# The Viral Recurring Decimal: Euler Problem 26

A few years ago a fraction broke the internet. What happens when you divide 1 by 998001?

$\frac{1}{998001} = 0.000001002003004005006007008009010011012013014015 \ldots$

What is special about this fraction is that it lists every three-decimal number except for 998. Look carefully at the sequence to see that is 000, 001, 0002, 003, 004, 005 and so on. After it has reached 999, the sequence continues from the start. This fraction thus has 2997 recurring decimals. James Grime from Numberphile explains this mathematical oddity with his usual enthusiasm.

The decimal fraction of 1/998001 is a recurring decimal. These are decimal numbers with periodic digits (repeating its values at regular intervals). Euler problem 26 asks us to analyse recurring decimals (reciprocal cycles). Euler Problem 33 also looks at fractions and Ford Circles.

## Euler Problem 26 Definition

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

$1/2 = 0.5$
$1/3 = 0.(3)$
$1/4 = 0.25$
$1/5 = 0.2$
$1/6 = 0.1(6)$
$1/7 = 0.(142857)$
$1/8 = 0.125$
$1/9 = 0.(1)$
$1/10 = 0.1$

Where 0.1(6) means 0.166666…, and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle. Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

## Solution

A051626 describes the length of the recurring numbers in 1/n in the On-Line Encyclopaedia of Integer Sequences. To solve Euler Problem 26, we need to generate the first 1000 numbers of this sequence and find out which number has the longest recurring cycle.

R can only display up to 22 decimals by using options(digits=22). The base R capability is unsuitable for solving this problem, so I wrote some code to perform long division the old-fashioned way.

The recur function divides 1 by any arbitrary integer. The code continues until the decimal terminates, for example 1/4 = 0.25, or when a recurring pattern emerges, e.g. 1/7 = 0.(142857).

The function has two arguments: n is the input number. The output argument determines the outcome of the function: “len” for the length of the recurring decimals. Any other value shows the result of the calculation. The output of the function is a string. Using the European notation, the recurring part of the decimals is shown between brackets, e.g. 1/14 = 0.0(714285).

recur <- function(x, output = "") {
# Prepare variable
if (x == 0) return(NaN)
if (x == 1) return(0)
x <- floor(abs(x))
# Initiate vectors to store decimals and remainders
dec <- vector()
rem <- vector()
# Initiate values
i <- 1
r <- 10
rem <- r
# Long division
repeat {
dec[i] <- floor(r / x)
r <- 10 * (r %% x)
# Test wether the number is terminating or repeating
if (r == 0 | r %in% rem) break
rem[i + 1] <- r
i <- i + 1
}
# Determine number of recurring digits
rep <- ifelse(r != 0, length(rem) - which(r == rem) + 1, 0)
# Output
if (output == "len")
return(rep)
else {
if (rep != 0) {
if (rep == length(dec))
l <- "("
else
l <- c(dec[1:(length(dec) - rep)], "(")
dec <- c(l, dec[(length(dec) - rep + 1):length(dec)], ")")
}
return(paste0("0.", paste0(dec, collapse = "", sep = "")))
}
}

A051626 <- sapply(1:1000, recur, "len")