# Double Counting explainer

The aim of this puzzle: Edit the classic for loop to print every odd number under 10.
Walkthrough of the solution: A classic for loop is used to repeat a block of code a specific number of times. A classic for loop has 3 components separated by semicolons:

1. A start condition that declares looping variable and gives it a value.
2. A test that keeps the classic for loop running as long as the test is true.
3. An update operation that changes the looping variable after each iteration (loop).

For example:

``````for (var i = 0; i < 10; i = 1 + 1) {
print(i)
}
``````
1. `var i = 0` sets up the looping variable `i` and gives it the value `0`.
2. `i < 10` is the test. The classic for loop will run as long as `i` is less than `10`.
3. `i = i + 1` updates the value of `i` after each iteration, adding `1` to the value each time.

In this puzzle, a classic for loop has been set up that sets `i` to `1` and counts up while `i < 10` is true. This prints out the numbers 1 through 9.

To complete the puzzle, edit the classic for loop to update `i` by `2` after each iteration, rather than `1`.

To do this, change `i = i + 1` to `i = i + 2`.

Sample code solution:
(Tap below to reveal)

``````for (var i = 1; i < 10; i = i + 2) {
print(i);
}
``````

JavaScript Concepts: Code Block (for loop), Calling Functions, Variable Declaration, Identifiers, Loops
Grasshopper Concepts: print()

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