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Grade 5

Numbers in other bases

We bundle in tens because we have ten fingers. What happens to place value if you bundle in fives, or in twos, instead?

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Expanded formGrade 2
Where you are 15 in Place value
Number9
Place value15
▴ 10 aboveRounding3Multiplying and dividing by ten, one hundred and one thousand
Numbers in other bases
Dividing a decimal by ten, one hundred or one thousandMultiplying a decimal by ten, one hundred or one thousand

Why this exists

Our numbers are built on bundles of ten. Ten ones make a ten, ten tens make a hundred, and a digit says how many of each bundle there are, as Tens and ones and Expanded form show. Ten is a natural choice because we have ten fingers. But nothing in the rule needs the number ten.

Chayton has a box of drinking straws and a bag of rubber bands. He puts a band around every five straws. When he has five bundles, he puts a bigger band around those, making one big bundle. He can then write down how many big bundles, bundles and loose straws he has, and that list of digits names the number of straws exactly. This is place value with a bundle size of five.

Other bundle sizes are in use all around us. A computer stores every number with only two signals, on and off, so it bundles in twos. A clock bundles sixty seconds into a minute and sixty minutes into an hour. Seeing place value with a different bundle size also shows what was special about ten all along: nothing, except that it is the size we chose.

Definition

The bundle size is called the base. The base can be any whole number 2 or more.

In base five, the places from right to left are worth:

  • 1 (a loose straw, the ones),
  • 5 (a bundle of five),
  • 25 = 5 \times 5 (a big bundle: five bundles of five),
  • 125 = 5 \times 5 \times 5 (five big bundles), and so on.

Each place is worth five of the place to its right. A digit tells how many of that place there are, and in base five the only digits are 0, 1, 2, 3 and 4.

To show which base is meant, the base is written small and low after the digits, as a word. So 23_{\text{five}} means 2 fives and 3 ones, which is

23_{\text{five}} = 2 \times 5 + 3 = 13.

A number written with no small word is in base ten, as always. In any base, the value of a numeral is its expanded form: each digit times the value of its place, all added.

In base two, also called binary, the places are 1, 2, 4, 8, 16, 32 and so on, each twice the one to its right, and the only digits are 0 and 1.

Intuition

Start with thirteen straws. Chayton bands five of them, then five more. Three are left, too few for another band. He has 2 bundles and 3 loose straws, and he writes 23_{\text{five}}.

2 fives 3 ones 2 3
Figure 1Thirteen straws bundled in fives. The digits read 23_{\text{five}}.

The same straws in base ten would be one ten and three ones, written 13. The straws did not change. Only the bundle size did, and so the digits changed.

Now take thirty-eight straws. Bundling in fives gives seven bundles and three loose straws. But seven is more than four, and five bundles must be tied into a big bundle. So Chayton ties five of the seven together. He has 1 big bundle, 2 bundles and 3 loose straws:

\begin{aligned} 38 &= 1 \times 25 + 2 \times 5 + 3 \\ &= 123_{\text{five}}. \end{aligned}
1 twenty-five 2 fives 3 ones 1 2 3
Figure 2Five bundles make one big bundle. Thirty-eight straws are 123_{\text{five}}.

A place value chart keeps the kinds of bundle in columns, just like the base ten The place value chart. The only change is the heading of each column.

125 25 2 5 1 1 4 ×5 ×5 ×5
Figure 3The base five chart. Each place is worth five of the place to its right.

Linh uses an abacus. On a base ten abacus each rod holds up to nine counted beads, because a tenth bead would be traded for one bead on the next rod. A base five abacus needs only four counted beads per rod. A base two abacus needs just one bead per rod: pushed up it counts, resting down it does not.

Results

Place values. In base b, the places from right to left are worth 1, b, b \times b, b \times b \times b, and so on. Each place is worth b of the place to its right.

Digits. In base b the digits are 0, 1, 2, \ldots up to one less than b. Base five uses 0 to 4. Base two uses only 0 and 1.

One way only. Every whole number can be written in a given base, and in only one way (leaving out zeros at the front).

Ten in its own base. In every base, 10 means one bundle and no ones, so it is the base itself: 10_{\text{five}} = 5 and 10_{\text{two}} = 2. In the same way 100 in any base is one bundle of bundles: 100_{\text{five}} = 25.

Why it is true

Why each place is b times the last. A new place is made by tying b of the previous kind together. In base five, a big bundle is five bundles, and each bundle is five straws, so a big bundle holds 5 \times 5 = 25 straws. The next band holds five big bundles, or 5 \times 25 = 125 straws.

Why no digit reaches the base. Bundling stops only when no place holds a full bundle. If a place held 5 or more in base five, five of them would be tied into one of the next place. So when bundling is finished, every place holds 0, 1, 2, 3 or 4. In base two a place holding 2 is tied up at once, so only 0 and 1 are left.

Why there is only one way. Write a number in base five with the biggest place first. The digit in the biggest place has to be as large as it can be. If it were smaller, the places to its right would have to make up a whole bundle of that biggest kind, and they cannot: even with every digit 4, they add to one less than that bundle. For example 44_{\text{five}} = 4 \times 5 + 4 = 24, one less than 25. The same reasoning fixes each digit in turn. So bundling a pile always ends the same way, whoever does it.

Worked examples

The lesson is taught in two parts. Examples 1 to 4 belong to the first: bundling in fives and moving between base five and base ten. Examples 5 to 7 belong to the second: base two, and bundles of other sizes such as sixty in time.

Example 1

Rodrigo reads a base five number

Rodrigo writes 214_{\text{five}} on the board. How many straws is that?

Solution

Read the places from the right: ones, fives, twenty-fives. The numeral says 2 big bundles of twenty-five, 1 bundle of five and 4 loose straws. Write it in expanded form and add:

\begin{aligned} 214_{\text{five}} &= 2 \times 25 + 1 \times 5 + 4 \\ &= 50 + 5 + 4 = 59. \end{aligned}

So 214_{\text{five}} is 59 straws. This is the number in the chart above.

Example 2

Bundling eighty-seven straws

Linh counts 87 straws. Write 87 in base five.

Solution

The places are 1, 5, 25, 125. Since 125 is more than 87, there are no bundles of 125, and the biggest place used is 25.

How many twenty-fives fit in 87? Three, since 3 \times 25 = 75 and 4 \times 25 = 100 is too much. That leaves 87 - 75 = 12 straws. Two fives fit in 12, leaving 12 - 10 = 2 loose. So

\begin{aligned} 87 &= 3 \times 25 + 2 \times 5 + 2 \\ &= 322_{\text{five}}. \end{aligned}

There is a second way that does the bundling exactly as Linh would. Band the straws in fives: 87 \div 5 = 17 remainder 2, so 17 bundles and 2 loose. Then tie the bundles in fives: 17 \div 5 = 3 remainder 2, so 3 big bundles and 2 bundles left over. The remainders, read from the last back to the first, with the final 3 in front, give 322_{\text{five}} again.

Check: 3 \times 25 + 2 \times 5 + 2 = 75 + 10 + 2 = 87.

Example 3

A zero in the middle

Write 55 in base five.

Solution

Two twenty-fives make 50, and three would be 75, too much. That leaves 55 - 50 = 5, which is exactly one bundle of five and no loose straws. So

\begin{aligned} 55 &= 2 \times 25 + 1 \times 5 + 0 \\ &= 210_{\text{five}}. \end{aligned}

The 0 matters. Without it, 21_{\text{five}} would mean 2 \times 5 + 1 = 11, a different number. A zero holds a place open, in any base.

Example 4

Counting on past a full place

Chayton has 344_{\text{five}} straws and finds one more. What is the new number in base five?

Solution

Add the one straw to the ones place: 4 + 1 = 5 loose straws. Five loose straws are a full bundle, so he bands them. The ones place becomes 0 and the fives place gains one bundle: 4 + 1 = 5 bundles. That is a full big bundle, so the fives place becomes 0 and the twenty-fives place gains one: 3 + 1 = 4. Four is allowed, so the bundling stops.

344_{\text{five}} + 1 = 400_{\text{five}}.

Check in base ten: 344_{\text{five}} = 75 + 20 + 4 = 99 and 400_{\text{five}} = 4 \times 25 = 100. This is just like 99 + 1 = 100 in base ten, where a full place rolls over to the next.

Example 5

Reading Linh’s abacus

Linh’s binary abacus shows 1011_{\text{two}}, as in the figure. What number is it?

8 1 4 0 2 1 1 1
Figure 4On a base two abacus each rod has one bead, up or down. This one shows 1011_{\text{two}}.
Solution

In base two the places from the right are 1, 2, 4, 8. A bead pushed up means one of that place; a bead down means none. The rods for 8, 2 and 1 are up, the rod for 4 is down:

\begin{aligned} 1011_{\text{two}} &= 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1 \\ &= 8 + 2 + 1 = 11. \end{aligned}

So the abacus shows 11. Since every digit is 0 or 1, reading a binary number is only adding up the places that have a 1.

Example 6

Twenty-six in base two

Write 26 in base two.

Solution

The places are 1, 2, 4, 8, 16, 32. Since 32 is too big, start at 16. Take 16, leaving 26 - 16 = 10. Take 8, leaving 2. Skip 4, since 4 is more than 2. Take 2, leaving 0. Skip 1. Write 1 for each place taken and 0 for each place skipped, biggest place first:

26 = 16 + 8 + 2 = 11010_{\text{two}}.

The bundling way works here too, with bundles of two. Halve and keep the remainders: 26 \div 2 = 13 remainder 0, 13 \div 2 = 6 remainder 1, 6 \div 2 = 3 remainder 0, 3 \div 2 = 1 remainder 1, and the last 1 goes in front. Read back from the end: 11010_{\text{two}}.

Base two takes five digits for a number that base ten writes with two. Small bundles mean many places.

Example 7

Sixty in a bundle: hours, minutes, seconds

A video is 7{,}384 seconds long. Write its length in hours, minutes and seconds.

Solution

Time bundles in sixties. Sixty seconds make a minute, and sixty minutes make an hour, so an hour is 60 \times 60 = 3{,}600 seconds. The places are 1 (seconds), 60 (minutes) and 3{,}600 (hours).

Two hours fit, since 2 \times 3{,}600 = 7{,}200 and three hours would be 10{,}800. That leaves 7{,}384 - 7{,}200 = 184 seconds. Three minutes fit in 184, since 3 \times 60 = 180, leaving 4 seconds. So the video is 2 hours, 3 minutes and 4 seconds:

7{,}384 = 2 \times 3{,}600 + 3 \times 60 + 4.

This is a base sixty numeral with digits 2, 3, 4, written the way clocks write it, 2{:}03{:}04. Each “digit” can run from 0 to 59, one less than the bundle size, just as in every base. Counting in sixties is very old: scribes in ancient Babylon wrote numbers in a system built on sixty, and our minutes and seconds still follow it. The Maya of southern Mexico and Central America wrote numbers in a system built mostly on twenty.

Boundaries

The base is a whole number, 2 or more. With a bundle size of one, every straw would be its own bundle and the bundling would never end; that is just a tally, with no places, as in tally marks.

A digit must be less than the base. The string 35_{\text{five}} is not a base five numeral: a place holding five would already have been tied into the next place. In base two the only digits are 0 and 1.

A base above ten needs new digits. Base twelve needs single symbols for ten and eleven, since “10” already means one twelve and no ones. The usual digits 0 to 9 only reach nine.

The same digits name different numbers in different bases. 101_{\text{two}} = 5, while 101_{\text{five}} = 26 and 101 in base ten is a hundred and one. Without the small base word the digits alone do not say which.

This lesson keeps to whole numbers. Places to the right of a point also exist in other bases, worth fifths and twenty-fifths in base five, just as base ten has tenths, hundredths and Thousandths: reading, writing and expanded form.

Where this goes next

The base ten places 1, 10, 100, 1{,}000 are each ten times the last, and that pattern is written with exponents in Powers of ten. The same exponents name the places in any base: 25 = 5^2, 125 = 5^3, 16 = 2^4. The rule that each place is worth the base times the place to its right is what makes multiplying by ten shift every digit one place left; in base five, writing a 0 on the end multiplies by five. Base two is the language every computer works in, and base sixteen is used to write long binary numbers in short form.

Exercises

Exercises 1 to 11 go with the first part of the lesson (Examples 1 to 4), Exercises 12 to 21 with the second (Examples 5 to 7), and the reasoning group draws on both.

Part 1: straws in base five

  1. Chayton has 4 bundles of five and 1 loose straw. Write the number in base five, then say how many straws he has.
  2. Linh has 3 big bundles of twenty-five, no bundles of five and 4 loose straws. Write the number in base five, then say how many straws she has.
  3. Write 32_{\text{five}} in base ten.
  4. Write 140_{\text{five}} in base ten.
  5. Write 1000_{\text{five}} in base ten. What does it mean in straws?
  6. Write 2431_{\text{five}} in base ten.
  7. Write 9 in base five.
  8. Write 24 in base five.
  9. Write 50 in base five.
  10. Write 130 in base five.
  11. Rodrigo has 1044_{\text{five}} straws and finds one more. Write the new number in base five, and check both numbers in base ten.

Part 2: base two, and other bundles

  1. Write 101_{\text{two}} in base ten.
  2. Write 1111_{\text{two}} in base ten.
  3. Write 100000_{\text{two}} in base ten.
  4. Write 110101_{\text{two}} in base ten.
  5. Write 13 in base two.
  6. Write 40 in base two.
  7. Write 63 in base two.
  8. Rodrigo’s binary abacus has five rods worth 16, 8, 4, 2 and 1. The beads are up on the rods for 16, 4 and 2. Write the number in base two and in base ten.
  9. A movie runs 250 minutes. Write its length in hours and minutes.
  10. Eggs are bundled in twelves: twelve eggs make a dozen and twelve dozen make a gross. A store has 1 gross, 3 dozen and 5 eggs. Write this as a base twelve numeral and find the number of eggs.

Reasoning

  1. Rodrigo writes 35_{\text{five}}. Explain why this is not a base five numeral, and write the number he probably meant.
  2. Linh says that 10 written in any base is always the base itself. Explain why she is right, and say what 100 means in base b.
  3. Explain why a number written in base two is even when its last digit is 0 and odd when its last digit is 1.
  4. Writing a 0 at the end of 23_{\text{five}} gives 230_{\text{five}}. Find both numbers in base ten, and explain why putting a 0 at the end of a base five numeral always multiplies the number by five.
Answers
  1. 41_{\text{five}}, which is 4 \times 5 + 1 = 21 straws.
  2. 304_{\text{five}}, which is 3 \times 25 + 0 \times 5 + 4 = 79 straws. The 0 holds the empty fives place.
  3. 17. 32_{\text{five}} = 3 \times 5 + 2 = 17.
  4. 45. 140_{\text{five}} = 1 \times 25 + 4 \times 5 + 0, which is 25 + 20 = 45.
  5. 125. 1000_{\text{five}} is one band around five big bundles, 5 \times 5 \times 5 = 125 straws, and nothing else.
  6. 366. 2431_{\text{five}} = 2 \times 125 + 4 \times 25 + 3 \times 5 + 1, which is 250 + 100 + 15 + 1 = 366.
  7. 14_{\text{five}}. One bundle of five fits in 9, leaving 4: 9 = 1 \times 5 + 4.
  8. 44_{\text{five}}. 24 = 4 \times 5 + 4, and four is the most any place can hold, so 24 is the largest two-digit base five number.
  9. 200_{\text{five}}. 50 = 2 \times 25, with nothing left for the fives or the ones.
  10. 1010_{\text{five}}. One 125 leaves 5; no twenty-fives fit; one five leaves 0. So 130 = 1 \times 125 + 0 \times 25 + 1 \times 5 + 0.
  11. 1100_{\text{five}}. The ones roll over (4 + 1 = 5, a full bundle), then the fives roll over (4 + 1 = 5, a full big bundle), and the twenty-fives place goes from 0 to 1. Check: 1044_{\text{five}} = 125 + 20 + 4 = 149 and 1100_{\text{five}} = 125 + 25 = 150.
  12. 5. 101_{\text{two}} = 4 + 1 = 5.
  13. 15. 1111_{\text{two}} = 8 + 4 + 2 + 1 = 15.
  14. 32. The places are 1, 2, 4, 8, 16, 32, and only the 32 place holds a 1.
  15. 53. 110101_{\text{two}} = 32 + 16 + 4 + 1 = 53.
  16. 1101_{\text{two}}. 13 = 8 + 4 + 1: take 8, take 4, skip 2, take 1.
  17. 101000_{\text{two}}. 40 = 32 + 8: take 32, skip 16, take 8, skip 4, 2 and 1.
  18. 111111_{\text{two}}. 63 = 32 + 16 + 8 + 4 + 2 + 1, one less than 64, so every place is used.
  19. 10110_{\text{two}}, which is 16 + 4 + 2 = 22.
  20. 4 hours and 10 minutes. Four hours are 4 \times 60 = 240 minutes, five would be 300, and 250 - 240 = 10.
  21. 135_{\text{twelve}}, which is 1 \times 144 + 3 \times 12 + 5 = 144 + 36 + 5 = 185 eggs. A gross is 12 \times 12 = 144.
  22. In base five a place can hold at most 4; a place holding five would have been tied into one of the next place. Three bundles and five loose straws is really four bundles and none loose, 40_{\text{five}} = 20, which is probably what he meant.
  23. In base b, 10 means one bundle of the base and no ones, so its value is 1 \times b + 0 = b. And 100 means one bundle of bundles: 1 \times b \times b, for example 100_{\text{five}} = 25 and 100_{\text{two}} = 4.
  24. Every place except the ones is worth 2, 4, 8 or more, all even, so those places add to an even number. The last digit then adds 0 (the number stays even) or 1 (the number becomes odd).
  25. 23_{\text{five}} = 13 and 230_{\text{five}} = 2 \times 25 + 3 \times 5 = 65, and 65 = 5 \times 13. The 0 pushes every digit one place left, and each place is worth five times the place to its right, so every digit is now worth five times as much, and so is the whole number.
kg1128 · updated 2026-10-19 · CC-BY-SA-4.0