# Modulo Operator (%%) in R: Explained + Practical Examples

The modulo operator (%% in R) returns the remainder of the division of 2 numbers.

Here are some examples:

• 5 %% 2 returns 1, because 2 goes into 5 two times and the remainder is 1 (i.e. 5 = 2 × 2 + 1).
• 4 %% 2 returns 0, since 4 = 2 × 2 + 0.
• 4 %% 1 returns 0, since 4 = 4 × 1 + 0.
• 2 %% 4 returns 2, since 2 = 0 × 4 + 2.
• 0 %% 2 returns 0, since nothing remains from dividing 0 by 2 (i.e. 0 = 0 × 2 + 0).
• 2 %% 0 returns NaN, since we cannot divide by zero.

The modulo operator also works with decimal numbers, but this is rarely used in practice:

• 5.6 %% 3.2 returns 2.4, since 5.6 = 1 × 3.2 + 2.4.
• 5.6 %% 3 returns 2.6.
• 5.6 %% 1 returns 0.6.

Below are 5 practical problems that can be solved using the modulo operator:

## 1. Checking if a number is even or odd

Problem: Is 27 even or odd?

Solution:

```n = 27

# if n is divisible by 2 then:
n %% 2 == 0 # must be TRUE

print(paste('n is', ifelse(n %% 2 == 0, 'even', 'odd')))
# this code prints: "n is odd"```

## 2. Checking if a number is prime

A prime number is any integer (other than 0 and 1) divisible only by 1 and itself without a remainder.

For example:

• 5 is a prime number since it is only divisible by 1 and 5.
• 4 is not a prime number since it can also be divided by 2.

Problem: Is 5 a prime number?

Solution:

```n = 5

# 5 is not prime if 5 %% 2, 5 %% 3, or 5 %% 4 is zero
notPrime = any(n %% 2:(n-1) == 0)

print(paste('n is', ifelse(notPrime, 'not a prime nb', 'a prime nb')))
# this code prints: "n is a prime nb"```

## 3. Systematic sampling from a population

Systematic sampling consists of select every kth person from a population of interest.

Problem: Select every 7th person from a list of 100 people. What is the sample size?

Solution:

```population = 1:100

sample = population[population %% 7 == 0]
print(sample)
# outputs: 7 14 21 28 35 42 49 56 63 70 77 84 91 98

# sample size
length(sample) # outputs: 14```

## 4. Creating equal subgroups from a larger group

Problem: A company has 507 employees. How many teams of equal size can we create?

Solution:

```# which numbers can 507 be divided by without a remainder?
which(507 %% 1:507 == 0)
# outputs: 1   3  13  39 169 507```

So, here are the possible groups:

• 1 group of 507 employees, or
• 3 groups of 169 (= 507/3) employees, or
• 13 groups of 39 (= 507/13) employees, or
• 39 groups of 13 employees, or
• 169 groups of 3 employees, or
• 507 groups of 1 employee.

## 5. Converting age from days to: years, months, and days

Problem: Report the age in the format: “years, months, days” of a person aged 500 days.

Solution:

```# age in days
days = 500

# how many years are there in 500 days?
years = floor(days / 365) # the floor function rounds down
print(years) # 1

remainder = days %% 365 # days remaining
print(remainder) # 135

# how many months are there in the remaining 135 days?
months = floor(remainder / 30)
print(months) # 4

remainder = remainder %% 30 # days remaining
print(remainder) # 15

# remaining days
days = remainder

# age
print(paste(years, 'year(s),', months, 'month(s),', days, 'day(s)'))
# this code prints: "1 year(s), 4 month(s), 15 day(s)"```