Divisibility Rules: Free Response

Everything above the ones column counts a whole number of tens. In $8{,}256$ that part is $8{,}250$, which is $825$ tens, so the cut gives

$8{,}256 = (825 \times 10) + 6.$

Because $10 = 2 \times 5$, the piece $825 \times 10$ is a multiple of $10$, a multiple of $5$ and a multiple of $2$ all at once, so in each of those three tests it has already passed and contributes a remainder of $0$. Only the leftover $6$ can still decide anything.

The digit $6$ is even, so $8{,}256$ is divisible by $2$. It is neither $0$ nor $5$, so $8{,}256$ is not divisible by $5$. It is not $0$, so $8{,}256$ is not divisible by $10$.

Checking by division: $8{,}256 \div 2 = 4{,}128$ exactly, while $8{,}256 \div 5$ gives $1{,}651$ with a remainder of $1$. That leftover $1$ is what the last digit predicted, and it is the whole reason the $5$ verdict is a no.

Call the missing digit $u$. The three digits in front of it make $275$ tens, so the number is

$(275 \times 10) + u.$

That first piece is a whole number of tens, hence a multiple of $2$, of $5$ and of $10$ at the same time, and it contributes remainder $0$ to all three tests. So $u$ alone settles every one of them, and the digits $2$, $7$ and $5$ never enter the question.

For divisibility by $2$ the leftover digit must be even, so $u$ is one of $0, 2, 4, 6, 8$.

For divisibility by $5$ the leftover digit must itself be a multiple of $5$, and the only single digits that are, are $0$ and $5$.

A digit that does both jobs has to appear on both lists. Of $0$ and $5$, only $0$ is even, so $u = 0$ is the single answer, and the number it builds is $2{,}750 = 2 \times 1{,}375 = 5 \times 550$. Notice that this is also the only digit that passes the $10$ test, which is another way of saying that a number divisible by both $2$ and $5$ is divisible by $10$.

Write any whole number as $N = T + u$, where $u$ is the ones digit and $T$ is everything to the left of it. Because $T$ is built only from the tens, hundreds, thousands and higher columns, it counts a whole number of tens:

$T = 10 \times k \quad \text{for some whole number } k.$

Now use $10 = 2 \times 5$. That one equation makes $T$ a multiple of $10$, a multiple of $5$ (it is $5 \times 2k$) and a multiple of $2$ (it is $2 \times 5k$), whatever $k$ happens to be. So in all three tests the entire piece $T$ has already passed, contributing a remainder of $0$, and the only thing that can still move a verdict is $u$. That is why the test reads one digit, and why the length of the number makes no difference to how long the test takes.

Run the same opening move for $3$ and it stalls at once. We would need $T = 10 \times k$ to be a multiple of $3$ for every $k$, and it is not, because $10$ is not a multiple of $3$. The tens part keeps a leftover of its own, and how big that leftover is depends on $k$, which is to say on the digits we were trying to ignore.

One pair of numbers shows the damage. Both $21$ and $41$ end in the same digit, yet

$21 = 3 \times 7, \qquad 41 = (3 \times 13) + 2.$

The first is divisible by $3$ and the second is not. A rule that looks only at the final digit cannot possibly separate them, so no last-digit test for $3$ can exist. The tens have to be dealt with rather than discarded, and dealing with them is exactly what the digit-sum rule does.

Add every digit, not just the last one:

$8 + 7 + 6 + 9 = 30.$

Now put the sum itself on trial. It is a multiple of $3$, since $30 = 3 \times 10$, so $8{,}769$ is divisible by $3$. It is not a multiple of $9$: the multiples of $9$ run $27$, then $36$, stepping straight over $30$. So $8{,}769$ is not divisible by $9$.

If $30$ is still awkward to read, fold it once more, which is allowed because the rule applies to the sum exactly as it applied to the original: $3 + 0 = 3$, a multiple of $3$ and not of $9$, the same pair of verdicts.

Checking by division, $8{,}769 = 3 \times 2{,}923$, while dividing by $9$ leaves a remainder of $3$. That remainder is the folded digit sum, which is not a coincidence and is the subject of the last question in this set.

Call the missing digit $b$. The three known digits add to

$4 + 1 + 5 = 10,$

so the digit sum of the whole number is $10 + b$. Writing it that way, with the unknown still in it, is what makes the search finite: a single digit runs from $0$ to $9$, so $10 + b$ can only land somewhere between $10$ and $19$.

For the $9$ test, $10 + b$ must be a multiple of $9$. The only multiple of $9$ between $10$ and $19$ is $18$, so

$10 + b = 18 \;\Rightarrow\; b = 8.$

That builds $4{,}815$, and indeed $4{,}815 = 9 \times 535$.

For the $3$ test, $10 + b$ must be a multiple of $3$. Between $10$ and $19$ those are $12$, $15$ and $18$, which ask for $b = 2$, $b = 5$ and $b = 8$ in turn.

Three digits clear the $3$ test where one clears the $9$ test, and the spacing explains it: multiples of $3$ sit three apart and multiples of $9$ nine apart, so a run of ten consecutive sums meets three or four of the first and one or two of the second.

The student's premise is correct: both tests start from the same digit sum. What the argument misses is that the sum is then compared against two different divisors, and those two divisors do not sit symmetrically.

The first half. Suppose the digit sum is a multiple of $9$, so it is $9 \times m$ for some whole number $m$. Since $9 = 3 \times 3$,

$9 \times m = 3 \times (3m),$

which is a multiple of $3$. So a digit sum that clears the $9$ test clears the $3$ test automatically, and a number that passes the $9$ test passes the $3$ test. That half holds for every whole number, with no exception to look for.

The second half. One number retires it. The digits of $1{,}527$ add to

$1 + 5 + 2 + 7 = 15,$

which is a multiple of $3$, so $1{,}527$ passes the $3$ test, and $1{,}527 = 3 \times 509$. But $15$ is not a multiple of $9$, so $1{,}527$ fails the $9$ test: dividing it by $9$ leaves a remainder of $6$, which is exactly the remainder $15$ itself leaves.

So the implication runs one way only. Passing the harder test guarantees the easier one, because $9$ is itself a multiple of $3$. Passing the easier one guarantees nothing about the harder, because $3$ is not a multiple of $9$.

For $4$, cut the count into the hundreds and above plus the last two digits:

$5{,}976 = (59 \times 100) + 76.$

Since $100 = 4 \times 25$, the piece $59 \times 100$ is a multiple of $4$ and leaves nothing over, so only $76$ is still on trial. And $76 = 4 \times 19$, so rows of $4$ work. There are $5{,}976 \div 4 = 1{,}494$ of them.

For $8$, cut one column further out:

$5{,}976 = (5 \times 1000) + 976.$

Since $1000 = 8 \times 125$, the piece $5 \times 1000$ is a multiple of $8$, so only the last three digits matter. Testing $976$ is far quicker than testing $5{,}976$: take $8 \times 100 = 800$, which leaves $176$, and $8 \times 22 = 176$, so $976 = 8 \times 122$. Rows of $8$ work too, and there are $5{,}976 \div 8 = 747$ of them.

Notice what each cut bought. The $4$ rule threw away two of the four digits of the count and the $8$ rule threw away one, and in both cases the thrown-away part was a multiple of the divisor for reasons that had nothing to do with this particular count.

Rows of $5$. The rule reads the final digit, which is $6$. It is neither $0$ nor $5$, so $5{,}976$ is not divisible by $5$ and some chairs would be left standing. Dividing confirms it: $5{,}976 = (5 \times 1{,}195) + 1$, so exactly one chair is stranded.

Rows of $9$. Add the digits.

$5 + 9 + 7 + 6 = 27,$

and $27 = 9 \times 3$, so $5{,}976$ is divisible by $9$. That gives $5{,}976 \div 9 = 664$ rows and nothing left over.

Rows of $6$. There is no rule of its own to run for $6$, so run the tests for $2$ and for $3$ instead. That substitution is legitimate because $6 = 2 \times 3$ and those two share no common factor above $1$. The last digit $6$ is even, so the $2$ test passes. The digit sum $27$ is $3 \times 9$, so the $3$ test passes. Both halves clear, so $5{,}976$ is divisible by $6$, giving $996$ rows.

One digit sum settled the $9$ question and half of the $6$ question at once. That reuse is why chaining tests beats chaining divisions: the work overlaps, and the divisions would not.

Start from what the two passes actually give you. The count is a multiple of $8$, so

$N = 8 \times q \quad \text{for some whole number } q,$

and it is a multiple of $9$ as well.

Now ask where that factor of $9$ can be hiding. It cannot come out of the $8$. Any whole number above $1$ that divided both $8$ and $9$ would have to be a divisor of $8$ other than $1$, so it would be $2$, $4$ or $8$, all even, and no even number divides the odd number $9$. So $8$ and $9$ share no factor above $1$, and the whole factor of $9$ has to sit inside $q$. That is, $q = 9 \times r$, and

$N = 8 \times (9 \times r) = 72 \times r.$

The count is a multiple of $72$, so rows of $72$ come out even, with $r$ of them.

Now try the same move on $2$ and $4$. Suppose a count passes the $4$ test, so $N = 4 \times s$, and we want the $2$ test to force a factor of $2$ into $s$. It cannot: $4$ is itself even, so $4 \times s$ is even for every whole number $s$, and the $2$ test is passed for free by anything that passed the $4$ test. Nothing is forced, and the argument collapses. It collapses in a way you can see in one number: $68$ is divisible by $2$ and by $4$, yet $68 = (8 \times 8) + 4$, a remainder of $4$, so it is not divisible by $8$.

The condition that separates the two cases is whether the two divisors share a factor above $1$. For $8$ and $9$ they do not, so the tests multiply. For $2$ and $4$ they do, so one of them is quietly doing the other's work.

Start from the place-value form:

$N = (a \times 1000) + (b \times 100) + (c \times 10) + d.$

Each place value is one more than a string of nines:

$1000 = 999 + 1, \qquad 100 = 99 + 1, \qquad 10 = 9 + 1.$

Substitute those in and multiply out. Each digit now appears twice, once against its string of nines and once on its own:

$N = a(999 + 1) + b(99 + 1) + c(9 + 1) + d = 999a + a + 99b + b + 9c + c + d.$

Now gather the terms carrying nines into one group and the bare digits into another:

$N = \underbrace{(999a + 99b + 9c)}_{\text{a multiple of } 9} + \underbrace{(a + b + c + d)}_{\text{the digit sum}}.$

The first group is a multiple of $9$ for every choice of digits, and not merely for convenient ones, because $999 = 9 \times 111$, $99 = 9 \times 11$ and $9 = 9 \times 1$, so it can be written with the $9$ pulled out front:

$999a + 99b + 9c = 9 \times (111a + 11b + c).$

So whatever $a$, $b$ and $c$ are, that group contributes a remainder of $0$ in the $9$ test, and the entire verdict falls on the second group, which is the digit sum. The same substitution with $9 = 3 \times 3$, $99 = 3 \times 33$ and $999 = 3 \times 333$ runs the argument for $3$, and a longer numeral only adds more terms of the same shape, each still a multiple of $9$.

Part A showed that a number equals a multiple of $9$ plus its digit sum. A multiple of $9$ leaves remainder $0$, so the number and its digit sum leave the same remainder when divided by $9$. That is more than the yes-or-no verdict the rule usually gets used for.

Add the digits:

$6 + 1 + 4 + 2 = 13.$

Now $13$ is much smaller than $6{,}142$ but still larger than $9$, so fold it once more, which is legitimate for exactly the same reason: $1 + 3 = 4$. Since $4$ is less than $9$, it cannot be reduced further, and it is the remainder itself.

Checking directly, $9 \times 682 = 6{,}138$ and $6{,}142 - 6{,}138 = 4$, so

$6{,}142 = (9 \times 682) + 4.$

The digit sum did not merely report whether the remainder was zero. It handed over the remainder, which is the extra information the split in Part A was carrying all along.

An "exactly when" claim makes two promises at once, so it has to be read forwards and backwards. Here neither reading survives.

Forwards: does divisibility by $6$ force a digit sum divisible by $6$? Take $78$, which is $6 \times 13$. Its digit sum is

$7 + 8 = 15,$

and $15$ is not a multiple of $6$. So a number can pass the $6$ test while its digit sum fails, and the first promise is broken.

Backwards: does a digit sum divisible by $6$ force the number to be? Take $105$. Its digit sum is $1 + 0 + 5 = 6$, a multiple of $6$. But $105$ is odd, so it is not divisible by $2$, let alone by $6$. The second promise is broken too.

It is worth seeing why the digit sum works for $3$ and $9$ and for nothing else on the list. Part A showed that a number is a multiple of $9$ plus its digit sum, and that split only helps against a divisor that also divides the part being thrown away, the strings of nines. The whole numbers that divide $9$ are $1$, $3$ and $9$, so $3$ and $9$ are the only divisors that get anything out of it. Because $6$ does not divide $9$, the pieces $999a$, $99b$ and $9c$ are not multiples of $6$ for every choice of digits (with $a = 1$ the piece $999a$ already is not), so they cannot be discarded and a digit-sum test for $6$ has nothing to stand on.

The correct test for $6$ does still use a digit sum, but only for its $3$ half, alongside a separate look at the last digit.

Ending in $0$ or $5$ is exactly the $5$ test, so the number is a multiple of $5$:

$N = 5 \times q \quad \text{for some whole number } q.$

The $3$ test also passed, so $N$ is a multiple of $3$ as well. The question is where that factor of $3$ can be.

It cannot come out of the $5$. A common factor of $5$ and $3$ above $1$ would have to divide $5$, so it could only be $5$ itself, and $5$ does not divide $3$. So $5$ and $3$ share no factor above $1$, and the factor of $3$ has nowhere to go but inside $q$. That is, $q = 3 \times r$ for some whole number $r$, and substituting back,

$N = 5 \times q = 5 \times (3 \times r) = 15 \times r.$

So $N$ is a multiple of $15$. Nothing in that argument named a particular number, so it holds for every $N$ that passes both tests, which is what makes it an argument rather than a pattern.

The step that did the work was "the factor of $3$ cannot come out of the $5$". That is where the requirement that the two divisors share no factor above $1$ enters, and it is the step that fails when they do.

Take $N = 68$ and run the three checks.

The $2$ test. The last digit $8$ is even, so $68$ is divisible by $2$.

The $4$ test. The last two digits are the whole number here, and $68 = 4 \times 17$, so $68$ is divisible by $4$.

The $8$ test. The last three digits are again the whole number. Now $8 \times 8 = 64$, with $4$ still to go:

$68 = (8 \times 8) + 4.$

That remainder of $4$ is not zero, so $68$ is not divisible by $8$, and the claim is dead.

Now set this beside Part A. There the argument ran: $N = 5 \times q$, the factor of $3$ cannot come out of the $5$, so it is forced into $q$. Repeat it here with $N = 4 \times s$. We know $N$ is even and would like to conclude that $s$ must be even too. But $4$ is itself even, so $4 \times s$ is even for every whole number $s$, and the $2$ test therefore tells us nothing at all about $s$. With $68$ we get $s = 17$, which is odd, and the forcing step has plainly failed.

The general condition is the one Part A depended on: a combined test works only when the two divisors share no factor above $1$. Here $2$ and $4$ share the factor $2$, so the pair cannot reach $8$. To settle $8$ you have to run the rule for $8$ itself, on the last three digits.

Both proposals pick two numbers whose product is $12$, so the product is not what separates them. The only question is whether the two chosen divisors share a factor above $1$.

The $3$ and $4$ proposal. These share nothing above $1$: a common factor would have to divide $3$, so it could only be $3$ itself, and $3$ does not divide $4$. So the Part A argument runs without change. If $N$ passes the $4$ test then $N = 4 \times t$, and the factor of $3$ owed by the $3$ test cannot come out of the $4$, so it sits inside $t$, giving $t = 3 \times u$ and

$N = 4 \times (3 \times u) = 12 \times u.$

The proposal is a genuine test for $12$, and a fast one: two digits for the $4$ test and one digit sum for the $3$ test.

The $2$ and $6$ proposal. These share the factor $2$, so there is nothing for the argument to force. Every multiple of $6$ is already even, so anything that passes the $6$ test passes the $2$ test for free, and running the $2$ test adds no information whatever. The pair is really the $6$ test wearing two hats.

One number ends it. Take $42$. It is even, so the $2$ test passes. Its digit sum is $4 + 2 = 6$, a multiple of $3$, and its last digit is even, so the $6$ test passes as well, and indeed $42 = 6 \times 7$. Yet

$42 = (12 \times 3) + 6,$

a remainder of $6$, so $42$ is not divisible by $12$. The second proposal would declare $42$ divisible by $12$, so it is wrong.

The moral is not that one pairing got lucky. A combined test is only as good as the independence of its two halves: the two divisors must share no factor above $1$, or one of them is quietly doing the other's work and the test is weaker than it looks.