Divisibility tests

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 9s:


                            \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 9s 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.