Sometimes in life, you need to find out weird facts, such as how many times the digit 1 occurs in a number range sequence of 0 to 10 (the answer is 2 times, appearing as "1" and the "1" in "10"). This came up a couple of years ago in a Reddit discussion thread that I was involved in (will post the link if I can find it).

Once again, JavaScript is a very lightweight tool that we can use to perform calculation.

The logic behind the looniness

Goal: Find out how many times the number 1 occurs in a number sequence range of 0 to 10.

We have 3 variables in this calculation:

• a: This is the starting number of the range e.g. 0.
• b: This is the ending number of the range e.g. 10.
• c: This is the number to find within the range e.g. 1.

The JavaScript to do the work for us is only 5 lines long!

var result = 0;
for (var i = startNumberA; i <= endNumberB; i++) {
    result += (i.toString().match(new RegExp(numToFindC.toString(), "g")) || []).length;
}
return result;

We are using a simple loop and counting from the start of the range e.g. 0, right through to the final number in the range e.g. 10 (inclusive).

For each number in the range, a Regular Expression match is run, which in simple terms, counts how many times the search c variable occurs between a and b, and then adds that quantity to the existing counter.

Calculation result

Using the above counter, we can accurately determine, that the number 1 appears twice between and including 0 and 10:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 2

Our answer is 2.

Working Result from JSFiddle

Jump straight to the Result tab to see the calculator.