solution to the other authors on this blog. I seem to recall that there were some…

## Problem 27 – Prime Generating Quadratics

September 27, 2009 by

With my freshly minted PrimeOracle, I’m ready to attack problem 27, in which we are asked to find polynomials that generate long strings of consecutive primes starting when you plug in 0.

Basically my idea for this problem was to iterate through all of the choices for the coefficients $a$ and $b$ in $n^2+an+b$, and then for each polynomial plug in values until we hit a composite. Record maximums as appropriate.

There are a few easy optimizations to make. First, $b$ itself has to be prime, because we test $n=0$. Next we test $n=1$, so we need $1+a+b\geq 2$, that is $a\geq 1-b$. Those are the only optimizations I made, though I’m sure there are some more advanced ones out there.

So here’s my code:

# assume PrimeOracle.py is in the same directory as this file
from PrimeOracle import PrimeOracle

def solve(n):
# seed our oracle to 2*n+2, since n=1 gives 1+a+b
isprime = PrimeOracle(2*n + 2)
maxp, maxn = 41, 40
for b in [m for m in xrange(2, n) if isprime[m] ]:
for a in xrange(1-b, n):
idx = 1 # we already tesetd 0 with isprime[b]
while(isprime[idx*idx+idx*a+b]):
idx += 1
if(idx - 1 > maxn):
maxp, maxn = a*b, idx - 1
return maxp


I had my oracle also tell me what the biggest number it was asked about was, and the answer was up around 12000. So we could change the starting seed in the oracle to be big enough to never have to extend, and probably save some time.

## Prime Oracle

September 27, 2009 by

I sat down to work on problem 27, and realized it might be handy to have a class that’ll tell me when numbers are prime. Something that’ll keep some memory of tested values, and sieve when it needs to know about bigger numbers than have been tested. Here’s what I threw together:

""" A class that'll help you decide if things are prime

Make an object of this class, I'll call it oracle, and
oracle[n] is True or False depending on if n is prime or not

Assumes negatives are not prime
"""

class PrimeOracle:
# Takes a starting maximum to sieve to
def __init__(self, startmax = 2000):
# self.isprime[n] = True if 2n+1 is prime
self.isprime = [True for n in xrange(0,startmax//2 + 1)]
self.isprime[0] = False
# run the sieve
for val in xrange(3, startmax + 1, 2):
if(self.isprime[val // 2]):
for mul in xrange(val*val, startmax + 1, 2*val):
self.isprime[mul // 2] = False

# is n prime?
def __getitem__(self, n):
if n == 2:
return True
if n<2 or n % 2 == 0:
return False
nidx = n//2
while(nidx >= len(self.isprime)):
# extend the sieve
# currently, we know the primality of 2*(len-1)+1=2*len-1
# we can extend that to its square, 4*len^2 - 4len + 1
# which would be index (4*len^2-4*len+1) // 2 = 4len(len-1)+1 // 2
# so the new length should be one more than this
# we already have length len, so we need to add (prev line) - len
curlen = len(self.isprime)
newlen = (4*curlen*(curlen-1) + 1) // 2
self.isprime += [True for n in xrange(0, newlen+1-curlen)]
maxknown = 2*curlen - 1
maxtoknow = 2*newlen - 1
for idx in xrange(1, curlen):
if(self.isprime[idx]):
val = 2*idx + 1
# start crossing off multiples larger than maxknown
sfact = 1 + (maxknown // val)
if(sfact % 2 == 0):
sfact += 1 # otherwise our indexing is wrong
startfact = max(val, sfact)
for mul in xrange(startfact * val, maxtoknow + 1, 2*val):
self.isprime[mul // 2] = False
return self.isprime[nidx]

# the largest index we know about
def __len__(self):
return 2*len(self.isprime) - 1

def __str__(self):
return "\n".join(["%d: %s" % (2*n+1, self.isprime[n])
for n in xrange(0,len(self.isprime))])


It’s basically just the usual sieve, and then it extends itself whenever necessary.

There’s room for improvement. We could remember the highest value we’ve sieved to, and then take our array and just tease out the primes from it. Then we only store the list of primes. When we go to extend, we make a new array to hold whatever the new values would be, sieve appropriately, and then just append to our list of primes. Here’s my modified __getitem__ method:

def __getitem__(self, n):
if n == 2:
return True
if n<2 or n % 2 == 0:
return False
while(n > self.sievedto):
# we have sieved to sst = self.sievedto (assume odd)
# we could sieve to sst^2 (will also be odd)
# that'd be [sst+1 .. sst^2]
# but really we might as well do [sst+2 .. sst^2]
# sst=2s+1 => [2s+3 .. 4s^2+4s+1] //2 is [s+1 .. 2s(s+1)]
# which has length 2s^2+s
# and index i corresponds to 2(s+1+i)+1
# so that an odd integer m is at m//2 - (s+1)
s = self.sievedto // 2
newsst = self.sievedto * self.sievedto
isprime = [True for idx in xrange(0,2*s*s+s+1)]
for pr in self.primes.keys():
sfact = 1 + (self.sievedto // pr)
if(sfact % 2 == 0):
sfact += 1
for mul in xrange(max(pr, sfact) * pr, newsst + 1, 2 * pr):
isprime[mul//2 - (s+1)] = False
for m in [2*(s+1+i)+1 for i in xrange(0,2*s*s+s+1) if isprime[i]]:
self.primes[m] = True
self.sievedto = newsst
return self.primes.has_key(n)


My first guess was that the second version would be better, because it would be storing less. However, it seems to be a little slower. I guess filtering out the primes at each extension has its cost. Or my implementation could use some tweaking? I wish I knew a good way to compare the size requirements of these scripts, in addition to how long they take to run.

By the way, I would be a little shocked to find that I’m not off by one somewhere (many somewheres) in the above code. Little tests seem to work out. But use are your own peril.

## Problem 28 – Spirals

September 27, 2009 by

In problem 28 we construct a spiral starting with 1 in the center, and build our way outwards clockwise. Then we are supposed to find the sum of the diagonals.

I figured there should be some formula for the sum of the diagonal entries, depending on the size of the grid you want to stop at. If we only compute diagonal elements when we have a square, the square will always have sides of odd length. So we should take a number $n$, and compute the appropriate sum of the elements in the $2n+1$ by $2n+1$ grid. I’ll write down the sum for the case $n=3$ in the following manner:

  1^2 + (1^2+2) + (1^2+2*2) + (1^2+3*2)
+ 3^2 + (3^2+4) + (3^2+2*4) + (3^2+3*4)
+ 5^2 + (5^2+6) + (5^2+2*6) + (5^2+3*6)
+ 7^2


That seems to make a pattern pretty clear. We’re adding up odd squares (the rising diagonal coming off of the central 1) and then some offsets of those squares. The formula for the sum of squares is

$\displaystyle \sum_{i=1}^{m} i^2 = \frac{m(m+1)(2m+1)}{6}$

and we can write the sum of odd squares as

$\displaystyle \sum_{k=1}^{m} (2k-1)^2 = \sum_{k=1}^{2m}k^2 - \sum_{k=1}^{m} (2k)^2 = \sum_{k=1}^{2m} k^2 - 4\sum_{k=1}^{m}k^2.$

Pushing some symbols around, I got down to the following formula for the sum of the diagonal elements in a $2n+1$ by $2n+1$ square:

$\frac{16}{3}n^3 + 10n^2 + \frac{26}{3}n + 1.$

Of course, those thirds make me nervous. The answer is clearly supposed to be an integer. But it’s ok. Gathering up those thirds we get

$\frac{2}{3}n(8n^2+13)+(10n^2+1)$

for the sum. Now, if $n$ is divisible by 3, we’ll get an integer in that first term. And if $n$ is not divisible by 3, it is either 1 or 2 mod 3, and so $8n^2+13\equiv -1(1)+1=0\pmod{3}$. So that third isn’t going to cause problems.

It’s pretty easy to code up the solution when you’ve got such a formula. By using an if-switch, I took into account the mod-3 thing above so that I was never working with floats. Here’s my code:

def solve(n):
ret = 10*n*n + 1
factor = 8*n*n+13
if(n % 3 == 0):
ret += 2*(n//3)*factor
else:
ret += 2*n*(factor//3)
return ret


Then I decided it would be a fun little exercise to actually cook up some code to generate the spiral. I knew this would have slower run-time, and require more resources, but I wanted to write down the code anyway.

The way I thought about it, I’ll keep track of my current coordinates $(posx,posy)$, starting at $(0,0)$. To go to the next place, I look to my right, and see if that spot is filled. If it is, I just continue on in whichever direction I was going; otherwise, I move to that spot to my right. This means I need to keep track of which direction I’m facing. There are only 4 directions, and they cycle, so I’ll think of direction as an integer mod 4. If I make 0,1,2,3 correspond to Up, Right, Down, Left, then (my direction)+1 is the direction to my right. After writing down a little grid of input directions, and what they meant about where to move next, I came up some little formulas, and was good to go.

I had originally thought I’d wrap up an implementation into a class, so that you could index via $[x][y]$, but then I realized that wouldn’t quite work, and you’d need to just index as $[(x,y)]$. At that point, I figured I might as well just use a dictionary type. And away I went:

def dx(dir):
""" interpret dir as 0,1,2,3 = N,E,S,W, return what dx is in that dir """
return (dir%2) * (2-dir)

def dy(dir):
""" dir as 0,1,2,3 = N,E,S,W return what dy is in that dir """
return ((dir+1)%2) * (1-dir)

def solve(n):
""" Build up a 2n+1 by 2n+1 grid, then sum diagonals """
grid = {}
posx, posy = 0, 0
idx = 1
dir = 0 # dir in 0,1,2,3 = N,E,S,W
while(posx <= n):
grid[(posx,posy)] = idx
idx += 1
# find the position to my right
rsposx,rsposy = posx + dx((dir+1)%4), posy + dy((dir+1)%4)
if(grid.has_key((rsposx,rsposy))):
# keep going the direction I was going
posx,posy = posx + dx(dir), posy + dy(dir)
else:
# turn right
posx,posy = rsposx, rsposy
dir = (dir+1)%4
# sum the diagonals
ret = sum([grid[(x,x)] for x in xrange(-n,n+1)])
ret += sum([grid[(x,-x)] for x in xrange(-n,n+1)])
ret -= 1 # the origin got counted twice
return ret


Only two days behind my artificial deadline too.

## Problem 25 – More Fibonacci Numbers

September 19, 2009 by

In problem 25 we are asked to find the first Fibonacci number with a given number of digits. Actually, we are only asked to find which Fibonacci number it is (its index in the sequence of Fibonacci numbers), instead of the number itself. We’ve already visited Fibonacci numbers in problem 2, and had useful work there. For this problem, it will be helpful to note that the number of digits of $n$ can be found as $\lfloor \log_{10} n\rfloor + 1$, where $\lfloor x\rfloor$ denotes the “floor” function, equal to the greatest integer less than $x$ (on a not-particularly-related note, this isn’t the formula I would have used a week ago, glad I got that cleared up).

I decided to do some performance comparisons with this problem, with a few different versions of code. [GRR. WordPress seems to not be doing the sourcecode “tag” like I thought it should below. Or I’m messing something up…]

In version 1, I’ll compute each Fibonacci number successively, using the recurrence relation, then calculate the number of digits until I’ve found a big enough number. My function looks like:

def solve(n):
ret = 1
prev, cur = 0, 1
while(cur &lt; 10**(n-1)):
prev, cur = cur, prev + cur
ret += 1
return ret

In the second version, I’ll compute the $n$-th Fibonacci number using the formula
$F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^2-\left(\frac{1-\sqrt{5}}{2}\right)^2\right)$
that was used last time we saw Fibonacci numbers. In fact, I’ll wrap that up into its own function, “digs_fib” that takes in an “n” and gives the number of digits of the appropriate Fibonacci number. Now my code is basically:
def fib(n):
""" The nth fibonacci number """
alpha = (1 + 5**.5) / 2
beta = (1 - 5**.5) / 2
return (1/5**.5) * (alpha**n - beta**n)
def digs_fib(n):
""" Number of digits of fib(n) """
return int(math.log(fib(n), 10)) + 1
def solve(n):
ret = 1
while(digs_fib(ret) &lt; n):
ret += 1
return ret

Of course, that second term in the formula is smallish, and raised to big powers it gets smaller, so we could maybe ignore it. And then we’re taking the log of products and powers, so we can use log rules to re-write the expression for the number of digits, without directly calculating the Fibonacci number. My “digs_fib” function changes to the following:
def digs_fib(n):
""" The approximate number of digits of the nth fibonum """
alpha = (1+5**.5)/2
return int(n*math.log(alpha,10)-.5*math.log(5, 10))+1

Finally, given a target number of digits $d$, we could just about solve for $n$ in
$d \approx \lfloor n\log_{10}((1+\sqrt{5})/2) - \frac{1}{2}\log_{10}(5)\rfloor + 1$
and obtain
$n\approx \dfrac{d-1+\frac{1}{2}\log_{10}(5)}{\log_{10}((1+\sqrt{5})/2)}$.
This means that given $d$ we can get an approximation to which Fibonacci number gives us that many digits in basically constant time. Now my “solve” is just
def solve(n):
num = n - 1 + .5*math.log(5, 10)
num /= math.log((1+5**.5)/2, 10)
return int(num+1) # round up, approximately

I ran each of these through my little performance evaluation script and came up with a reasonable graph. I could only go up to asking for 300 digits, because after that the second version fails, since the numbers are too big. The other versions have no problem at all going up to 1000 digits (and well beyond). But anyway, here’s my graph:
You can barely see that constant time algorithm there along the $x$-axis. The scale of the $y$-axis is some linear change in actual time taken, but the shape of these graphs is what’s important anyway. Right?


## Problem 26 – Recurring Decimals

September 19, 2009 by

In problem 26, we are asked to find the integer less than a given bound whose reciprocal has the longest decimal period among such integers. That is, among all fractions $1/d$ where $d$ is less than some bound, which fraction has the longest periodic part in its decimal representation?

According to Hardy and Wright, An Introduction to the Theory of Numbers (6th ed, Theorem 135),

The decimal for a rational number $p/q$ between 0 and 1 is terminating or recurring, and any terminating or recurring decimal is equal to a rational number. If $(p,q)=1$, $q=2^{\alpha}5^{\beta}$, and $\max(\alpha,\beta)=\mu$, then the decimal terminates after $\mu$ digits. If $(p,q)=1$, $q=2^{\alpha}5^{\beta}Q$, where $Q>1$, $(Q,10)=1$, and $\nu$ is the order of 10 (mod $Q$), then the decimal contains $\mu$ non-recurring and $\nu$ recurring digits.

Let’s trim that down a bit. Our $p=1$, so automatically $(p,q)=1$ (this notation is for the greatest common divisor, as usual). If the denominator, $q$, is divisible by either 2 or 5, then either (1) the decimal terminates, or (2) the decimal repeats, with the same length recurring part as the reciprocal of a smaller integer. In either case, we will suppose that this $q$ is not the answer. Either (1) its recurring part has length 1, being a string of 0s, or (2) we could return the smaller integer with the same length recurring part.

So, suppose our $q$ is not divisible by 2 or 5. Then according to the theorem above, we may compute the length of the repeating part of the decimal for $1/q$ by finding the “order of 10 (mod $q$)”. That is, find the smallest $p$ so that $10^p\equiv 1\pmod{q}$. This is easy enough to find, just keep raising 10 to bigger powers, and taking the answer mod $q$, until you find 1.

Here’s my code

def solve(n):
""" Return the d with largest decimal period, d&lt;n """
max_d, max_p = 2, 1
for d in xrange(3,n,2):
if(d % 5 != 0):
# now find the order of 10 mod Q
# that is, the smallest p for which 10^p == 1 mod Q
p = 1
tmd = 10 % d # tmd = 10 mod d
while(tmd != 1):
p += 1
tmd = (tmd * 10) % d
if (p &gt; max_p):
max_d, max_p = d, p
return max_d

Pretty straightforward, I think. [Except, as of right now, WordPress doesn’t seem to be handling < and > correctly in that code. Grr]
There’s almost certainly more math we could do here. But this seems to work well enough for now.