Chapter 4: Number Representation and CalculationContent check — Contemporary Math, Chapter 4: Number Representation and Calculation

$3\times 2^5+5\times 8^2$

Use the order of expressions (PEMDAS), which means

do the exponents first,

$3\times 32+5\times 64$

then multiplications,

$96+320$

then additions.

416

416

$5\times 7^3+2\times 7^2+5\times 7^1+3\times 7^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$5\times 343+2\times 49+5\times 7+3\times 1$

then multiplications,

$1,715+98+35+3$

then additions.

1,851

1,851

$1\times {10}^4+7\times {10}^3+4\times {10}^2+8\times {10}^1+8\times {10}^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$1\times 10,000+7\times 1,000+4\times 100+8\times 10+8\times 1$

then multiplications,

$10,000+7,000+400+80+8$

then additions.

17,488

17,488

Because there are three digits in 924, n is 3. So, start with the first digit (9) times 10 raised to n – 1, which is the second power.

Add the next digit times 10 raised to one less than the previous power.

Continue until you end with the last digit times 10⁰.

$9\times {10}^2+2\times {10}^1+4\times {10}^0$

$9\times {10}^2+2\times {10}^1+4\times {10}^0$

Because there are four digits in 1,279, n is 4. So, start with the first digit (1) times 10 raised to n – 1, which is the third power.

Add the next digit times 10 raised to one less than the previous power.

Continue until you end with the last digit times 10⁰.

$1\times {10}^3+2\times {10}^2+7\times {10}^1+9\times {10}^0$

$1\times {10}^3+2\times {10}^2+7\times {10}^1+9\times {10}^0$

Because there are seven digits in 4,130,045, n is 7. So, start with the first digit (4) times 10 raised to n – 1, which is the sixth power.

Add the next digit times 10 raised to one less than the previous power.

Continue until you end with the last digit times 10⁰.

$4\times {10}^6+1\times {10}^5+3\times {10}^4+0\times {10}^3+0\times {10}^2+4\times {10}^1+5\times {10}^0$

$4\times {10}^6+1\times {10}^5+3\times {10}^4+0\times {10}^3+0\times {10}^2+4\times {10}^1+5\times {10}^0$

$6\times {10}^2+2\times {10}^1+1\times {10}^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$6\times 100+2\times 10+1\times 1$

then multiplications,

$600+20+1$

then additions.

621

621

$3\times {10}^3+2\times {10}^2+0\times {10}^1+3\times {10}^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$3\times 1,000+2\times 100+0\times 10+3\times 1$

then multiplications,

$3,000+200+0+3$

then additions.

3,203

3,203

$4\times {10}^7+0\times {10}^6+6\times {10}^5+3\times {10}^4+0\times {10}^3+8\times {10}^2+9\times {10}^1+1\times {10}^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$4\times 10,000,000+0\times 1,000,000+6\times 100,000+3\times 10,000+0\times 1,000+8\times 100+9\times 10+1\times 1$

then multiplications,

$40,000,000+0+600,000+30,000+0+800+90+1$

then additions.

40,630,891

40,630,891

A system in which the position of a numeral determines the value associated with that numeral.

$4\times 8^2+2\times 8^1+7\times 8^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$4\times 64+2\times 8+7\times 1$

then multiplications,

$256+16+7$

then additions.

279

279

Because there are five digits in 45,209, n is 5. So, start with the first digit (4) times 10 raised to n – 1, which is the fourth power.

Add the next digit times 10 raised to one less than the previous power.

Continue until you end with the last digit times 10⁰.

$4\times {10}^3+5\times {10}^3+2\times {10}^2+0\times {10}^1+9\times {10}^0$

$4\times {10}^4+5\times {10}^3+2\times {10}^2+0\times {10}^1+9\times {10}^0$

10

A numeral is a symbol representing a number. A number is a quantity or amount.

$6\times {10}^5+0\times {10}^4+1\times {10}^3+9\times {10}^2+4\times {10}^1+7\times {10}^0$

Remember that any base raised to the exponent 0 is 1.

Use the order of expressions (PEMDAS), which means

do the exponents first,

$6\times 100,000+0\times 10,000+1\times 1,000+9\times 100+4\times 10+7\times 1$

then multiplications,

$600,000+0+1,000+900+40+7$

then additions.

601,947

601,947

This is a two-digit number.

The first digit: 21

The second digit: 9

21(60) + 9 = 1,269

1269

This is a three-digit number.

The first digit: 11

The second digit: 42

The third digit: 16

11(60²) + 42(60) + 16 = 42,136

42,136

This is a three-digit number.

The first digit: 29

The second digit: 16

The third digit: 43

29(60²) + 16(60) + 43 = 105,403

6,105,643

The top symbol represents 12.

The next symbol represents 17.

12 × 20 + 17 = 257

257

The top symbol represents 15.

The next symbol represents 2.

The next symbol represents 14.

15 × 20² + 2 × 20 + 14 = 6,054

6,054

The top symbol represents 7.

The next symbol represents 16.

The next symbol represents 0.

The next symbol represents 3.

The next symbol represents 13.

7 × 20⁴ + 16 × 20³ + 0 × 20² + 3 × 20 + 13 = 1,248,073

1,248,073

Unless a smaller digit precedes a larger digit, add the digit’s values.

L	X	X	V	I	I
50	10	10	5	1	1

Add the digit values: 50 + 10 + 10 + 5 + 1 + 1 = 77

77

Unless a smaller digit precedes a larger digit, add the digit’s values.

C	C	X	L
100	100	10	50
		Subtract 10 from 50 50 – 10 = 40

100 + 100 + 40

240

240

Unless a smaller digit precedes a larger digit, add the digit’s values.

M	M	M	C	D	X	L	V	I	I
1,000	1,000	1,000	100	500	10	50	5	1	1
			Subtract 100 from 500. 500 – 100 = 400		Subtract 10 from 50. 50 – 10 = 40

1,000 + 1,000 + 1,000 + 400 + 40 + 5 + 1 + 1

3,447

3,447

The maximum number of symbols in a row is three.

Special combinations: IV = 4, XL = 40, XC = 90, CD = 400, CM = 900

Write the larger symbols first.

27

Write 27 as a sum where there is an equivalent Roman numeral or special combination.

27 = 10 + 10 + 5 + 1 + 1

Translate to Roman numerals: XXVII

XXVII

The maximum number of symbols in a row is three.

Special combinations: IV = 4, IX = 9, XL = 40, XC = 90, CD = 400, CM = 900

Write the larger symbols first.

Write 49 as a sum where there is an equivalent Roman numeral or special combination.

49 = 10 + 10 + 10 + 10 + 5 + 1 + 1 + 1 + 1

This would have more than three symbols in a row. Look at the special combinations. You can use the ones for 40 and 9.

49 = 40 + 9

Translate to Roman numerals: XLIX

XLIX

The maximum number of symbols in a row is three.

Special combinations: IV = 4, IX = 9, XL = 40, XC = 90, CD = 400, CM = 900

Write the larger symbols first.

Write 739 as a sum where there is an equivalent Roman numeral or special combination.

739 = 500 + 100 + 100 + 10 + 10 + 10 + 9

Translate to Roman numerals: DCCXXXIX

DCCXXXIX

The maximum number of symbols in a row is three.

Special combinations: IV = 4, IX = 9, XL = 40, XC = 90, CD = 400, CM = 900

Write the larger symbols first.

Write 3,647 as a sum where there is an equivalent Roman numeral or special combination.

3,647 = 1,000 + 1,000 + 1,000 + 500 + 100 + 40 + 5 + 1 + 1

Translate to Roman numerals: MMMDCXLVII

MMMDCXLVII

60

20

Roman numerals do not use place value.

This is a two-digit number.

The first digit: 5

The second digit: 41

5(60) + 41 = 341

341

209

The top symbol represents 10.

The next symbol represents 9.

10 × 20 + 9 = 209

Unless a smaller digit precedes a larger digit, add the digit’s values.

C	C	X	L	V	I	I
100	100	10	50	5	1	1
		Subtract 10 from 50. 50 – 10 = 40

Add the digit values: 100 + 100 + 40 + 5 + 1 + 1 = 247

247

The maximum number of symbols in a row is three.

Special combinations: IV = 4, IX = 9, XL = 40, XC = 90, CD = 400, CM = 900

Write the larger symbols first.

Write 479 as a sum where there is an equivalent Roman numeral or special combination.

479 = 400 + 50 + 10 + 10 + 9

Translate to Roman numerals: CDLXXIX

CDLXXIX

0, 1, 2, 3

$0,1,2,3,4,5,6,7,8,9,\mathrm{A},\mathrm{B}$

A represents the digit 10.

B represents the digit 11.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B

The place values are powers of 6 in base 6.

${421}_6=4\times 6^2+2\times 6^1+1\times 6^0$

$4\times 36+2\times 6+1\times 1$

$144+12+1$

$157$

157

In base 14,

A represents the digit 10.

B represents the digit 11.

C represents the digit 12.

D represents the digit 13.

The place values are powers of 14 in base 14.

$\mathrm{A}3{\mathrm{C}}_{14}=10\times {14}^2+3\times {14}^1+12\times {14}^0$

$10\times 196+3\times 14+12\times 1$

$1,960+42+12$

$2,014$

2,014

In base 12,

A represents the digit 10.

B represents the digit 11.

The place values are powers of 12 in base 12.

$5\mathrm{A}{\mathrm{B}}_{12}=5\times {12}^2+10\times {12}^1+11\times {12}^0$

$5\times 144+10\times 12+11\times 1$

$720+120+11$

$851$

851

The place values are powers of 2 in base 2.

${11011}_2=1\times 2^4+1\times 2^3+0\times 2^2+1\times 2^1+1\times 2^0$

$1\times 16+1\times 8+0\times 4+1\times 2+1\times 1$

$16+8+0+2+1$

$27$

27

0, 1, 2, 3
10, 11, 12, 13
20, 21, 22, 23
30, 31, 32, 33
100

There are four digits in the base four system:

$0,1,2,3.$

When you run out of digits, you need two digits. Because the place value is 4 in the base 4 system, start with a 1 in the 4s place.

$10,11,12,13$

Now, add 2 in the 4s place.

$20,21,22,23$

Now, write 3 in the 4s place.

$30,31,32,33$

Now, you have run out of digits in the 4s place. Now, you move over to the $4^2$ place.

$100$

This is where you were told to stop.

There are 12 digits in the base four system:

$0,1,2,3,5,6,7,8,9,\mathrm{A},\mathrm{B}$

A represents the digit 10.

B represents the digit 11.

When you run out of digits in the 1s place, you need two digits. Because the place value is 12 in the base 12 system, start with a 1 in the 12s place.

$10,11,12,13,14,15,16,17,18,19,1\mathrm{A},1\mathrm{B}$

Now, add 2 in the 12s place.

$20,21,22,23,24,25,26,27,28,29,2\mathrm{A},2\mathrm{B}$

Now, write 3 in the 12s place.

$30,31,32,33,34,35,36,37,38,39,3\mathrm{A},3\mathrm{B}$

Now, write 4 in the 12s place.

$40,41,42,43,44,45,46,47,48,49,4\mathrm{A},4\mathrm{B}$

Now, write 5 in the 12s place.

$50,51,52,53,54,55,56,57,58,59,5\mathrm{A},5\mathrm{B}$

Now, write 6 in front.

$60,61,62,63,64,65,66,67,68,69,6\mathrm{A},6\mathrm{B}$

Now, write 7 in front.

$70,71,72,73,74,75,76,77,78,79,7\mathrm{A},7\mathrm{B}$

Now, write 8 in front.

$80,81,82,83,84,85,86,87,88,89,8\mathrm{A},8\mathrm{B}$

Now, write 9 in front.

$90,91,92,93,94,95,96,97,98,99,9\mathrm{A},9\mathrm{B}$

Now, write A in front.

$\mathrm{A}0,\mathrm{A}1,\mathrm{A}2,\mathrm{A}3,\mathrm{A}4,\mathrm{A}5,\mathrm{A}6,\mathrm{A}7,\mathrm{A}8,\mathrm{A}9,\mathrm{A}\mathrm{A},\mathrm{A}\mathrm{B}$

Now, write B in front.

$\mathrm{B}0,\mathrm{B}1,\mathrm{B}2,\mathrm{B}3,\mathrm{B}4,\mathrm{B}5,\mathrm{B}6,\mathrm{B}7,\mathrm{B}8,\mathrm{B}9,\mathrm{B}\mathrm{A},\mathrm{B}\mathrm{B}$

You’ve run out of digits for two-digit numbers.

Now, you move over to the ${12}^2$ place.

$100$

This is where you were told to stop.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B

20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 2A, 2B

30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 3A, 3B

40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 4A, 4B

50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 5A, 5B

60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 6A, 6B

70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 7A, 7B

80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 8A, 8B

90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 9A, 9B

A0, A1, A2, A3, A4, A5, A6, A7, A8, A9, AA, AB

B0, B1, B2, B3, B4, B5, B6, B7, B8, B9, BA, BB

100

There are three digits in the base three system:

$0,1,2.$

When you run out of digits, you need two digits. Because the place value is 3 in the base 3 system, start with a 1 in the 3s place.

$10,11,12$

Now, add 2 in the 3s place.

$20,21,22$

Now, you have run out of digits in the 3s place. Now, you move over to the $3^2$ place.

$100$

This is where you were told to stop.

0, 1, 2, 10, 11, 12, 20, 21, 22, 100

Step 1: Divide the number by 7.

$693÷7=99$ with no remainder.	$99\mathrm{R}0$
Step 2: Divide the quotient by 7.
$99÷7=14$ with a remainder. What is it?
$7\times 14=98$
$99-98=1$
The remainder is 1.	$14\mathrm{R}1$
Step 3: Divide the quotient by 7.
$14÷7=2$ with no remainder.	$2\mathrm{R}0$
Step 4: Divide the quotient by 7.
$2÷7=0$ with remainder 2.	$0\mathrm{R}2$

The base 7 number has digits equal to the remainders in reverse order: ${2010}_7$.

2010₇

Step 1: Divide the number by 12.

$9,275÷12=772$ with a remainder. What is it?

$12\times 772=9,264$

$9,275-9,264=-11$	$772\mathrm{R}11\left(\mathrm{B}\mathrm{i}\mathrm{n}\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}12\right)$
Step 2: Divide the quotient by 2.
$772÷12=64$ with a remainder. What is it?
$12\times 64=768$
$772-768=4$
The remainder is 4.	$64\mathrm{R}4$
Step 3: Divide the quotient by 12.
$64÷12=5$ with a remainder. What is it?
$12\times 5=60$
$64-60=4$	$5\mathrm{R}4$
Step 4: Divide the quotient by 12.
$5÷12=0$ with remainder 5.	$0\mathrm{R}5$

The base 12 number has digits equal to the remainders in reverse order: $544{\mathrm{B}}_{12}$.

554B₁₂

Step 1: Divide the number by 2.

$137÷2=68$ with a remainder. What is it?

$2\times 68=136$

$137-136=1$

The remainder is 1.	$68\mathrm{R}1$
In base two, any remainder will be 1.
Step 2: Divide the quotient by 2.
$68÷2=34$ with no remainder.	$34\mathrm{R}0$
Step 3: Divide the quotient by 2.
$34÷2=17$ with remainder 0.	$17\mathrm{R}0$
Step 4: Divide the quotient by 2.
$17÷2=8$ with remainder 1.	$8\mathrm{R}1$
Step 5: Divide the quotient by 2.
$8÷2=4$ with no remainder.	$4\mathrm{R}0$
Step 6: Divide the quotient by 2.
$4÷2=2$ with no remainder.	$2\mathrm{R}0$
Step 7: Divide the quotient by 2.
$2÷2=1$ with no remainder.	$1\mathrm{R}0$
Step 8: Divide the quotient by 2.
$1÷2=0$ with remainder 1.	$0\mathrm{R}1$

The base 2 number has digits equal to the remainders in reverse order: ${10001001}_2$.

10001001₂

Convert the base 10 number into base 20.

Step 1: Divide the number by 20.

$137÷20=6$ with a remainder. What is it?

$20\times 6=120$

$137-120=17$	$6\mathrm{R}17$
Step 2: Divide the quotient by 20.
$6÷20=0$ with remainder 6.	$0\mathrm{R}6$

The base 20 number has digits equal to the remainders in reverse order.

$6\times {20}^1+17\times {20}^0$

The Mayan number would have 6 in the top row and 17 in the bottom row.

In the top row, the symbol would have one bar and one dot.

In the bottom row, the symbol would have three bars and two dots.

Convert the base 10 number into base 20.

Step 1: Divide the number by 20.

$2,171÷20=108$ with a remainder. What is it?

$20\times 108=2,160$

$2,171-2,160=11$	$108\mathrm{R}11$
Step 2: Divide the quotient by 20.
$108÷20=5$ with a remainder. What is it?
$20\times 5=100$
$108-100=8$
The remainder is 8.	$5\mathrm{R}8$
Step 2: Divide the quotient by 20.
$5÷20=0$ with remainder 5.	$0\mathrm{R}5$

The base 20 number has digits equal to the remainders in reverse order,

$5\times {20}^2+8\times {20}^1+11\times {20}^0$

The Mayan number would have 5 in the top row, 8 in the middle, and 11 in the bottom row.

In the top row, the symbol would have one bar.

In the middle row, the symbol would have one bar and three dots.

In the bottom row, the symbol would have two bars and one dot.

The result has the digit 7 in it. In base 4, the 7 is an illegal symbol.

The remainders include 10, which in base 6 is an illegal symbol.

Since 12 is larger than 10, the base 10 number cannot have less digits than the base 12 number. Since it did, we know an error has been made.

A base 25 system would require 25 symbols.

Since the 4 is the second digit, its place value is 18¹ times 4, or 72.

The place values are powers of 5 in base 5.

${2304}_5=2\times 5^3+3\times 5^2+0\times 5^1+4\times 5^0$

$2\times 125+3\times 25+0\times 5+4\times 1$

$250+75+0+4$

$329$

329

There is no 9 digit in base 9, so you increase from 38₉ by moving over to the 9s place and increasing 3 to 4. The number after 38 (base 9) is 40 (base 9). 40₉

Step 1: Divide the number by 8.

$329÷8=41$ with a remainder. What is it?

$41\times 8=328$

$329-328=1$

The remainder is 1.	$41\mathrm{R}1$
Step 2: Divide the quotient by 8.
$41÷8=5$ with a remainder. What is it?
$8\times 5=40$
$41-40=1$
The remainder is 1.	$5\mathrm{R}1$
Step 3: Divide the quotient by 8.
$5÷8=0$ with remainder 5.	$0\mathrm{R}5$

The base 8 number has digits equal to the remainders in reverse order: ${511}_8$.

511₈

In base 14,

A represents the digit 10.

B represents the digit 11.

C represents the digit 12.

D represents the digit 13.

The place values are powers of 14 in base 14.

$\mathrm{A}\mathrm{B}{\mathrm{C}}_{14}=10\times {14}^2+11\times {14}^1+12\times {14}^0$

$10\times 196+11\times 14+12\times 1$

$1,960+154+12$

$2,126$

2,126

In base 4, 5 is not a valid symbol. So, a mistake has been made.

1242₆

Step 1: Do the one’s place first.

In base 10: 3 + 5 = 8 = 6 + 2

In base 6: 12

One’s place: 2

Carry: 1

Carry:		1
	4	5	3
+	3	4	5
=			2

Next step: Do the next place to the left.

In base 10: 1 + 5 + 4 = 10 = 6 + 4

In base 6: 14

This column: 4

Carry: 1

Carry:	1	1
	4	5	3
+	3	4	5
=		4	2

Next step: Do the next place to the left.

In base 10: 1 + 4 + 3 = 8 = 6 + 2

In base 6: 12

This column: 12

Carry: You don’t need to carry because there are no more columns.

Carry:	1	1
	4	5	3
+	3	4	5
=	12	4	2

${1242}_6$

There are three digits in base 4: 0, 1, 2, 3. Fill in your table headers.

+		0	1	2	3
0
1
2
3

Add what you can that does not use more than those digits.

+	0	1	2	3
0	0	1	2	3
1	1	2	3
2	2	3
3	3

For the rest, you need to use base 4 math.

1 + 3 in base 10 is 4. To write this in base 4, you need to use the four’s place: 10.

2 + 2 in base 10 is 4. That will also be 10 in base 4.

3 + 1 in base 10 is 4. That will also be 10 in base 4.

Write 10 in those three cells.

+	0	1	2	3
0	0	1	2	3
1	1	2	3	10
2	2	3	10
3	3	10

2 + 3 in base 10 is 5. To write this in base 4, you need one 4 and one 1. That is 11 in base 4.

3 + 2 in base 10 is 5. That will also be 11 in base 4.

Write 11 in those cells.

+	0	1	2	3
0	0	1	2	3
1	1	2	3	10
2	2	3	10	11
3	3	10	11

3 + 3 = 6 in base 10. To write this in base 4, you need one 4 and two 1s. That is 12 in base 4.

Write 12 in that cell.

+	0	1	2	3
0	0	1	2	3
1	1	2	3	10
2	2	3	10	11
3	3	10	11	12

+	0	1	2	3
0	0	1	2	3
1	1	2	3	10
2	2	3	10	11
3	3	10	11	12

Base 4 Addition Table

Step 1: Do the one’s place first.

In base 10: 1 + 2 = 3

In base 7: 3

One’s place: 3

Carry: No need

Carry:
	4	6	1
+	1	4	2
=			3

Next step: Do the next place to the left.

In base 10: 6 + 4 = 10 = 7 + 3

In base 7: 13

This column: 3

Carry: 1

Carry:	1
	4	6	1
+	1	4	2
=		3	3

Next step: Do the next place to the left.

In base 10: 1 + 4 + 1 = 6

In base 7: 6

This column: 6

Carry: You don’t need to carry because there are no more columns.

Carry:	1
	4	6	1
+	1	4	2
=	6	3	3

${633}_7$

633₇

+	0	1	2	3	4	5	6	7	8	9	A	B	C	D
0	0	1	2	3	4	5	6	7	8	9	A	B	C	D
1	1	2	3	4	5	6	7	8	9	A	B	C	D	10
2	2	3	4	5	6	7	8	9	A	B	C	D	10	11
3	3	4	5	6	7	8	9	A	B	C	D	10	11	12
4	4	5	6	7	8	9	A	B	C	D	10	11	12	13
5	5	6	7	8	9	A	B	C	D	10	11	12	13	14
6	6	7	8	9	A	B	C	D	10	11	12	13	14	15
7	7	8	9	A	B	C	D	10	11	12	13	14	15	16
8	8	9	A	B	C	D	10	11	12	13	14	15	16	17
9	9	A	B	C	D	10	11	12	13	14	15	16	17	18
A	A	B	C	D	10	11	12	13	14	15	16	17	18	19
B	B	C	D	10	11	12	13	14	15	16	17	18	19	1A
C	C	D	10	11	12	13	14	15	16	17	18	19	1A	1B
D	D	10	11	12	13	14	15	16	17	18	19	1A	1B	1C

Base 14 Addition Table

A represents the digit 10.

B represents the digit 11.

Step 1: Do the one’s place first.

In base 10: 3 + 6 = 9

In base 12: 9

One’s place: 9

Carry: No need

Carry:
	4	B	3
+	B	0	6
=			9

Next step: Do the next place to the left.

In base 10: 11 + 0 = 11

In base 12: B

This column: B

Carry: No need

Carry:
	4	B	3
+	B	0	6
=		B	9

Next step: Do the next place to the left.

In base 10: 4 + 11 = 15 = 12 + 3

In base 12: 13

This column: 13

Carry: You don’t need to carry because there are no more columns.

Carry:
	4	B	3
+	B	0	6
=	13	B	9

$13\text{B}{9}_{12}$

13B9₁₂

Step 1: Do the one’s place first.

In base 10: 1 + 1 = 2

In base 2: 10

One’s place: 0

Carry: 1

Carry:						1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=							0

Next step: Do the next place to the left.

1 + 1 + 1 = 3 (base 10)

In base 2: 11

This column: 1

Carry: 1

Carry:					1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=						1	0

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

Carry:				1	1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=					0	1	0

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

Carry:			1	1	1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=				0	0	1	0

Next step: Do the next place to the left.

1 + 0 + 0 = 1 (base 10)

In base 2: 1

This column: 1

Carry: No need

Carry:			1	1	1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=			1	0	0	1	0

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

Carry:	1		1	1	1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=		0	1	0	0	1	0

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 10

Carry: You don’t need to carry because there are no more columns.

Carry:	1		1	1	1	1
		1	0	1	1	1	1
+	1	1	0	0	0	1	1
=	10	0	1	0	0	1	0

${10010010}_2$

10010010₂

Step 1: Do the one’s place first.

In base 10: 5 – 3 = 2

In base 6: 2

	1	1	5
–		4	3
=			2

Next step: Do the next place to the left.

Borrow from next column.

In base 10: When you borrow from next column, (6 + 1) – 4 = 3

In base 6: 3

This column: 3

	$1$ 0	1	5
–		4	3
=		3	2

Next step: Do the next place to the left.

This column is just the remaining 0. There is nothing to write.

${1242}_6$

	$1$ 0	1	5
–		4	3
=		3	2

${32}_6$

32₆

227₁₂

A represents the digit 10.

B represents the digit 11.

Step 1: Do the one’s place first.

You need to borrow from the next place to the left.

In base 10: (12 + 6) – 11 = 18 – 11 = 7

In base 12: 7

One’s place: 7

	7	$1$ 0	6
–	4	A	B
=			7

Next step: Do the next place to the left.

You need to borrow from the next place to the left.

In base 10: (12 + 0) – 10 = 12 – 10 = 2

In base 12: 2

This column: 2

	$7$ 6	$1$ 0	6
–	4	A	B
=		2	7

Next step: Do the next place to the left.

In base 10: 6 – 4 = 2

In base 12: 2

This column: 2

	$7$ 6	$1$ 0	6
–	4	A	B
=	2	2	7

${227}_{12}$

The symbols 4 and 5 are not used in base 4.

Step 1: Do the one’s place first.

In base 10: 3 + 2 = 5

In base 4: 11

One’s place: 1

Carry: 1

Carry:		1
	1	3	3
+	1	1	2
=			1

Next step: Do the next place to the left.

In base 10: 1 + 3 + 1 = 5

In base 4: 11

This column: 1

Carry: 1

Carry:	1	1
	1	3	3
+	1	1	2
=	3	1	1

${311}_4$

The symbols 4 and 5 are not legal symbols in base 4. Careful use of the base 4 addition table would correct this error.

	1	3	3
+	1	1	2
	3	1	1

The correct answer is 311₄.

The answer is wrong because it shows the answer in base 10 but has a base indicated as base 14. In base 14,

A represents the digit 10.

B represents the digit 11.

C represents the digit 12.

D represents the digit 13.

Step 1: Do the one’s place first.

In base 10: 9 + 9 = 18 = 14 + 4

In base 14: 14

One’s place: 4

Carry: 1

Carry:		1
	1	4	9
+		1	9
=			4

Next step: Do the next place to the left.

In base 10: 1 + 4 + 1 = 6

In base 14: 6

This column: 6

Carry: No need

Carry:		1
	1	4	9
+		1	9
=		6	4

Next step: Do the next place to the left.

In base 10: Bring down the 1.

In base 14: Bring down the 1.

This column: 1

Carry: You don’t need to carry because there are no more columns.

Carry:		1
	1	4	9
+		1	9
=	1	6	4

${164}_{14}$

This is correct if the numbers are base 10 numbers, but these numbers are base 14 numbers. In base 14, 9 + 9 is not 18, but instead is 13. Careful use of the base 14 addition table generates the correct answer, ${163}_{14}$.

Step 1: Do the one’s place first.

In base 10: 4 + 3 = 7 = 6 + 1

In base 6: 11

One’s place: 1

Carry: 1

Carry:	1
	2	4
+	5	3
=		1

Next step: Do the next place to the left.

In base 10: 1 + 2 + 5 = 8 = 6 + 2

In base 6: 12

This column: 12

Carry: You don’t need to carry because there are no more columns.

Carry:	1
	2	4
+	5	3
=	12	1

${121}_6$

121₆

Step 1: Do the one’s place first.

Borrow from the next column. Change 3 in the next column to 2.

In base 10: (8 + 5) – 6 = 7

In base 8: 7

One’s place: 7

	$3$ 2	5
–	2	6
=		1

Next step: Do the next place to the left.

In base 10: 2 – 2 = 0

In base 8: 0

This column: Leave blank as it is the last column.

	$3$ 2	5
–	2	6
=		1

$7_8$

7₈

A represents the digit 10.

B represents the digit 11.

C represents the digit 12.

D represents the digit 13.

Step 1: Do the one’s place first.

In base 10: B + 5 = 11 + 5 = 16 = 14 + 2

In base 14: 12

One’s place: 2

Carry: 1

Carry:	1
	3	B
+	4	5
=		2

Next step: Do the next place to the left.

In base 10: 1 + 3 + 4 = 8

In base 14: 8

This column: 8

Carry:	1
	3	B
+	4	5
=	8	2

${82}_{14}$

82₁₄

A represents the digit 10.

B represents the digit 11.

Step 1: Do the one’s place first.

Borrow from the next column over. Change A to 9.

In base 10: (12 + 4) – B = 16 – 11 = 5

In base 12: 5

One’s place: 5

	$\text{A}$ 9	4
–	9	B
=		5

Next step: Do the next place to the left.

In base 10: 9 – 9 = 0

In base 12: 0

This column: Leave blank.

	$\text{A}$ 9	4
–	9	B
=		5

$5_{12}$

5₁₂

In base 8, 8 is not a valid symbol. So, a mistake has been made.

A common base 14 error is performing the operation in base 10.

*	0	1	2	3
0	0	0	0	0
1	0	1	2	3
2	0	2	10	12
3	0	3	12	21

Base 4 Multiplication Table

First step: You can use the base 6 multiplication table.

In base 6:

3 times 2 is 10.

4 times 2 is 12.

3 times 5 is 23.

4 times 5 is 32.

Remember to pad the entries by the appropriate number of zeros.

			4	3
		×	5	2
			1	0
Carry:
		1	2	0
		2	3	0
	3	2	0	0
Product:

Next step: In the rightmost column:

base 10: 0 + 0 + 0 + 0 = 0

base 6: 0

In the rightmost product column, write 0.

			4	3
		×	5	2
Carry:
			1	0
		1	2	0
		2	3	0
	3	2	0	0
Product:				0

Next step: In the next column:

base 10: 1 + 2 + 3 = 6

base 6: 10

In this product column, write 0.

Carry 1.

			4	3
		×	5	2
Carry:		1
			1	0
		1	2	0
		2	3	0
	3	2	0	0
Product:			0	0

Next step: In the next column:

base 10: 1 + 1 + 2 + 2 = 6

base 6: 10

In this column, write 0.

Carry 1.

			4	3
		×	5	2
Carry:	1	1
			1	0
		1	2	0
		2	3	0
	3	2	0	0
Product:		0	0	0

Next step: In the next column:

base 10: 1 + 3 = 4

base 6: 4

In this column, write 4.

			4	3
		×	5	2
Carry:	1	1
			1	0
		1	2	0
		2	3	0
	3	2	0	0
Product:	4	0	0	0

${4000}_6$

4000₆

0 × 0 = 0

0 × 1 = 0

1 × 1 = 1

			1	1	1	0	1
					×	1	1
			1	1	1	0	1
		1	1	1	0	1	0
Product:

Next step: In the rightmost column:

base 10: 1 + 0 = 1

base 2: 1

In the rightmost product column, write 1.

			1	1	1	0	1
					×	1	1
			1	1	1	0	1
		1	1	1	0	1	0
Product:							1

Next step: In the next column:

base 10: 0 + 1 = 1

base 2: 1

In this column, write 1.

			1	1	1	0	1
					×	1	1
			1	1	1	0	1
		1	1	1	0	1	0
Product:						1	1

Next step: In the next column:

base 10: 1 + 0 = 1

base 2: 1

In this column, write 1.

			1	1	1	0	1
					×	1	1
			1	1	1	0	1
		1	1	1	0	1	0
Product:					1	1	1

Next step: In the next column:

base 10: 1 + 1 = 2

base 2: 10

In this column, write 0.

Carry 1.

			1	1	1	0	1
					×	1	1
Carry:			1
			1	1	1	0	1
		1	1	1	0	1	0
Product:				0	1	1	1

Next step: In the next column:

base 10: 1 + 1 + 1 = 3

base 2: 11

In this column, write 1.

Carry 1.

			1	1	1	0	1
					×	1	1
Carry:
			1	1	1	0	1
		1	1	1	0	1	0
Product:		10	1	0	1	1	1

Next step: In the next column:

base 10: 1 + 1 = 2

base 2: 10

In this column, write 10.

You can just write it in because that is the last column.

${1010111}_2$

1010111₂

First step: You can use the base 12 multiplication table created as an example.

3 × 7 = 19

B × 7 = 65

3 × 4 = 10

B × 4 = 38

Remember to pad the entries by the appropriate number of zeros.

			B	3
		×	4	7
Carry:
			1	9
		6	5	0
		1	0	0
	3	8	0	0
Product:				9

Next step: In the rightmost column:

In base 10: 9 + 0 + 0 + 0 = 9

In base 12: 9

Write 9 in the rightmost column.

			B	3
		×	4	7
Carry:
			1	9
		6	5	0
		1	0	0
	3	8	0	0
Product:				9

Next step: In the next column over:

In base 10: 1 + 5 + 0 + 0 = 6

In base 12: 6

Write 6 in this column.

			B	3
		×	4	7
Carry:
			1	9
		6	5	0
		1	0	0
	3	8	0	0
Product:			6	9

Next step: In the next column:

In base 10: 6 + 1 + 8 = 15 = 12 + 3

In base 12: 3

Carry: 1

			B	3
		×	4	7
Carry:	1
			1	9
		6	5	0
		1	0	0
	3	8	0	0
Product:		3	6	9

Next step: In the next column:

In base 10: 1 + 3 = 4

In base 12: 4

Carry: No need

			B	3
		×	4	7
Carry:	1
			1	9
		6	5	0
		1	0	0
	3	8	0	0
Product:	4	3	6	9

${4369}_{12}$

4369₁₂

Use the multiplication table for base 6. Find 10 in the product part of the table where 3 is the column header. The answer is the row header, 2, for that cell.

The answer is 2₆.

${10}_6÷3_6=2_6$

Use the multiplication table for base 12. Find 50 in the product part of the table where A is the column header. The answer is the row header, 6, for that cell.

The answer is 6₁₂.

${50}_{12}÷{\text{A}}_{12}=6_{12}$

The symbols 4 and 5 are not legal symbols in base 4. Careful use of the base 4 multiplication table would correct this error. The correct answer is 333₄.

This is correct if the numbers are base 10 numbers, but these numbers are base 14 numbers. In base 14, ${49}_{14}\times 9_{14}=2\text{D}{\text{B}}_{14}$ is not 81, but instead is 5B. Careful use of the base 14 addition table (Table 4.9) generates the correct answer, ${49}_{14}\times 9_{14}=2\text{D}{\text{B}}_{14}$.

You can use either an addition table if you have it or repeated addition. the addition table

The process is the same, except the multiplication table for the base is used instead of the familiar base 10 rules.

the multiplication table for the base

First step: You can use the base 6 multiplication table.

In base 6:

4 times 3 is 20.

2 times 3 is 10.

4 times 5 is 32.

2 times 5 is 14.

Remember to pad the entries by the appropriate number of zeros.

			2	4
		×	5	3
Carry:
			2	0
		1	0	0
		3	2	0
	1	4	0	0
Product:

Next step: In the rightmost column:

base 10: 0 + 0 + 0 + 0 = 0

base 6: 0

Write 0 in the rightmost column.

			2	4
		×	5	3
Carry:
			2	0
		1	0	0
		3	2	0
	1	4	0	0
Product:				0

Next step: In the next column over:

In base 10: 2 + 0 + 2 + 0 = 4

In base 6: 4

In this column, write 4.

			2	4
		×	5	3
Carry:
			2	0
		1	0	0
		3	2	0
	1	4	0	0
Product:			4	0

Next step: In the next column over:

In base 10: 1 + 3 + 4 = 8 = 6 + 2

In base 6: 12

In this column, write 2.

Carry: 1

			2	4
		×	5	3
Carry:	1
			2	0
		1	0	0
		3	2	0
	1	4	0	0
Product:		2	4	0

Next step: In the rightmost column:

base 10: 1 + 1 = 2

base 6: 2

In the rightmost column, write 2.

			2	4
		×	5	3
Carry:	1
			2	0
		1	0	0
		3	2	0
	1	4	0	0
Product:	2	2	4	0

${2240}_6$

2240₆