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Place Value and the Number System: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Finding a contribution

The digit $6$ appears twice in $681{,}462$ . What does each $6$ contribute to the number?
1. Hint 1
  A digit's contribution is the digit multiplied by the value of its place.
2. Hint 2
  Name the place of each $6$ by counting places from the right.
Answer

The left $6$ contributes $600{,}000$ ; the right $6$ contributes $60$ .

Full solution

From the right, the places are ones, tens, hundreds, thousands, ten-thousands and hundred-thousands.
The left $6$ occupies the hundred-thousands place.
$6 \times 100{,}000 = 600{,}000$

The right $6$ occupies the tens place.
$6 \times 10 = 60$
The same digit contributes different amounts because it sits in different places.

Answer

The left $6$ contributes $600{,}000$ ; the right $6$ contributes $60$ .

Key idea

A digit's contribution equals the digit multiplied by its place value.
Problem 1: Mark complete
Problem 2 Writing a large count

Write three billion, fourteen as a numeral with commas separating the periods.
1. Hint 1
  The periods, from right to left, are ones, thousands, millions and billions.
2. Hint 2
  Keep three digits for every period after the leading period, even when an entire period has no nonzero digits.
Answer

$3{,}000{,}000{,}014$ .

Full solution

The billions period is $3$ .
The words name no millions and no thousands, so the millions and thousands periods are each written as $000$ .
Fourteen fills the ones period as $014$ .

Writing the periods in order gives
$3{,}000{,}000{,}014$
The nonzero periods read as three billion and fourteen, matching the words.

Answer

$3{,}000{,}000{,}014$ .

Key idea

Each period after the leading period uses three digits, including zeros for empty places.
Problem 2: Mark complete
Problem 3 Comparing two counts

Compare $605{,}090$ and $650{,}009$ by writing a true statement with $<$ , $>$ or $=$ .
1. Hint 1
  For whole numbers with the same number of digits, compare places from left to right.
2. Hint 2
  The hundred-thousands digits match; inspect the next place before looking at the smaller places.
Answer

$605{,}090 < 650{,}009$ .

Full solution

Both numbers have six digits and have $6$ in the hundred-thousands place.
The first difference is in the ten-thousands place: the first number has $0$ , and the second has $5$ .

The smaller digit in that first differing place determines the comparison.
$605{,}090 < 650{,}009$
The digits farther right do not change that decision.

Answer

$605{,}090 < 650{,}009$ .

Key idea

The first differing place from the left decides the order of two whole numbers with the same digit count.
Problem 3: Mark complete
Problem 4 Restoring the separators

A file records a whole-number count as $31006020$ , with no commas. Insert commas between its periods, then write the number in words.
1. Hint 1
  Period boundaries are found by counting groups of three digits from the right.
2. Hint 2
  Name the rightmost group as ones and the next group as thousands before reading the leading group.
Answer

$31{,}006{,}020$ ; thirty-one million, six thousand, twenty.

Full solution

Starting at the right, the first group is $020$ , the next is $006$ , and the remaining leading group is $31$ .
They are the ones, thousands and millions periods, respectively.

The numeral with separators is
$31{,}006{,}020$
Read the nonzero amounts in those periods as thirty-one million, six thousand, twenty.
Removing the commas reproduces the original digit sequence.

Answer

$31{,}006{,}020$ ; thirty-one million, six thousand, twenty.

Key idea

Grouping digits into periods of three from the right connects a large numeral to its spoken name.
Problem 4: Mark complete
Problem 5 Rebundling a stockroom count

A stockroom has $3$ sealed cartons of $1{,}000$ tokens each, $17$ packets of $100$ tokens each, $4$ strips of $10$ tokens each, and $6$ loose tokens. Write the total number of tokens as a numeral and in expanded form.
1. Hint 1
  Ten groups of one place value can be exchanged for one group of the next larger place value.
2. Hint 2
  Exchange ten of the hundred-token packets for one thousand-token group, then count how many groups remain at each place.
Answer

$4{,}746$ tokens; expanded form: $4{,}000 + 700 + 40 + 6$ , or $(4 \times 1{,}000) + (7 \times 100) + (4 \times 10) + (6 \times 1)$ .

Full solution

Ten of the $17$ hundred-token packets make one thousand-token group.
$10 \times 100 = 1{,}000$
This gives $4$ thousands in total and leaves $7$ hundreds, along with $4$ tens and $6$ ones.

Add the contributions of the four places.
$4{,}746 = 4{,}000 + 700 + 40 + 6$
The stockroom has $4{,}746$ tokens.

Check the rebundling: the extra thousand and the remaining $7$ hundreds account for all $17$ original packets.
The strips and loose tokens have not changed.

Answer

$4{,}746$ tokens; expanded form: $4{,}000 + 700 + 40 + 6$ , or $(4 \times 1{,}000) + (7 \times 100) + (4 \times 10) + (6 \times 1)$ .

Key idea

Rebundling ten units of one place as one unit of the next place preserves the total.
Problem 5: Mark complete
Problem 6 Two reported totals

One counter reports $6{,}040{,}080$ visits. Another report gives six million, four hundred thousand, eighty visits. Write the second count as a numeral, decide which report gives the larger count, and name the first place from the left where the counts differ.
1. Hint 1
  Put both counts in the same written form before comparing their places.
2. Hint 2
  Write the spoken count in millions, thousands and ones periods, giving every period after the first exactly three digits.
3. Hint 3
  After comparing the millions digits, inspect the leftmost digit in the thousands period.
Answer

Second count: $6{,}400{,}080$ visits. The second report is larger; the first difference is in the hundred-thousands place.

Full solution

The spoken count has millions period $6$ , thousands period $400$ , and ones period $080$ .
Its numeral is $6{,}400{,}080$ .

The millions digits match.
The next place, hundred-thousands, contains $0$ in the first count and $4$ in the second.
$6{,}040{,}080 < 6{,}400{,}080$
The second report gives the larger count.
Reading its periods back gives six million, four hundred thousand, eighty, as required.

Answer

Second count: $6{,}400{,}080$ visits. The second report is larger; the first difference is in the hundred-thousands place.

Key idea

Translate number words into place-value periods before comparing them with a numeral.
Problem 6: Mark complete
Problem 7 An incomplete counter reading

A six-digit counter begins with the digits $284$ , but its last three digits are unreadable. What are the smallest and largest possible readings? Could its reading be greater than $285{,}000$ ? Explain using place value.
1. Hint 1
  The known digits fix the higher places, while each unreadable digit can range from $0$ to $9$ .
2. Hint 2
  Fill the last three places with the smallest digits for one endpoint and the largest digits for the other.
3. Hint 3
  Compare the largest possible reading with $285{,}000$ , starting at the left.
Answer

Smallest: $284{,}000$ . Largest: $284{,}999$ . A reading greater than $285{,}000$ is not possible.

Full solution

The first three digits contribute $284{,}000$ .
The last three places contribute at least $0$ and at most $999$ .
$\begin{aligned} 284{,}000 + 0 &= 284{,}000 \\ 284{,}000 + 999 &= 284{,}999 \end{aligned}$

Compare the largest reading with $285{,}000$ .
Their first two digits match, and the thousands digits are $4$ and $5$ .
$284{,}999 < 285{,}000$
So no possible reading is greater than $285{,}000$ .
Both endpoints have six digits and begin with $284$ , as required.

Answer

Smallest: $284{,}000$ . Largest: $284{,}999$ . A reading greater than $285{,}000$ is not possible.

Key idea

Fixed leading digits limit how much the unknown lower places can change a whole number.
Problem 7: Mark complete
Problem 8 An empty middle period

A student reads $8{,}000{,}045$ as eight thousand, forty-five, saying that the middle group can be skipped since it contains only zeros. Is the reading correct? Give the correct reading and explain the role of the middle group.
1. Hint 1
  A period with no spoken amount still occupies three places in the written numeral.
2. Hint 2
  Name the periods from the right and locate the period containing the $8$ .
Answer

No; eight million, forty-five. The middle group marks an empty thousands period.

Full solution

From the right, $045$ is the ones period, $000$ is the thousands period, and $8$ is the millions period.
The number therefore reads as eight million, forty-five.
$8{,}000{,}045 = 8{,}000{,}000 + 45$

The empty thousands period keeps the $8$ in the millions period.
It has no amount to say aloud, but its positions still count.
Deleting its three digits would produce $8{,}045$ , which does read as eight thousand, forty-five.
$8{,}045 = 8{,}000 + 45$
That is a different number.

Answer

No; eight million, forty-five. The middle group marks an empty thousands period.

Key idea

An empty period contributes no separate amount but preserves the positions of higher periods.
Problem 8: Mark complete
Problem 9 Bundles, trays and boxes

A collector packs cards in bundles of ten cards. Ten bundles fill a tray, and ten trays fill a box. A student says a box holds $20$ times as many cards as a bundle, because a box is two packing steps above a bundle and $10 + 10 = 20$ . Is the student correct? State how many cards a bundle and a full box hold, and explain how their sizes compare.
1. Hint 1
  Work out what one packing step does to the number of cards, then do it again for the second step.
2. Hint 2
  Count the cards in a full tray, then in a full box, and compare the box with a single bundle.
Answer

No. A bundle holds $10$ cards; a full box holds $1{,}000$ cards. A box holds $100$ times as many cards as a bundle.

Full solution

A tray holds ten bundles of ten cards.
$10 \times 10 = 100$
A full tray holds $100$ cards.

A box holds ten of those trays.
$10 \times 100 = 1{,}000$
A full box holds $1{,}000$ cards.

A bundle holds $10$ cards, and $1{,}000$ cards make $100$ bundles.
$100 \times 10 = 1{,}000$
So a box holds $100$ times as many cards as a bundle, not $20$ times.

Each packing step multiplies the count by ten, so two steps multiply it by ten twice.
The student added the two factors of ten instead of multiplying them.

Answer

No. A bundle holds $10$ cards; a full box holds $1{,}000$ cards. A box holds $100$ times as many cards as a bundle.

Key idea

Two steps of bundling in tens multiply a size by ten twice, which is $100$ times, not $20$ times.
Problem 9: Mark complete
Problem 10 Does the final digit decide

Two whole numbers each have six digits and begin with $63$ . One ends in $9$ , and the other ends in $0$ . Ari says the number ending in $9$ must be greater. Is that guaranteed? If it is, explain why; if it is not, give a pair of numbers meeting the conditions for which Ari is wrong, and justify their comparison.
1. Hint 1
  The shared leading digits and the specified final digits still leave other places available.
2. Hint 2
  Try making the thousands digits different, then fill unused places with zeros.
3. Hint 3
  Compare your numbers from the left; the first differing place decides whether the ones digits matter.
Answer

No; for example, $630{,}009 < 631{,}000$ .

Full solution

Choose $630{,}009$ for the number ending in $9$ , and $631{,}000$ for the number ending in $0$ .
Both have six digits and begin with $63$ , so they meet every condition.

The hundred-thousands and ten-thousands places match.
In the thousands place, the number ending in $9$ has $0$ , while the other has $1$ .
$630{,}009 < 631{,}000$
The lower places do not overturn this difference.

This pair makes the number ending in $9$ smaller, so Ari's claim is not guaranteed.
Other pairs can work as counterexamples too.

Answer

No; for example, $630{,}009 < 631{,}000$ .

Key idea

A larger ones digit does not determine the larger whole number when a higher place differs.
Problem 10: Mark complete

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