Euler Problem 23: Non-Abundant Sums

A demonstration of the abundance of the number 12 using Cuisenaire rods (Wikipedia).

Euler problem 23 asks us to solve a problem with abundant or excessive numbers.

These are numbers for which the sum of its proper divisors is greater than the number itself.

12 is an abundant number because the sum of its proper divisors (the aliquot sum) is larger than 12: (1 + 2 + 3 + 4 + 6 = 16).

All highly composite numbers or anti-primes greater than six are abundant numbers. These are numbers that have so many divisors that they are considered the opposite of primes, as explained in the Numberphile video below.

Euler Problem 23 Definition

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis, even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution

This solution repurposes the divisors function that determines the proper divisors for a number, introduced for Euler Problem 21. The first code snippet creates the sequence of all abundant numbers up to 28123 (sequence A005101 in the OEIS). An abundant number is one where its aliquot sum is larger than n.

# Generate abundant numbers (OEIS A005101)
A005101 <- function(x){
abundant <- vector()
a <- 1
for (n in 1:x) {
aliquot.sum <- sum(proper.divisors(n)) - n
if (aliquot.sum > n) {
abundant[a] <- n
a <- a + 1
}
}
return(abundant)
}

abundant <- A005101(28123)


The solution to this problem is also a sequence in the Online Encyclopedia of Integer Sequences (OEIS A048242). This page states that the highest number in this sequence is 20161, not 28123 as stated in the problem definition.

The second section of code creates a list of all potential numbers not the sum of two abundant numbers. The next bit of code sieves any sum of two abundant numbers from the list. The answer is determined by adding remaining numbers in the sequence.

# Create a list of potential numbers that are not the sum of two abundant numbers
A048242 <- 1:20161

# Remove any number that is the sum of two abundant numbers
for (i in 1:length(abundant)) {
for (j in i:length(abundant)) {
if (abundant[i] + abundant[j] <= 20161) {
A048242[abundant[i] + abundant[j]] <- NA
}
}
}
A048242 <- A048242[!is.na(A048242)]


Euler Problem 21: Amicable Numbers

Euler problem 21 takes us to the realm of amicable numbers, which are listed in sequence A259180 in the OEIS. Amicable, or friendly, numbers are the most romantic numbers known to maths. Amicable numbers serve absolutely no practical purpose, other than mathematical entertainment.

A related concept is a perfect number, which is a number that equals the sum of its proper divisors. Mathematicians have also defined sociable numbers and betrothed numbers which are similar to amicable numbers. But perhaps these are for another Euler problem.

Euler Problem 21 Definition

Let $d(n)$ be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If $d(a) = b$ and $d(b) = a$, where $a \neq b$, then $a$ and $b$ are an amicable pair and each of $a$ and $b$ are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore $d(220) = 284$. The proper divisors of 284 are 1, 2, 4, 71 and 142; so, $d(284) = 220$.

Solution

The first part of the code provides for a function to list all proper divisors for a given integer x. The loop determines the divisors for the numbers 220 to 10,000, calculates their sum and then checks if these numbers are amicable. When the code finds an amicable number, the counter jumps to the sum of the divisors to check for the next one.

proper.divisors <- function(x) {
divisors <- vector()
d <- 1
for (i in 1:floor(sqrt(x))) {
if (x %% i == 0) {
divisors[d] <- i
if (i != x/i) {
d <- d + 1
divisors[d] <- x / i
}
d <- d + 1
}
}
return(divisors)
}

n <- 220
while (n <= 10000) {
div.sum <- sum(proper.divisors(n)) - n
if (n == sum(proper.divisors(div.sum)) - div.sum & n != div.sum) {
print(paste0("(", n, ",", div.sum, ")"))
n <- div.sum
}
n <- n + 1
}


Amicable numbers were known to the Pythagoreans, who credited them with many mystical properties. Before we had access to computers, finding amicable numbers was a task that required a lot of patience. No algorithm can systematically generate all amicable numbers, and until 1946 only 390 pairs were known. Medieval Muslim mathematicians developed several formulas to create amicable numbers, but the only way to be complete is using brute force.

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 5: Smallest Multiple

Euler Problem 5 relates to the divisibility of numbers.

Euler Problem 5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Solution

The solution will also be divisible by the number 1 to 10 so we can start at 2520 and increment by 2520. The loop checks whether the number is divisible by the numbers 1 to 20.

# Start as high as possible
i <- 2520
# Check consecutive numbers for divisibility by 1:20
while (sum(i%%(1:20)) != 0) {
i <- i + 2520 # Increase by smallest number divisible by 1:10
}


Euler Problem 1: Multiples of 3 or 5

Euler Problem 1 Definition

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

Solution

There are four ways to solve this problem in R.

1. Brute force loop through all numbers from 1 to 999 and test whether they are divisible by 3 or by 5 using the modulus function.
2. Vector arithmetics.
3. Sequences of 3 and 5, excluding duplicates (numbers divisible by 15).
4. Using an arithmetic approach.

By the way, the problem definition on the Project Euler website is not consistent: the title mentions multiples of 3 AND 5, while the description asks for multiples of 3 OR 5.

# Solution 1
for (i in 1:999) {
if (i%%3 == 0 | i%%5 == 0)
}

# Solution 2
sum((1:999)[((1:999)%%3 == 0) | ((1:999)%%5 == 0)])

# Solution 3
sum(unique(c(seq(3, 999, 3), seq(5, 999, 5))))


These three brute-force solutions all take time to process, especially when using targets much igher than 999. An analytical solution will significantly reduce the processing time.

Analytical Solution

The sum of an arithmetic progression , where n is the number of elements and a1 and an are the lowest and highest value, is:

$\mathrm{sum}= \frac{n(a_{1} + a_n)}{2}$

The numbers divisible by $n=3$ can be expressed as:

$\mathrm{sum}_3(999)=3+6+9+12+ \ldots +999 = 3(1+2+3+4+ \ldots +333)$

We can now calculate the sum of all divisors by combining the above progression with the formula for arithmetic progressions as expressed in the above code, where m is the divisor and n the extent of the sequence.

p is the highest number less than n divisible by m. In the case of 5, this number is 995.

$p = n \lfloor (m/n) \rfloor$

Substitution gives:

$\mathrm{sum}_m(n) = p \frac{1+(p/m)}{2}$

# Solution 4
SumDivBy <- function(m, n) {
p <- floor(n/m)*m # Round to multiple of n
return (p*(1+(p/m))/2)
}

answer <- SumDivBy(3, 999) + SumDivBy(5, 999) - SumDivBy(15, 999)