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## Solution: Using a Combination of `if` and `else`


testVariable <- 45

if(testVariable %% 3 == 0 && testVariable %% 5 == 0)
{
  cat("foo bar")
} else {
  if(testVariable %% 3 == 0)
  {
    cat("foo")
  }
  if(testVariable %% 5 == 0)
  {
    cat("bar")
  }
}

if(testVariable %% 3 == 0 && testVariable %% 5 == 0)
{
  cat("foo bar")
} else {
  if(testVariable %% 3 == 0)
  {
    cat("foo")
  }
  if(testVariable %% 5 == 0)
  {
    cat("bar")
  }
}

### Explanation
We first check whether the `testVariable` is a multiple of both $3$ and $5$. For this, we use an `if` condition 
```
testVariable %% 3 == 0 && testVariable %% 5 == 0
```
The modulus sign `%%` helps us figure out whether the `testVariable` is a multiple of the number. The remainder after division of `testVariable` with $3$ would be $0$ if it is a multiple of $3$. The remainder after division of `testVariable` with $5$ would also be $0$ if it is a multiple of $5$. 

We then **AND** the results of both the tests.

If this condition is not satisfied, we move towards the `else` block. In the else block we have two more `if` conditions: one for only multiple of $3$ and one for checking only multiple of $5$.

# Solution: Using a Combination of `if` and `else`


## Explanation
We first check whether the `testVariable` is a multiple of both $3$ and $5$. For this, we use an `if` condition 
```
testVariable %% 3 == 0 && testVariable %% 5 == 0
```
The modulus sign `%%` helps us figure out whether the `testVariable` is a multiple of the number. The remainder after division of `testVariable` with $3$ would be $0$ if it is a multiple of $3$. The remainder after division of `testVariable` with $5$ would also be $0$ if it is a multiple of $5$. 

We then **AND** the results of both the tests.

If this condition is not satisfied, we move towards the `else` block. In the else block we have two more `if` conditions: one for only multiple of $3$ and one for checking only multiple of $5$.

In this review, we provide a detailed analysis of the solution to this problem.

Introduction to R

R variables

Data Structures in R

Operator in R

Conditional Statements in R

Loops in R

Function in R

Input/Output in R

Exception Handling in R

Classes in R

R Programming Challenges

Conclusion

Solution Review: if/else Statements

Solution: Using a Combination of `if` and `else`

Explanation