Determine a value by incrementing a number in a loop

+---+---+---+
| 1 | 0 | 0 |
+---+---+---+
| 2 | 1 | 0 |
+---+---+---+
| 3 | 2 | 0 |
+---+---+---+
| 4 | 0 | 1 |
+---+---+---+
| 5 | 1 | 1 |
+---+---+---+
| 6 | 2 | 1 |
+---+---+---+
| 7 | 0 | 2 |
+---+---+---+
| 8 | 1 | 2 |
+---+---+---+
| 9 | 2 | 2 |
+---+---+---+

The code in question:

var loop = 1;
while(loop < 10) {
    console.log(loop, loop%3, "I cannot perform this calculation.")
    loop++;
}

In the above script, a variable loop is counting from 1 to 9.

The objective is to calculate two numbers based on the incremental variable: one determines the remainder of dividing the loop number by 3 (loop % 3), which is achievable. However, the second calculation involves generating either all 0s or all 1s.

Essentially, the goal is to associate a value based on the remainder of the division.

Answer №1

Here is a corrected version of your code ((loop-1)%3) that includes division with rounding down:

var loop = 1;
while(loop < 10) {
    console.log(loop, (loop-1)%3, Math.floor((loop-1)/3))
    loop++;
}

It's worth noting that some languages have built-in support for "integer division", eliminating the need for Math.floor.

Answer №2

Here is the code snippet you can use:

var x = 10;
var y = 5;

while (x < 20) {
    console.log(x, (x - 1) % y, Math.floor((x - 1) / y));
    x++;
}

It's really simple!

Answer №3

If you're looking to convert numbers into a different decimal system, specifically base three with a unique starting point of one instead of zero, then this is the guide for you. The challenge here lies in reversing the result by swapping the lower and higher indicators.

Let's start with the straightforward conversion process:

let num;

for (num = 0; num < 9; num++) {
    console.log(num, num.toString(3).padStart(2, '0'));
}

Next, we'll explore the shifted result achieved by adding one to the output and reversing the sequence:

let num;

for (num = 0; num < 9; num++) {
    console.log(num + 1, Array.from(num.toString(3).padStart(2, '0')).reverse().join(''));
}

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