QuestionCalculate the following in the given bases: a. $31_{\text {five }} \cdot 3_{\text {five }}$ b. $31_{\text {five }}+3_{\text {five }}$ c. $32_{\text {six }} \cdot 23_{\text {six }}$

Studdy Solution

STEP 1

Assumptions1. We are performing operations in different bases (base five and base six). . The operations include multiplication and addition.

STEP 2

For the first operation, we need to convert the numbers from base five to base ten. The number $31_{\text {five }}$ is converted as follows $31_{\text {five }} = \times5^1 +1 \times5^0$

STEP 3

Calculate the value of $31_{\text {five }}$ in base ten.
$31_{\text {five }} =3 \times5 +1 \times1 =15 +1 =16$

STEP 4

Similarly, convert $3_{\text {five }}$ from base five to base ten.
$3_{\text {five }} =3 \times^0$

STEP 5

Calculate the value of $3_{\text {five }}$ in base ten.
$3_{\text {five }} =3 \times1 =3$

STEP 6

Now, perform the multiplication operation in base ten.
$16 \times3 =48$

STEP 7

Convert the result back to base five. To do this, divide48 by5 and keep track of the remainders.
$48 \div5 =9 \text{ remainder }3$

STEP 8

Then divide the quotient by5 again and keep track of the remainder.
$\div5 =1 \text{ remainder }4$

STEP 9

Finally, divide the last quotient by5.
$\div5 = \text{ remainder }$

STEP 10

The remainders in reverse order give the result in base five.
$48_{\text {ten }} =143_{\text {five }}$ So, $31_{\text {five }} \cdot3_{\text {five }} =143_{\text {five }}$ .

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QuestionCalculate the following in the given bases: a. $31_{\text {five }} \cdot 3_{\text {five }}$ b. $31_{\text {five }}+3_{\text {five }}$ c. $32_{\text {six }} \cdot 23_{\text {six }}$

Studdy Solution

STEP 1

Assumptions1. We are performing operations in different bases (base five and base six). . The operations include multiplication and addition.

STEP 2

For the first operation, we need to convert the numbers from base five to base ten. The number $31_{\text {five }}$ is converted as follows $31_{\text {five }} = \times5^1 +1 \times5^0$

STEP 3

Calculate the value of $31_{\text {five }}$ in base ten.
$31_{\text {five }} =3 \times5 +1 \times1 =15 +1 =16$

STEP 4

Similarly, convert $3_{\text {five }}$ from base five to base ten.
$3_{\text {five }} =3 \times^0$

STEP 5

Calculate the value of $3_{\text {five }}$ in base ten.
$3_{\text {five }} =3 \times1 =3$

STEP 6

Now, perform the multiplication operation in base ten.
$16 \times3 =48$

STEP 7

Convert the result back to base five. To do this, divide48 by5 and keep track of the remainders.
$48 \div5 =9 \text{ remainder }3$

STEP 8

Then divide the quotient by5 again and keep track of the remainder.
$\div5 =1 \text{ remainder }4$

STEP 9

Finally, divide the last quotient by5.
$\div5 = \text{ remainder }$

STEP 10

The remainders in reverse order give the result in base five.
$48_{\text {ten }} =143_{\text {five }}$ So, $31_{\text {five }} \cdot3_{\text {five }} =143_{\text {five }}$ .

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QuestionCalculate the following in the given bases: a. 31five ⋅3five 31_{\text {five }} \cdot 3_{\text {five }}31five ​⋅3five ​ b. 31five +3five 31_{\text {five }}+3_{\text {five }}31five ​+3five ​ c. 32six ⋅23six 32_{\text {six }} \cdot 23_{\text {six }}32six ​⋅23six ​

Studdy Solution

STEP 1

Assumptions1. We are performing operations in different bases (base five and base six). . The operations include multiplication and addition.

STEP 2

For the first operation, we need to convert the numbers from base five to base ten. The number 31five 31_{\text {five }}31five ​ is converted as follows31five =×51+1×5031_{\text {five }} = \times5^1 +1 \times5^031five ​=×51+1×50

STEP 3

Calculate the value of 31five 31_{\text {five }}31five ​ in base ten.31five =3×5+1×1=15+1=1631_{\text {five }} =3 \times5 +1 \times1 =15 +1 =1631five ​=3×5+1×1=15+1=16

STEP 4

Similarly, convert 3five 3_{\text {five }}3five ​ from base five to base ten.3five =3×03_{\text {five }} =3 \times^03five ​=3×0

STEP 5

Calculate the value of 3five 3_{\text {five }}3five ​ in base ten.3five =3×1=33_{\text {five }} =3 \times1 =33five ​=3×1=3

STEP 6

Now, perform the multiplication operation in base ten.16×3=4816 \times3 =4816×3=48

STEP 7

Convert the result back to base five. To do this, divide48 by5 and keep track of the remainders.48÷5=9 remainder 348 \div5 =9 \text{ remainder }348÷5=9 remainder 3

STEP 8

Then divide the quotient by5 again and keep track of the remainder.÷5=1 remainder 4 \div5 =1 \text{ remainder }4÷5=1 remainder 4

STEP 9

Finally, divide the last quotient by5.÷5= remainder \div5 = \text{ remainder }÷5= remainder

STEP 10

The remainders in reverse order give the result in base five.48ten =143five 48_{\text {ten }} =143_{\text {five }}48ten ​=143five ​So, 31five ⋅3five =143five 31_{\text {five }} \cdot3_{\text {five }} =143_{\text {five }}31five ​⋅3five ​=143five ​.

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QuestionCalculate the following in the given bases: a. $31_{\text {five }} \cdot 3_{\text {five }}$ b. $31_{\text {five }}+3_{\text {five }}$ c. $32_{\text {six }} \cdot 23_{\text {six }}$

For the first operation, we need to convert the numbers from base five to base ten. The number $31_{\text {five }}$ is converted as follows $31_{\text {five }} = \times5^1 +1 \times5^0$

Calculate the value of $31_{\text {five }}$ in base ten.
$31_{\text {five }} =3 \times5 +1 \times1 =15 +1 =16$

Similarly, convert $3_{\text {five }}$ from base five to base ten.
$3_{\text {five }} =3 \times^0$

Calculate the value of $3_{\text {five }}$ in base ten.
$3_{\text {five }} =3 \times1 =3$

Now, perform the multiplication operation in base ten.
$16 \times3 =48$

Convert the result back to base five. To do this, divide48 by5 and keep track of the remainders.
$48 \div5 =9 \text{ remainder }3$

Then divide the quotient by5 again and keep track of the remainder.
$\div5 =1 \text{ remainder }4$

Finally, divide the last quotient by5.
$\div5 = \text{ remainder }$

The remainders in reverse order give the result in base five.
$48_{\text {ten }} =143_{\text {five }}$ So, $31_{\text {five }} \cdot3_{\text {five }} =143_{\text {five }}$ .