QuestionCalculate the following in specified bases: a. $31_{\text{five}} \cdot 3_{\text{five}}$ b. $31_{\text{five}} \div 3_{\text{five}}$ c. $32_{\text{six}} \cdot 23_{\text{six}}$ d. $124_{\text{five}} \div 3_{\text{five}}$ e. $10100_{\text{two}} \div 10_{\text{two}}$

Studdy Solution

STEP 1

Assumptions1. We are working with numbers in different bases. . The operations are multiplication and division.
3. The results are also expected to be in the same base as the input numbers.

STEP 2

First, we will solve the multiplication problem in base five.
$31_{\text {five }} \cdot_{\text {five }}$

STEP 3

Convert the base five numbers to base ten to perform the multiplication.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 4

Perform the multiplication in base ten.
$16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}$

STEP 5

Convert the result back to base five.
$48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}$

STEP 6

Now, we will solve the division problem in base five.
$31_{\text {five }} \div3_{\text {five }}$

STEP 7

Convert the base five numbers to base ten to perform the division.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 8

Perform the division in base ten.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}$

STEP 9

Convert the result and the remainder back to base five.
$5_{\text {ten }} = \cdot5^ =_{\text {five }}$ $_{\text {ten }} = \cdot5^ =_{\text {five }}$ So, the result is $_{\text {five }}$ with a remainder of $_{\text {five }}$ .

STEP 10

Now, we will solve the multiplication problem in base six.
$32_{\text {six }} \cdot23_{\text {six }}$

STEP 11

Convert the base six numbers to base ten to perform the multiplication.
$32_{\text {six }} =3 \cdot6^ + \cdot6^0 =20_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 =15_{\text {ten }}$

STEP 12

Perform the multiplication in base ten.
$20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}$

STEP 13

Convert the result back to base six.
$300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}$

STEP 14

Now, we will solve the division problem in base five.
$124_{\text {five }} \div3_{\text {five }}$

STEP 15

Convert the base five numbers to base ten to perform the division.
$124_{\text {five }} = \cdot5^2 +2 \cdot5^ +4 \cdot5^0 =39_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 16

Perform the division in base ten.
$39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}$

STEP 17

Convert the result and the remainder back to base five.
$13_{\text {ten }} =2 \cdot5^ +3 \cdot5^0 =23_{\text {five }}$ $0_{\text {ten }} =0 \cdot5^0 =0_{\text {five }}$ So, the result is $23_{\text {five }}$ with a remainder of $0_{\text {five }}$ .

STEP 18

Now, we will solve the division problem in base two.
$10100_{\text {two }} \div10_{\text {two }}$

STEP 19

Convert the base two numbers to base ten to perform the division.
$10100_{\text {two }} =1 \cdot^4 + \cdot^3 +1 \cdot^ + \cdot^1 + \cdot^ =_{\text {ten }}$ $10_{\text {two }} =1 \cdot^1 + \cdot^ =_{\text {ten }}$

STEP 20

Perform the division in base ten.
$20_{\text {ten }} \div_{\text {ten }} =10_{\text {ten }} \text { remainder }0_{\text {ten }}$

STEP 21

Convert the result and the remainder back to base two.
$10_{\text {ten }} =1 \cdot^3 +0 \cdot^ +1 \cdot^1 +0 \cdot^0 =1010_{\text {two }}$ $0_{\text {ten }} =0 \cdot^0 =0_{\text {two }}$ So, the result is $1010_{\text {two }}$ with a remainder of $0_{\text {two }}$ .

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Studdy Solution

STEP 1

Assumptions1. We are working with numbers in different bases. . The operations are multiplication and division.3. The results are also expected to be in the same base as the input numbers.

STEP 2

First, we will solve the multiplication problem in base five.31five ⋅five 31_{\text {five }} \cdot_{\text {five }}31five ​⋅five ​

STEP 3

STEP 4

Perform the multiplication in base ten.16ten ⋅3ten =48ten 16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}16ten ​⋅3ten ​=48ten ​

STEP 5

Convert the result back to base five.48ten =1⋅53+4⋅51+3⋅50=143five 48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}48ten ​=1⋅53+4⋅51+3⋅50=143five ​

STEP 6

Now, we will solve the division problem in base five.31five ÷3five 31_{\text {five }} \div3_{\text {five }}31five ​÷3five ​

STEP 7

Convert the base five numbers to base ten to perform the division.31five =3⋅51+1⋅50=16ten 31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =16_{\text {ten }}31five ​=3⋅51+1⋅50=16ten ​3five =3⋅50=3ten 3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}3five ​=3⋅50=3ten ​

STEP 8

Perform the division in base ten.16ten ÷3ten =5ten remainder 1ten 16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}16ten ​÷3ten ​=5ten ​ remainder 1ten ​

STEP 9

STEP 10

Now, we will solve the multiplication problem in base six.32six ⋅23six 32_{\text {six }} \cdot23_{\text {six }}32six ​⋅23six ​

STEP 11

Convert the base six numbers to base ten to perform the multiplication.32six =3⋅6+⋅60=20ten 32_{\text {six }} =3 \cdot6^ + \cdot6^0 =20_{\text {ten }}32six ​=3⋅6+⋅60=20ten ​23six =⋅6+3⋅60=15ten 23_{\text {six }} = \cdot6^ +3 \cdot6^0 =15_{\text {ten }}23six ​=⋅6+3⋅60=15ten ​

STEP 12

Perform the multiplication in base ten.20ten ⋅15ten =300ten 20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}20ten ​⋅15ten ​=300ten ​

STEP 13

Convert the result back to base six.300ten =⋅63+2⋅62+2⋅60=1220six 300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}300ten ​=⋅63+2⋅62+2⋅60=1220six ​

STEP 14

Now, we will solve the division problem in base five.124five ÷3five 124_{\text {five }} \div3_{\text {five }}124five ​÷3five ​

STEP 15

STEP 16

Perform the division in base ten.39ten ÷3ten =13ten remainder 0ten 39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}39ten ​÷3ten ​=13ten ​ remainder 0ten ​

STEP 17

STEP 18

Now, we will solve the division problem in base two.10100two ÷10two 10100_{\text {two }} \div10_{\text {two }}10100two ​÷10two ​

STEP 19

STEP 20

Perform the division in base ten.20ten ÷ten =10ten remainder 0ten 20_{\text {ten }} \div_{\text {ten }} =10_{\text {ten }} \text { remainder }0_{\text {ten }}20ten ​÷ten ​=10ten ​ remainder 0ten ​

STEP 21

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Assumptions1. We are working with numbers in different bases. . The operations are multiplication and division.
3. The results are also expected to be in the same base as the input numbers.

First, we will solve the multiplication problem in base five.
$31_{\text {five }} \cdot_{\text {five }}$

Perform the multiplication in base ten.
$16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}$

Convert the result back to base five.
$48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}$

Now, we will solve the division problem in base five.
$31_{\text {five }} \div3_{\text {five }}$

Convert the base five numbers to base ten to perform the division.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

Perform the division in base ten.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}$

Now, we will solve the multiplication problem in base six.
$32_{\text {six }} \cdot23_{\text {six }}$

Convert the base six numbers to base ten to perform the multiplication.
$32_{\text {six }} =3 \cdot6^ + \cdot6^0 =20_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 =15_{\text {ten }}$

Perform the multiplication in base ten.
$20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}$

Convert the result back to base six.
$300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}$

Now, we will solve the division problem in base five.
$124_{\text {five }} \div3_{\text {five }}$

Perform the division in base ten.
$39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}$

Now, we will solve the division problem in base two.
$10100_{\text {two }} \div10_{\text {two }}$

Perform the division in base ten.
$20_{\text {ten }} \div_{\text {ten }} =10_{\text {ten }} \text { remainder }0_{\text {ten }}$