# Euler Problem 19: Counting Sundays — When does the week start?

Euler Problem 19 is so trivial it is almost not worth writing an article about. One interesting aspect of this problem is the naming of weekdays and deciding which day the week starts. This issue is more complex than it sounds because data science is in essence not about data but about people.

## Euler Problem 19 Definition

• 1 Jan 1900 was a Monday.
• Thirty days has September, April, June and November.
• All the rest have thirty-one,
• Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

## Solution

The problem can be quickly solved with R base code and a tiny bit faster when using the lubridate package.

# Base R-code
dates <- seq.Date(as.Date("1901/01/01"), as.Date("2000/12/31"), "days")
days <- rep(1:7, length.out = length(dates))
answer <- sum(days[substr(dates, 9, 10) == "01"] == 1)

#Using Lubridate
library(lubridate, quietly = TRUE)
answer <- sum(wday(dates[substr(dates, 9, 10) == "01"]) == 1)


To draw out this post a little bit further I wrote some code to solve the problem without using the calendar functions in R.

week.day <- 0
for (y in 1901:2000) {
for (m in 1:12) {
max.day <- 31
if (m %in% c(4, 6, 9, 11)) max.day <- 30
# Leap years
if (m == 2) {
if (y %% 4 == 0 & y %% 100 != 0 | y %% 400 == 0) max.day <- 29
else max.day <- 28
}
for (d in 1:max.day) {
week.day <- week.day + 1
if (week.day == 8) week.day <- 1
if (week.day == 1 & d == 1) answer <- answer + 1
}
}
}


## Which day does the week start?

The only aspect remotely interesting about this problem is the conversion from weekdays to numbers. In R, Monday is considered day one, which makes sense in the Christian context of Western culture. Saturday and Sunday are the weekend, the two last days of the week so they are day 6 and 7. According to international standard ISO 8601, Monday is the first day of the week. Although this is the international standard, several countries, including the United States and Canada, consider Sunday to be the first day of the week.

The international standard is biased towards Christianity. The Christian or Western world marks Sunday as their day of rest and worship. Muslims refer to Friday as their day of rest and prayer. The Jewish calendar counts Saturday—the Sabbath—as the day of rest and worship. This idea is also shared by Seventh-Day Adventists.

this example shows that data science is not only about data: it is always about people and how they interpret the world.

# Euler Problem 18 & 67: Maximum Path Sums

An example of a pedigree. Source: Wikimedia.

Euler Problem 18 and 67 are exactly the same besides that the data set in the second version is larger than in the first one. In this post, I kill two Eulers with one code.

These problems deal with binary trees, which is a data structure where each node has two children. A practical example of a binary tree is a pedigree chart, where each person or animal has two parents, four grandparents and so on.

## Euler Problem 18 Definition

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

As there are only 16,384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

## Solution

This problem seeks a maximum path sum in a binary tree. The brute force method, as indicated in the problem definition, is a very inefficient way to solve this problem. The video visualises the quest for the maximum path, which takes eleven minutes of hypnotic animation.

A more efficient method is to define the maximum path layer by layer, starting at the bottom. The maximum sum of 2+8 or 2+5 is 10, the maximum sum of 4+5 or 4+9 is 13 and the last maximum sum is 15. These numbers are now placed in the next row. This process cycles until only one number is left. This algorithm solves the sample triangle in four steps:

Step 1:

3
7 4
2 4 6
8 5 9 3

Step 2:

3
7 4
10 13 15

Step 3:

3
20 19

Step 4:

23

In the code below, the data is triangle matrix. The variables rij (row) and kol (column) drive the search for the maximum path. The triangle for Euler Problem 18 is manually created and the triangle for Euler Problem 67 is read from the website.

path.sum <- function(triangle) {
for (rij in nrow(triangle):2) {
for (kol in 1:(ncol(triangle)-1)) {
triangle[rij - 1,kol] <- max(triangle[rij,kol:(kol + 1)]) + triangle[rij - 1, kol]
}
triangle[rij,] <- NA
}
return(max(triangle, na.rm = TRUE))
}

# Euler Problem 18
triangle <- matrix(ncol = 15, nrow = 15)
triangle[1,1] <- 75
triangle[2,1:2] <- c(95, 64)
triangle[3,1:3] <- c(17, 47, 82)
triangle[4,1:4] <- c(18, 35, 87, 10)
triangle[5,1:5] <- c(20, 04, 82, 47, 65)
triangle[6,1:6] <- c(19, 01, 23, 75, 03, 34)
triangle[7,1:7] <- c(88, 02, 77, 73, 07, 63, 67)
triangle[8,1:8] <- c(99, 65, 04, 28, 06, 16, 70, 92)
triangle[9,1:9] <- c(41, 41, 26, 56, 83, 40, 80, 70, 33)
triangle[10,1:10] <- c(41, 48, 72, 33, 47, 32, 37, 16, 94, 29)
triangle[11,1:11] <- c(53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14)
triangle[12,1:12] <- c(70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57)
triangle[13,1:13] <- c(91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48)
triangle[14,1:14] <- c(63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31)
triangle[15,1:15] <- c(04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23)



## Euler Problem 67

The solution for problem number 67 is exactly the same. The data is read directly from the Project Euler website.

# Euler Problem 67
triangle.67 <- matrix(nrow = 100, ncol = 100)
for (i in 1:100) {
triangle.67[i,1:i] <- as.numeric(unlist(strsplit(triangle.file[i,], " ")))
}


# Euler Problem 17: Number Letter Counts

Euler Problem 17 asks to count the letters in numbers written as words. This is a skill we all learnt in primary school mainly useful when writing cheques—to those that still use them.

Each language has its own rules for writing numbers. My native language Dutch has very different logic to English. Both Dutch and English use compound words after the number twelve.

Linguists have theorised this is evidence that early Germanic numbers were duodecimal. This factoid is supported by the importance of a “dozen” as a counting word and the twelve hours in the clock. There is even a Dozenal Society that promotes the use of a number system based on 12.

The English language changes the rules when reaching the number 21. While we say eight-teen in English, we do no say “one-twenty”. Dutch stays consistent and the last number is always spoken first. For example, 37 in English is “thirty-seven”, while in Dutch it is written as “zevenendertig” (seven and thirty).

## Euler Problem 17 Definition

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.

## Solution

The first piece of code provides a function that generates the words for numbers 1 to 999,999. This is more than the problem asks for, but it might be a useful function for another application. The last line concatenates all words together and removes the spaces.

numword.en <- function(x) { if (x > 999999) return("Error: Oustide my vocabulary")
# Vocabulary
single <- c("one", "two", "three", "four", "five", "six", "seven", "eight", "nine")
teens <- c( "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen")
tens <- c("ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety")
# Translation
numword.10 <- function (y) {
a <- y %% 100
if (a != 0) {
and <- ifelse(y > 100, "and", "")
if (a < 20)
return (c(and, c(single, teens)[a]))
else
return (c(and, tens[floor(a / 10)], single[a %% 10]))
}
}
numword.100 <- function (y) {
a <- (floor(y / 100) %% 100) %% 10
if (a != 0)
return (c(single[a], "hundred"))
}
numword.1000 <- function(y) {
a <- (1000 * floor(y / 1000)) / 1000
if (a != 0)
return (c(numword.100(a), numword.10(a), "thousand"))
}
numword <- paste(c(numword.1000(x), numword.100(x), numword.10(x)), collapse=" ")
return (trimws(numword))
}

answer <- nchar(gsub(" ", "", paste0(sapply(1:1000, numword.en), collapse="")))


## Writing Numbers in Dutch

I went beyond Euler Problem 17 by translating the code to spell numbers in Dutch. Interesting bit of trivia is that it takes 307 fewer characters to spell the numbers 1 to 1000 in Dutch than it does in English.

It would be good if other people can submit functions for other languages in the comment section. Perhaps we can create an R package with a multi-lingual function for spelling numbers.

numword.nl <- function(x) {
if (x > 999999) return("Error: Getal te hoog.")
single <- c("een", "twee", "drie", "vier", "vijf", "zes", "zeven", "acht", "negen")
teens <- c( "tien", "elf", "twaalf", "dertien", "veertien", "fifteen", "zestien", "zeventien", "achtien", "negentien")
tens <- c("tien", "twintig", "dertig", "veertig", "vijftig", "zestig", "zeventig", "tachtig", "negengtig")
numword.10 <- function(y) {
a <- y %% 100
if (a != 0) {
if (a < 20)
return (c(single, teens)[a])
else
return (c(single[a %% 10], "en", tens[floor(a / 10)]))
}
}
numword.100 <- function(y) {
a <- (floor(y / 100) %% 100) %% 10
if (a == 1)
return ("honderd")
if (a > 1)
return (c(single[a], "honderd"))
}
numword.1000 <- function(y) {
a <- (1000 * floor(y / 1000)) / 1000
if (a == 1)
return ("duizend ")
if (a > 0)
return (c(numword.100(a), numword.10(a), "duizend "))
}
numword<- paste(c(numword.1000(x), numword.100(x), numword.10(x)), collapse="")
return (trimws(numword))
}

antwoord <- nchar(gsub(" ", "", paste0(sapply(1:1000, numword.nl), collapse="")))
print(antwoord)



# Euler Problem 16: Power Digit Sum

Euler Problem 16 is reminiscent of the famous fable of wheat and chess. Lahur Sessa invented the game of chess for King Iadava. The king was very pleased with the game and asked Lahur to name his reward.

Lahur asked the king to place one grain of rice on the first square of a chessboard, two on the next square, four on the third square and so on until the board is filled. The king was happy with his humble request until his mathematicians worked out that it would take millions of tonnes of grain. Assuming there are 25 grains of wheat in a gramme, the last field will contain more than 461,168,602,000 tonnes of grain.

## Euler Problem 16 Definition

$2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$. What is the sum of the digits of the number $2^{1000}$?

## Solution

The most straightforward solution uses the GMP package for Multiple Precision Arithmetic to calculate big integers. The as.bigz function results in a special class of arbitrarily large integer numbers

# Raise 2 to the power 1000
library(gmp)
digits <- as.bigz(2^1000) # Define number
# Sum all digits


We can also solve this problem in base-r with the bigg.add function which I developed for Euler Problem 13. This function uses basic string operations to add to arbitrarily large numbers. Raising a number to the power of two can also be written as a series of additions:

$2^4 = 2 \times 2 \times 2 \times 2 = ((2+2)+(2+2)) + ((2+2)+(2+2))$

The solution to this problem is to add 2 + 2 then add the outcome of that equation to itself, and so on. Repeat this one thousand times to raise the number two to the power of one thousand.

# Raise 2 to the power 1000
pow <- 2
for (i in 2:1000)
# Sum all digits


# Euler Problem 15: Pathways Through a Lattice

Euler Problem 15 analyses taxicab geometry. This system replaces the usual distance function with the sum of the absolute differences of their Cartesian coordinates. In other words, the distance a taxi would travel in a grid plan. The fifteenth Euler problem asks to determine the number of possible routes a taxi can take in a city of a certain size.

## Euler Problem 15 Definition

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner. How many possible routes are there through a 20×20 grid?

## Solution

The defined lattice is one larger than the number of squares. Along the edges of the matrix, only one pathway is possible: straight to the right or down. We can calculate the number of possible pathways for the remaining number by adding the number to the right and below the point.

$p_{i,j}=p_{i,j{+1}}+p_{{i+1},j}$

For the two by two lattice the solution space is:

6  3  1
3  2  1
1  1  0

The total number of pathways from the upper left corner to the lower right corner is thus 6. This logic can now be applied to a grid of any arbitrary size using the following code.

The code defines the lattice and initiates the boundary conditions. The bottom row and the right column are filled with 1 as there is only one solution from these points. The code then calculates the pathways by working backwards through the matrix. The final solution is the number is the first cell.

# Define lattice
nLattice <- 20
lattice = matrix(ncol=nLattice + 1, nrow=nLattice + 1)

# Boundary conditions
lattice[nLattice + 1,-(nLattice + 1)] <- 1
lattice[-(nLattice + 1), nLattice + 1] <- 1

# Calculate Pathways
for (i in nLattice:1) {
for (j in nLattice:1) {
lattice[i,j] <- lattice[i+1, j] + lattice[i, j+1]
}
}



# Euler Problem 14: Longest Collatz Sequence

Euler Problem 14 looks at the Collatz Conjecture. These playful sequences, named after German mathematician Lothar Collatz (1910–1990), cause mathematicians a lot of headaches. This video introduces the problem much better than I can describe it.

## Euler Problem 14 Definition

The following iterative sequence is defined for the set of positive integers:

$n \rightarrow n/2$ ( $n$ is even)
$n \rightarrow 3n + 1$ ( $n$ is odd)

Using the rule above and starting with 13, we generate the following sequence:

$13 \rightarrow 40 \rightarrow 20 \rightarrow 10 \rightarrow 5 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1$

This sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? Note: Once the chain starts the terms are allowed to go above one million.

## Solution

This problem is highly computationally intensive and it highlights R’s lack of speed. Generating one million Collatz sequences and finding the longest one requires a lot more than a minute of processing time allowed for in Project Euler.

collatz.chain <- function(n) {
chain <- vector()
i <- 1
while (n! = 1) {
if (n%%2 == 0)
n <- n / 2
else
n <- 3 * n + 1
chain[i] <- n
i <- i + 1
}
return(chain)
}
collatz.max <- 0
for (n in 1:1E6) {
collatz.length <- length(collatz.chain(n))
if (collatz.length > collatz.max) {
collatz.max <- collatz.length
}
}


The second version of the code contains some optimisations. The code stores the length of all sequences in an array. When the code generates a sequence and lands on a number already analysed, then it adds that previous number to the current one and moves on. This approach requires more memory but saves a lot of computation time. A minor tweak to the code optimises the rule for uneven numbers. Tripling an uneven number and adding one will always result in an even number so we can skip one step. This solution is more than twice as fast as the first version.

collatz.length <- vector(length=1e6)
collatz.length[1] <- 0
for (n in 2:1e6) {
x <- n
count <- 0 while (x != 1 & x >= n) {
if (x %% 2 == 0) {
x <- x / 2
count <- count + 1
}
else {
x <- (3 * x + 1) / 2
count <- count + 2
}
}
count <- count + collatz.length[x]
collatz.length[n] <- count
}


## Visualising Collatz Sequences

The Collatz sequence is an example of a simple mathematical rule that can create an unpredictable pattern. The number of steps required to reach 1 is listed in A006577 of the Online Encyclopedia of Integer Sequences.

The image below visualises the number of steps for the first 1000 positive numbers. The scatterplot shows some interesting patterns. Does this visualisation show that the Collatz Sequence does have a pattern after all?

## Collatz Chains

The Collatz sequences can also be visualised using networks. Each step between two numbers is an edge and the numbers are the vertices. For example, the network for the Collatz sequence for number 10 is 5–16, 16–8, 8–4, 4–2, 2–1. When generating subsequent sequences the network will start to overlap and a tree of sequences appears. The tree below combines the Collatz sequences for the numbers 2 to 26. Number 27 has a very long sequence, making the tree much harder to read.

Network of Collatz sequences n=2-26

edgelist <- data.frame(a = 2, b = 1)
for (n in 3:26) {
chain <- as.character(c(n, collatz.chain(n)))
chain <- data.frame(a = chain[-length(chain)], b = chain[-1])
edgelist <- rbind(edgelist, chain)
}
library(igraph)
g <- graph.edgelist(as.matrix(edgelist))
g <- simplify(g)
par(mar=rep(0,4))
V(g)$color <- degree(g, mode = "out") + 1 plot(g, layout=layout.kamada.kawai, vertex.color=V(g)$color,
vertex.size=6,
vertex.label.cex=.7,
vertex.label.color="black",
edge.arrow.size=.1,
edge.color="black"
)


# Euler Problem 13: Large Sum of 1000 Digits

Euler Problem 13 asks to add one hundred numbers with fifty digits. This seems like a simple problem where it not that most computers are not designed to deal with numbers with a lot of integers. For example:

$2^{64} = 18446744073709551616$

When asking R to compute this value we get 1.844674e+19, losing most of the digits and limiting the accuracy of the results. Computers solve this problem using Arbitrary-precision Arithmetic. There are many software libraries that can process long integers without loosing accuracy. Euler Problem 13 requires this type of approach.

## Euler Problem 13 Definition

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

## Solution

The easy way to solve this problem is to use the gmp package for working with very large integers. This package uses a special number types such as Big Rational and Big Integer. The number of digits in these number types is only limited by the size of the memory.

library(gmp)
digits <- sum(as.bigz(numbers))


# Using Base-R

To find the solution to this problem using only base R, I wrote a function to add numbers using strings instead of integers. The function adds leading zeros to the smallest number to make them both the same length. The function then proceeds to add numbers in the same way we were taught in primary school. This function can in principle be used for several other Euler Problems using large integers.

# Add numbers with many digits
if (nchar(a) < nchar(b))
a <- paste0(paste(rep(0, nchar(b) - nchar(a)), collapse = ""), a)
if (nchar(a) > nchar(b))
b <- paste0(paste(rep(0, nchar(a) - nchar(b)), collapse = ""), b)
solution <- vector()
remainder <- 0
for (i in nchar(b):1) {
p <- as.numeric(substr(a, i, i))
q <- as.numeric(substr(b, i, i))
r <- p + q + remainder
if (r >= 10 & i!=1) {
solution <- c(solution, r %% 10)
remainder <- (r - (r %% 10))/10
} else {
solution <- c(solution, r)
remainder <- 0
}
}
return(paste(rev(solution), collapse = ""))
}


With this function, the problem is easy to solve. The second part of the code runs this function over the one hundred numbers provided on the Euler Problem page and calculates the answer.

numbers <- readLines("Euler/p013_numbers.txt")
for (i in numbers) {
}


## Multiplying Big Numbers

You can expand this function to multiply a very large number with a smaller number using the Reduce function. This function adds the number a to itself, using the big.add function. The outcome of the addition is used in the next iteration until it has been repeated b times. The number b in this function needs to be a ‘low’ number because it uses a vector of the length b.

big.mult <- function(a, b) {
}


# Euler Problem 12: Highly Divisible Triangular Number

The divisors of 10 illustrated with Cuisenaire rods: 1, 2, 5, and 10 (Wikipedia).

Euler Problem 12 takes us to the realm of triangular numbers and proper divisors.

The image on the left shows a hands-on method to visualise the number of divisors of an integer. Cuisenaire rods are learning aids that provide a hands-on way to explore mathematics.

## Euler Problem 12 Definition

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. The first ten terms would be: $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots$ Let us list the factors of the first seven triangle numbers:

1: 1

3: 1, 3

6: 1, 2, 3, 6

10: 1, 2, 5, 10

15: 1, 3, 5, 15

21: 1, 3, 7 ,21

28: 1, 2, 4, 7, 14, 28

We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?

## Solution

Vishal Kataria explains a simple method to determine the number of divisors using prime factorization as explained by in his video below. The prime factorization of $n$ is given by:

$n = p^{\alpha_1}_1 \times p^{\alpha_2}_2 \times p^{\alpha_k}_k$

The number of proper divisors is:

$d = (\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_k + 1)$

The code reuses the prime factorisation function developed for Euler Problem 3. This function results in a vector of all prime factors, e.g. the prime factors of 28 are 2, 2 and 7.

The code to solve this problem determines the values for alpha using the run length function. This function counts the number of times each element in a sequence is repeated. The outcome of this function is a vector of the values and the number of times each is repeated. The prime factors of 28 are 2 and 7 and their run lengths are 2 and 1. The number of divisors can now be determined.

$28 = 2^2 \times 7^1$

$d = (2+1)(1+1) = 6$

The code to solve Euler Problem 12 is shown below. The loop continues until it finds a triangular number with 500 divisors. The first two lines increment the index and create the next triangular number. The third line in the loop determines the number of times each factor is repeated (the run lengths). The last line calculates the number of divisors using the above-mentioned formula.

i <- 0
divisors <- 0
while (divisors < 500) {
i <- i + 1
triangle <- (i * (i+1)) / 2
pf <- prime.factors(triangle)
alpha <- rle(pf)
divisors <- prod(alpha\$lengths+1)
}


# Euler Problem 11: Largest Product in a Grid

## Euler Problem 11 Definition

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

The product of these numbers is 26 × 63 × 78 × 14 = 1,788,696. What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20 by 20 grid?

## Solution

The solution applies straightforward vector arithmetic. The product of all verticals is an array of the product of rows 1 to 4, rows 2 to 5 and so on. The code uses a similar logic for the horizontals and the diagonals.

#Read and convert data
square <- as.numeric(unlist(lapply(square, function(x){strsplit(x, " ")})))
square <- matrix(square, ncol=20)

# Define products
prod.vert <- square[1:17, ] * square[2:18, ] * square[3:19, ] * square[4:20, ]
prod.hori <- square[,1:17] * square[,2:18] * square[,3:19] * square[,4:20]
prod.dia1 <- square[1:17, 1:17] * square[2:18, 2:18] * square[3:19, 3:19] * square[4:20, 4:20]
prod.dia2 <- square[4:20, 1:17] * square[3:19, 2:18] * square[2:18, 3:19] * square[1:17, 4:20]

answer <- max(prod.vert, prod.hori, prod.dia1, prod.dia2)


# Euler Problem 10: Summation of Primes

Euler Problem 10 asks for the summation of primes. Computationally this is a simple problem because we can re-use the prime sieve developed for problem 3.

When generating a large number of primes the erratic pattern at which they occur is much more interesting than their sum. Mathematicians consider primes the basic building blocks of number theory. No matter how hard we look, however, they do not seem to obey any logical sequence. This Euler Problem is simply

## Euler Problem 10 Definition

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.

## Solution

The sieve of Eratosthenes function used in Euler Problem 3 can be reused once again to generate the prime numbers between two and two million. An interesting problem occurs when I run the code. When I sum all the primes without the as.numeric conversion, R throws an integer overflow error and recommends the conversion.

primes <- esieve(2e6)