Place Value and the Number System: Free Response

Count the columns from the right. The final $7$ sits in the ones place, the $2$ before it in the tens place, the next $7$ in the hundreds place, and the leading $2$ in the thousands place.

A digit's value is the digit multiplied by the value of its column:

$2 \times 1{,}000 = 2{,}000, \qquad 7 \times 100 = 700, \qquad 2 \times 10 = 20, \qquad 7 \times 1 = 7.$

The two $7$s are the same symbol, yet one contributes $700$ and the other $7$. Nothing about them differs except the column they stand in.

Each stated contribution names a column, because a contribution is a digit multiplied by that column's place value:

$3{,}000 = 3 \times 1{,}000, \qquad 500 = 5 \times 100, \qquad 9 = 9 \times 1.$

So the $3$ goes in the thousands column, the $5$ in the hundreds column and the $9$ in the ones column. The tens column contributes nothing, but a column that contributes nothing still has to be written down, and $0$ is what holds it open. Reading the four columns off in order and checking the contributions back,

$3{,}509 = 3{,}000 + 500 + 0 + 9.$

Leave that $0$ out and the numeral reads $359$, in which the $3$ contributes $300$ instead of $3{,}000$ and the $5$ contributes $50$ instead of $500$.

The student has answered each question with the other kind of thing. A place is the name of a column (ones, tens, hundreds, thousands); a value is an amount, namely the digit multiplied by that column's place value.

Count the columns of $9{,}140$ from the right: $0$ ones, $4$ tens, $1$ hundred, $9$ thousands. So the digit $1$ is in the hundreds place, and its value is

$1 \times 100 = 100.$

The student's "$1$" is the digit itself, not a place, and "the hundreds" is the name of a column, not an amount. It is worth noticing that simply exchanging the two answers does not repair the statement: that would give the place as the hundreds, which is right, but it would leave the value as $1$, and the value is $100$.

There are four digits, so there are four columns. From the right: $0$ ones, $6$ tens, $0$ hundreds, $9$ thousands. Adding only the contributions that are not zero,

$9{,}060 = 9{,}000 + 60.$

Showing every column explicitly, zeros included,

$(9 \times 1{,}000) + (0 \times 100) + (6 \times 10) + (0 \times 1).$

The second form is the one that makes the zeros visible. They add nothing to the sum, and they still occupy columns.

Each product names one column together with the digit that belongs in it: $3 \times 10{,}000$ puts a $3$ in the ten-thousands column, $8 \times 100$ puts an $8$ in the hundreds column, and $2 \times 1$ puts a $2$ in the ones column.

The thousands and the tens are never mentioned, so nothing is contributed there, and "nothing" is written $0$:

$30{,}802 = 30{,}000 + 800 + 2.$

Closing the two gaps instead would give $382$, a completely different number: the $3$ drops from the ten-thousands column to the hundreds column and the $8$ drops from the hundreds column to the tens column, leaving only the $2$ where the description placed it.

Write each numeral as the sum of its digit values:

$9{,}060 = 9{,}000 + 60, \qquad 96 = 90 + 6.$

The two sums are not equal, so the two numerals do not name one number.

The student's premise is correct and the conclusion does not follow from it. A zero digit really does contribute nothing to the sum, but contributing is not its only job: it also holds its column, and that fixes where every digit to its left lands. Delete the two zeros from $9{,}060$ and the $9$ falls from the thousands column to the tens column and the $6$ falls from the tens column to the ones column, so their contributions collapse from $9{,}000$ and $60$ to $90$ and $6$.

A stack is ten chips and a crate is ten stacks, so a crate is ten groups of ten chips:

$10 \times 10 = 100.$

One crate therefore holds $100$ chips, which is exactly what a hundreds column is worth. Now take each size on its own:

$2 \times 100 = 200, \qquad 6 \times 10 = 60, \qquad 5 \times 1 = 5.$

Those three contributions come to $265$ chips. Notice that the digits of $265$ are precisely the counts of crates, stacks and chips: the packing and the numeral are the same idea written two ways.

A crate is $100$ chips and a pallet is ten crates, so a pallet is ten groups of one hundred chips:

$10 \times 100 = 1{,}000.$

One pallet therefore holds $1{,}000$ chips, which is exactly what a thousands column is worth. Now take each size on its own:

$3 \times 1{,}000 = 3{,}000, \qquad 4 \times 100 = 400, \qquad 2 \times 10 = 20.$

Those contributions come to $3{,}420$ chips, and the digits of $3{,}420$ are once again the counts of pallets, crates, stacks and single chips.

Look at what one unit of a place is actually made of. The columns are built by bundling, so one ten is ten ones, one hundred is ten tens, and one thousand is ten hundreds. A single unit of any column is therefore a bundle of ten units of the column on its right, which is to say its value is ten times as much:

$10 = 10 \times 1, \qquad 100 = 10 \times 10, \qquad 1{,}000 = 10 \times 100.$

That reason did not depend on which two columns we picked, so it holds at every pair of neighbouring columns. Chips, stacks, crates and pallets are those same four columns packed in a warehouse.

The ten itself comes from the size of the bundle, not from the digits. We bundle in tens, so each column ends up worth ten of the column to its right, and that is exactly what the name base ten records.

Take the periods one at a time. The words name $406$ in the millions period, $90$ in the thousands period and $25$ in the ones period.

Every period except the leading one must be filled out to three digit positions, so $90$ is written $090$ and $25$ is written $025$:

$406{,}090{,}025.$

Without those placeholder zeros the digits would run together as $4{,}069{,}025$, which is a far smaller number: the $4$ would sit in the millions column instead of the hundred-millions column.

The commas have already marked the periods off in threes from the right: $18$ in the millions period, $304$ in the thousands period and $060$ in the ones period.

$18{,}304{,}060 = 18{,}000{,}000 + 304{,}000 + 60.$

Read each period as an ordinary number of at most three digits followed by its period name, and read the last period with no name at all: eighteen million, three hundred four thousand, sixty. The unit goes on the end, so the capacity is eighteen million, three hundred four thousand, sixty liters.

The middle group is the one attached to the word thousand, so $304$ occupies the thousands period, and its $3$ is a hundred-thousands digit.

The ones place is the anchor. Place values are built outward from it, one bundle of ten at a time, so the rightmost column is the only one whose position is fixed before you know how long the numeral is. Counting periods from the right therefore counts from the one column that every number has, whatever its length. Counting from the left has nothing fixed to start from, because the leading column changes as soon as the number gets longer.

On the given figure, counting from the left produces the groups $183$, $040$ and $60$:

$183{,}040{,}60$

whose last group holds two digits instead of three, which is the first sign that the grouping is wrong. Read as periods, those groups would announce one hundred eighty-three million, when the figure contains only eighteen million, because the groups no longer line up with the columns the digits actually occupy.

Digit count comes first, since none of these numerals is padded with leading zeros. $7{,}481$ has four digits and the other three have five, so it is the smallest of the four: every five-digit number is at least $10{,}000$, and $7{,}481$ is below that.

Among the five-digit numbers, compare columns from the left. All three begin with $7$ and then $0$, so the ten-thousands and the thousands columns tie and the hundreds column decides first: $1$ in $70{,}148$ against $4$ in each of the others, so $70{,}148$ is the least of the three. Between $70{,}418$ and $70{,}481$ the hundreds tie as well, so move one column right, to the tens: $1$ against $8$. That settles it, and the ones column is never needed:

$7{,}481 < 70{,}148 < 70{,}418 < 70{,}481.$

The two numerals have the same number of digits, so compare from the left. The thousands column is the first one that differs: $4$ against $5$. Split each number at that column:

$4{,}963 = 4{,}000 + 963, \qquad 5{,}008 = 5{,}000 + 8.$

The three columns below the thousands are the hundreds, the tens and the ones, and the largest digit any of them can hold is $9$. So the most those three columns can ever contribute together is

$900 + 90 + 9 = 999,$

which is less than $1{,}000$, the value of one unit of the deciding column. The $963$ carried by $4{,}963$ therefore cannot close the gap of $1{,}000$ that the thousands column opened, and no arrangement of lower digits could:

$4{,}963 < 5{,}008.$

The digits $9$, $6$ and $3$ look impressive beside $0$, $0$ and $8$, but they sit in columns that are worth too little to decide anything here.

Take the pair $99$ and $100$, and add the digits of each:

$9 + 9 = 18, \qquad 1 + 0 + 0 = 1.$

The shortcut compares $18$ with $1$ and announces that $99$ is the larger number. But $100$ has three digits and $99$ has two, so $100$ is larger, and one pair like this is enough to retire the shortcut.

What the shortcut throws away is the link between each digit and the column it stands in, and that link is what fixes how much the digit contributes. A digit sum treats a $9$ in the tens column and a $9$ in the ones column as the same contribution, when the first is worth ten times the second.