Place Value and the Number System: Practice

Read the columns from the right: $2$ ones, $8$ tens, $5$ hundreds. The $5$ sits in the hundreds place, so multiply it by $100$.

$5 \times 100 = 500$

Read from the right: $9$ is ones, $0$ is the next column, $3$ is hundreds.

$309 = (3 \times 100) + (0 \times 10) + (9 \times 1)$

The $0$ holds the tens place as a placeholder.

Multiply each digit by its place value: $7$ hundreds, $2$ tens, $5$ ones.

$725 = 700 + 20 + 5$

The comma marks the thousands period, so the $8$ sits in the thousands place.

$8 \times 1{,}000 = 8{,}000$

Read from the right: $1$ ones, $5$ tens, $9$ hundreds.

$951 = (9 \times 100) + (5 \times 10) + (1 \times 1)$

So the digit $5$ is in the tens place.

Read the columns from the right: $0$ ones, $7$ tens, $2$ hundreds. The $7$ sits in the tens place.

$7 \times 10 = 70$

They tie in the hundreds place ($5 = 5$), so compare the tens: $3$ against $0$.

$3 > 0 \;\Rightarrow\; 530 > 503$

Place each part in its column: $3$ thousands, $2$ hundreds, $5$ tens, $1$ one.

$3{,}000 + 200 + 50 + 1 = 3{,}251$

Read from the right: $6$ ones, $2$ tens, $8$ hundreds, $4$ thousands.

$4{,}826 = (4 \times 1{,}000) + (8 \times 100) + (2 \times 10) + (6 \times 1)$

So the digit $8$ is in the hundreds place.

The $0$ sits in the hundreds place, but a zero digit contributes nothing no matter where it sits.

$0 \times 100 = 0$

It still does the job of holding the column so the other digits stay put.

Each column is built by bundling ten units of the column to its right. Ten ones bundle into one ten, and ten tens bundle into one hundred.

$100 = 10 \times 10$

So one unit of the hundreds column is ten of the tens units, which is what ten times means. Ten more than ten would only be $20$, and where a column is printed does not set its value.

The three zeros hold the hundreds, tens, and ones places, so the $5$ sits in the thousands place.

$5 \times 1{,}000 = 5{,}000$