# Project Euler – Problem 78 Solution

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#### Problem

Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can separated into piles in exactly seven different ways, so p(5)=7.

OOOOO

OOOO   O

OOO   OO

OOO   O   O

OO   OO   O

OO   O   O   O

O   O   O   O   O

Find the least value of n for which p(n) is divisible by one million.

#### Solution

```// the pentagonal numbers sequence, i.e. 1, 2, 5, 7, 12, 15, 22, ...
let pentagonalNumbers =
Seq.unfold (fun state -> Some(state, state+1)) 1
|> Seq.collect (fun n -> [n; -1*n])
|> Seq.map (fun n -> int(0.5 * double(n) * double(3 * n - 1)))

// the coefficients sequence, i.e. +, +, -, -, +, +, -, -, ...
let pCoefficients =
Seq.unfold (fun state -> Some(state, -1*state)) 1 |> Seq.collect (fun n -> [n; n])

// cache results to improve performance
let mutable cache = Array.init 100000 (fun n -> if n = 0 then 1I else 0I)

// define the function p using the pentagonal numbers
let rec p k =
if cache.[k] <> 0I then cache.[k]
else
let pSeq =
pentagonalNumbers
|> Seq.map (fun n -> k - n)
|> Seq.takeWhile (fun n -> n >= 0)
|> Seq.map p
let pk =
pCoefficients
|> Seq.zip pSeq
|> Seq.sumBy (fun (pk, coe) -> pk * bigint(coe))
cache.[k] <- pk
pk

Seq.unfold (fun state -> Some(state, state+1)) 1
|> Seq.filter (fun k -> (p k) % 1000000I = 0I)
```

Well, this one took some thought and some time to come up with a solution which runs under a minute! A brute force approach simply wouldn’t have worked here as p(k) becomes very large very quickly (p(200) = 3972999029388…). Thankfully, as the wiki page on partition points out, you can build a recurrence for the function p such that:

p(k) = p(k – 1) + p(k – 2) – p(k – 5) – p(k – 7) + p(k – 12) + p(k – 15) – p(k – 22) – …

using pentagonal numbers in the form of for n running over positive and negative integers (n = 1, –1, 2, –2, 3, –3, …), generating the sequence 1, 2, 5, 7, 12, 15, 22, … The signs in the summation continue to alternate +, +, –, –, +, +, …

Using dynamic programming technique, I’m caching the value of p(k) as they’re generated and in doing so, enabled this solution to run in under 30 seconds. Hi, I’m Yan. I’m an AWS Serverless Hero and the author of Production-Ready Serverless.

I specialise in rapidly transitioning teams to serverless and building production-ready services on AWS.

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