0 randFromArray([true, false]) randRange(10000, 1000000) $._("Yes") $._("No")

Is `\large{NUMBER}` divisible by `FACTOR`?

DIVISIBLE ? YES : NO

- YES
- NO

2

Any even number is divisible by `2`.

`NUMBER` is even, so it is divisible by `2`.

`NUMBER` is odd, so it is not divisible by `2`.

3 (function(){ var steps = []; var colors = [PINK, PURPLE, GREEN, BLUE]; var num = NUMBER; var numColor = colors.pop(); while (num > FACTOR * 11) { var sumColor = colors.pop(); var digits = integerToDigits(num); var sum = _.reduce(digits, function(m, v) { return m + v; }, 0); steps.push({ num: "\\color{" + numColor + "}{" + num + "}", digits: _.map(digits, function(digit) { return "\\color{" + numColor + "}{" + digit + "}"; }), sum: "\\color{" + sumColor + "}{" + sum + "}", }); num = sum; numColor = sumColor; } return steps; })()

A number is divisible by `FACTOR` if the sum of its digits is divisible by `FACTOR`. [Why?]

First, we can break the number up by place value:

`\qquad\begin{eqnarray} \blue{NUMBER}= integerToDigits(NUMBER).map(function(v, p) { var placeValue = pow(10, integerToDigits(NUMBER).length - p - 1); return "&&\\blue{" + v + "}\\cdot" + placeValue; }).join("+ \\\\") \end{eqnarray}`

Next, we can rewrite each of the place values as `1` plus a bunch of `9`s:

`\qquad\begin{eqnarray} \blue{NUMBER}= integerToDigits(NUMBER).map(function(v, p) { var placeValue = pow(10, integerToDigits(NUMBER).length - p - 1); if (placeValue === 1) { return "&&\\blue{" + v + "}"; } return "&&\\blue{" + v + "}(" + (placeValue - 1) + "+1)"; }).join("+ \\\\") \end{eqnarray}`

Now if we distribute and rearrange, we get this:

`\qquad\begin{eqnarray} \blue{NUMBER}= integerToDigits(NUMBER).map(function(v, p) { var placeValue = pow(10, integerToDigits(NUMBER).length - p - 1); if (placeValue === 1) { return ""; } return "&&\\gray{" + v + "\\cdot" + (placeValue - 1) + "}"; }).join("+ \\\\")&& \blue{integerToDigits(NUMBER) .join("}+\\blue{")} \end{eqnarray}`

Any number consisting only of `9`s is a multiple of `FACTOR`, so the first cardinalThrough20(integerToDigits(NUMBER) .length - 1) terms must all be multiples of `FACTOR`.

That means that to figure out whether the original number is divisible by `FACTOR` , all we need to do is add up the digits and see if the sum is divisible by `FACTOR`. In other words, `\blue{NUMBER}` is divisible by `FACTOR` if `\blue{ integerToDigits(NUMBER).join("}+\\blue{") }` is divisible by `FACTOR`!

Add the digits of `STEP.num`:

`STEP.digits.join("+") = STEP.sum`

If `STEP.sum` is divisible by `FACTOR`, then `STEP.num` must also be divisible by `FACTOR`.

`_.last(STEPS).sum` is divisible by `FACTOR`, therefore `\blue{NUMBER}` must also be divisible by `FACTOR`.

`_.last(STEPS).sum` is not divisible by `FACTOR`, therefore `\blue{NUMBER}` must not be divisible by `FACTOR`.

4

A number is divisible by `4` if the last two digits are divisible by `4`. [Why?]

We can rewrite the number as a multiple of `100` plus the last two digits:

`\qquad \gray{NUMBER.toString().slice(0, -2)} \blue{("00" + (NUMBER % 100)).slice(-2)} = \gray{NUMBER.toString().slice(0, -2)} \gray{00} + \blue{("00" + (NUMBER % 100)).slice(-2)}`

Because `NUMBER.toString().slice(0, -2) 00` is a multiple of `100`, it is also a multiple of `4`.

So as long as the value of the last two digits, `\blue{NUMBER % 100}`, is divisible by `4`, the original number must also be divisible by `4`!

Is the value of the last two digits, `NUMBER % 100`, divisible by `4`?

Yes, `\blue{NUMBER % 100 \div 4 = NUMBER % 100 / 4}`, so `NUMBER` must also be divisible by `4`.

No, `NUMBER % 100` is not divisible by `4`, so `NUMBER` is also not divisible by `4`.

5

A number is divisible by `5` if the last digit is a `0` or a `5`.

The last digit of `NUMBER` is `NUMBER % 10`, so yes `NUMBER` is divisible by `5`.

The last digit of `NUMBER` is `NUMBER % 10`, so no `NUMBER` is not divisible by `5`.

9
10

A number is divisible by `10` if the last digit is a `0`.

The last digit of `NUMBER` is `NUMBER % 10`, so yes `NUMBER` is divisible by `10`.

The last digit of `NUMBER` is `NUMBER % 10`, so no `NUMBER` is not divisible by `10`.

`NUMBER` is divisible by `FACTOR` if it can be divided by `FACTOR` without leaving a remainder.