Chapter 4

$3\times <!--\; \times \; -->2^5+5\times <!--\; \times \; -->8^2$

Use the order of expressions (PEMDAS), which means

do the exponents first,

$3\times <!--\; \times \; -->32+5\times <!--\; \times \; -->64$

then multiplications,

$96+320$

then additions.

416

$5\times <!--\; \times \; -->7^3+2\times <!--\; \times \; -->7^2+5\times <!--\; \times \; -->7^1+3\times <!--\; \times \; -->7^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$5\times <!--\; \times \; -->343+2\times <!--\; \times \; -->49+5\times <!--\; \times \; -->7+3\times <!--\; \times \; -->1$

then multiplications,

$1,715+98+35+3$

then additions.

1,851

$1\times <!--\; \times \; -->{10}^4+7\times <!--\; \times \; -->{10}^3+4\times <!--\; \times \; -->{10}^2+8\times <!--\; \times \; -->{10}^1+8\times <!--\; \times \; -->{10}^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$1\times <!--\; \times \; -->10,000+7\times <!--\; \times \; -->1,000+4\times <!--\; \times \; -->100+8\times <!--\; \times \; -->10+8\times <!--\; \times \; -->1$

then multiplications,

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

then additions.

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.

$9\times <!--\; \times \; -->{10}^2+2\times <!--\; \times \; -->{10}^1+4\times <!--\; \times \; -->{10}^0$

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.

$1\times <!--\; \times \; -->{10}^3+2\times <!--\; \times \; -->{10}^2+7\times <!--\; \times \; -->{10}^1+9\times <!--\; \times \; -->{10}^0$

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.

$4\times <!--\; \times \; -->{10}^6+1\times <!--\; \times \; -->{10}^5+3\times <!--\; \times \; -->{10}^4+0\times <!--\; \times \; -->{10}^3+0\times <!--\; \times \; -->{10}^2+4\times <!--\; \times \; -->{10}^1+5\times <!--\; \times \; -->{10}^0$

$6\times <!--\; \times \; -->{10}^2+2\times <!--\; \times \; -->{10}^1+1\times <!--\; \times \; -->{10}^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$6\times <!--\; \times \; -->100+2\times <!--\; \times \; -->10+1\times <!--\; \times \; -->1$

then multiplications,

$600+20+1$

then additions.

621

$3\times <!--\; \times \; -->{10}^3+2\times <!--\; \times \; -->{10}^2+0\times <!--\; \times \; -->{10}^1+3\times <!--\; \times \; -->{10}^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$3\times <!--\; \times \; -->1,000+2\times <!--\; \times \; -->100+0\times <!--\; \times \; -->10+3\times <!--\; \times \; -->1$

then multiplications,

$3,000+200+0+3$

then additions.

3,203

$4\times <!--\; \times \; -->{10}^7+0\times <!--\; \times \; -->{10}^6+6\times <!--\; \times \; -->{10}^5+3\times <!--\; \times \; -->{10}^4+0\times <!--\; \times \; -->{10}^3+8\times <!--\; \times \; -->{10}^2+9\times <!--\; \times \; -->{10}^1+1\times <!--\; \times \; -->{10}^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$4\times <!--\; \times \; -->10,000,000+0\times <!--\; \times \; -->1,000,000+6\times <!--\; \times \; -->100,000+3\times <!--\; \times \; -->10,000+0\times <!--\; \times \; -->1,000+8\times <!--\; \times \; -->100+9\times <!--\; \times \; -->10+1\times <!--\; \times \; -->1$

then multiplications,

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

then additions.

40,630,891

This is a two-digit number.

The first digit: 21

The second digit: 9

21(60) + 9 = 1,269

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

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

The top symbol represents 12.

The next symbol represents 17.

12 × 20 + 17 = 257

The top symbol represents 15.

The next symbol represents 2.

The next symbol represents 14.

15 × 20² + 2 × 20 + 14 = 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

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

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

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

100 + 100 + 40

240

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

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

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

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

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

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

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

A represents the digit 10.

B represents the digit 11.

The place values are powers of 6 in base 6.

${421}_6=4\times <!--\; \times \; -->6^2+2\times <!--\; \times \; -->6^1+1\times <!--\; \times \; -->6^0$

$4\times <!--\; \times \; -->36+2\times <!--\; \times \; -->6+1\times <!--\; \times \; -->1$

$144+12+1$

$157$

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.

$\mathrm{A}3{\mathrm{C}}_{14}=10\times <!--\; \times \; -->{14}^2+3\times <!--\; \times \; -->{14}^1+12\times <!--\; \times \; -->{14}^0$

$10\times <!--\; \times \; -->196+3\times <!--\; \times \; -->14+12\times <!--\; \times \; -->1$

$1,960+42+12$

$2,014$

In base 12,

A represents the digit 10.

B represents the digit 11.

The place values are powers of 12 in base 12.

$5\mathrm{A}{\mathrm{B}}_{12}=5\times <!--\; \times \; -->{12}^2+10\times <!--\; \times \; -->{12}^1+11\times <!--\; \times \; -->{12}^0$

$5\times <!--\; \times \; -->144+10\times <!--\; \times \; -->12+11\times <!--\; \times \; -->1$

$720+120+11$

$851$

The place values are powers of 2 in base 2.

${11011}_2=1\times <!--\; \times \; -->2^4+1\times <!--\; \times \; -->2^3+0\times <!--\; \times \; -->2^2+1\times <!--\; \times \; -->2^1+1\times <!--\; \times \; -->2^0$

$1\times <!--\; \times \; -->16+1\times <!--\; \times \; -->8+0\times <!--\; \times \; -->4+1\times <!--\; \times \; -->2+1\times <!--\; \times \; -->1$

$16+8+0+2+1$

$27$

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.

Step 1: Divide the number by 7.

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

Step 1: Divide the number by 12.

$9,275÷<!--\; ÷\; -->12=772$ with a remainder. What is it?

$12\times <!--\; \times \; -->772=9,264$

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

Step 1: Divide the number by 2.

$137÷<!--\; ÷\; -->2=68$ with a remainder. What is it?

$2\times <!--\; \times \; -->68=136$

$137-<!--\; -\; -->136=1$

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

Convert the base 10 number into base 20.

Step 1: Divide the number by 20.

$137÷<!--\; ÷\; -->20=6$ with a remainder. What is it?

$20\times <!--\; \times \; -->6=120$

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

$6\times <!--\; \times \; -->{20}^1+17\times <!--\; \times \; -->{20}^0$

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?

$20\times <!--\; \times \; -->108=2,160$

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

$5\times <!--\; \times \; -->{20}^2+8\times <!--\; \times \; -->{20}^1+11\times <!--\; \times \; -->{20}^0$

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.

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

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

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.

${1242}_6$

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

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

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.

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.

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.

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

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

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.

${633}_7$

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

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

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.

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

Step 1: Do the one’s place first.

In base 10: 1 + 1 = 2

In base 2: 10

One’s place: 0

Carry: 1

Next step: Do the next place to the left.

1 + 1 + 1 = 3 (base 10)

In base 2: 11

This column: 1

Carry: 1

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

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

Next step: Do the next place to the left.

1 + 1 + 0 = 2 (base 10)

In base 2: 10

This column: 0

Carry: 1

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.

${10010010}_2$

Step 1: Do the one’s place first.

In base 10: 5 – 3 = 2

In base 6: 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

Next step: Do the next place to the left.

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

${1242}_6$

${32}_6$

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

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

Next step: Do the next place to the left.

In base 10: 6 – 4 = 2

In base 12: 2

This column: 2

${227}_{12}$

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

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

${311}_4$

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

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

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.

${164}_{14}$

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.

Next step: In the rightmost column:

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

base 6: 0

In the rightmost product column, write 0.

Next step: In the next column:

base 10: 1 + 2 + 3 = 6

base 6: 10

In this product column, write 0.

Carry 1.

Next step: In the next column:

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

base 6: 10

In this column, write 0.

Carry 1.

Next step: In the next column:

base 10: 1 + 3 = 4

base 6: 4

In this column, write 4.

${4000}_6$

0 × 0 = 0

0 × 1 = 0

1 × 1 = 1

Next step: In the rightmost column:

base 10: 1 + 0 = 1

base 2: 1

In the rightmost product column, write 1.

Next step: In the next column:

base 10: 0 + 1 = 1

base 2: 1

In this column, write 1.

Next step: In the next column:

base 10: 1 + 0 = 1

base 2: 1

In this column, write 1.

Next step: In the next column:

base 10: 1 + 1 = 2

base 2: 10

In this column, write 0.

Carry 1.

Next step: In the next column:

base 10: 1 + 1 + 1 = 3

base 2: 11

In this column, write 1.

Carry 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.

${1010111}_2$

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.

Next step: In the rightmost column:

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

In base 12: 9

Write 9 in the rightmost column.

Next step: In the next column over:

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

In base 12: 6

Write 6 in this column.

Next step: In the next column:

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

In base 12: 3

Carry: 1

Next step: In the next column:

In base 10: 1 + 3 = 4

In base 12: 4

Carry: No need

${4369}_{12}$

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.

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.

$4\times <!--\; \times \; -->8^2+2\times <!--\; \times \; -->8^1+7\times <!--\; \times \; -->8^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$4\times <!--\; \times \; -->64+2\times <!--\; \times \; -->8+7\times <!--\; \times \; -->1$

then multiplications,

$256+16+7$

then additions.

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.

$4\times <!--\; \times \; -->{10}^3+5\times <!--\; \times \; -->{10}^3+2\times <!--\; \times \; -->{10}^2+0\times <!--\; \times \; -->{10}^1+9\times <!--\; \times \; -->{10}^0$

$6\times <!--\; \times \; -->{10}^5+0\times <!--\; \times \; -->{10}^4+1\times <!--\; \times \; -->{10}^3+9\times <!--\; \times \; -->{10}^2+4\times <!--\; \times \; -->{10}^1+7\times <!--\; \times \; -->{10}^0$

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

Use the order of expressions (PEMDAS), which means

do the exponents first,

$6\times <!--\; \times \; -->100,000+0\times <!--\; \times \; -->10,000+1\times <!--\; \times \; -->1,000+9\times <!--\; \times \; -->100+4\times <!--\; \times \; -->10+7\times <!--\; \times \; -->1$

then multiplications,

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

then additions.

601,947

This is a two-digit number.

The first digit: 5

The second digit: 41

5(60) + 41 = 341

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.

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

The place values are powers of 5 in base 5.

${2304}_5=2\times <!--\; \times \; -->5^3+3\times <!--\; \times \; -->5^2+0\times <!--\; \times \; -->5^1+4\times <!--\; \times \; -->5^0$

$2\times <!--\; \times \; -->125+3\times <!--\; \times \; -->25+0\times <!--\; \times \; -->5+4\times <!--\; \times \; -->1$

$250+75+0+4$

$329$

329

Step 1: Divide the number by 8.

$329÷<!--\; ÷\; -->8=41$ with a remainder. What is it?

$41\times <!--\; \times \; -->8=328$

$329-<!--\; -\; -->328=1$

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

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.

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

$10\times <!--\; \times \; -->196+11\times <!--\; \times \; -->14+12\times <!--\; \times \; -->1$

$1,960+154+12$

$2,126$

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

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.

${121}_6$

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

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.

$7_8$

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

Next step: Do the next place to the left.

In base 10: 1 + 3 + 4 = 8

In base 14: 8

This column: 8

${82}_{14}$

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

Next step: Do the next place to the left.

In base 10: 9 – 9 = 0

In base 12: 0

This column: Leave blank.

$5_{12}$

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.

Next step: In the rightmost column:

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

base 6: 0

Write 0 in the rightmost column.

Next step: In the next column over:

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

In base 6: 4

In this column, write 4.

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

Next step: In the rightmost column:

base 10: 1 + 1 = 2

base 6: 2

In the rightmost column, write 2.

${2240}_6$

In base 14:

A represents the digit 10.

B represents the digit 11.

C represents the digit 12.

D represents the digit 13.

You can use the base 14 multiplication table created in an earlier exercise.

Use the multiplication table for base 14. Find 32 in the product part of the table where 4 is the column header. The answer is the row header, B, for that cell.

The answer is B₁₄.