QuestionCalculate the following in the 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. $124_{\text{five}} \div 3_{\text{five}}$

Studdy Solution

STEP 1

Assumptions1. We are working with base5 and base6 numbers. . The operations to be performed are multiplication and division.
3. The numbers are already given in the appropriate bases.

STEP 2

We start with the first operation $31_{\text {five }} \cdot_{\text {five }}$ .

STEP 3

Convert the base5 numbers to base10 to make the multiplication easier.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 4

Multiply the base10 numbers.
$16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}$

STEP 5

Convert the result back to base5.
$48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}$

STEP 6

Now, we move on to the second operation $31_{\text {five }} \div3_{\text {five }}$ .

STEP 7

We already converted $31_{\text {five }}$ and $3_{\text {five }}$ to base10 in the previous steps.

STEP 8

Divide the base10 numbers.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}$

STEP 9

Convert the quotient and remainder back to base5.
$5_{\text {ten }} = \cdot5^ =_{\text {five }}$ $_{\text {ten }} = \cdot5^ =_{\text {five }}$

STEP 10

Next, we perform the third operation $32_{\text {six }} \cdot23_{\text {six }}$ .

STEP 11

Convert the base6 numbers to base10.
$32_{\text {six }} =3 \cdot6^ + \cdot6^0 =18 + =20_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 = +3 =15_{\text {ten }}$

STEP 12

Multiply the base10 numbers.
$20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}$

STEP 13

Convert the result back to base6.
$300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}$

STEP 14

Finally, we perform the fourth operation $124_{\text {five }} \div3_{\text {five }}$ .

STEP 15

Convert the base5 numbers to base10.
$124_{\text {five }} = \cdot5^2 +2 \cdot5^ +4 \cdot5^0 =25 +10 +4 =39_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 16

Divide the base10 numbers.
$39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}$

STEP 17

Convert the quotient and remainder back to base5.
$13_{\text {ten }} =2 \cdot5^ +3 \cdot5^0 =23_{\text {five }}$ $0_{\text {ten }} =0_{\text {five }}$ So, the solutions area. $31_{\text {five }} \cdot3_{\text {five }}=143_{\text {five }}$ b. $31_{\text {five }} \div3_{\text {five }}=10_{\text {five }} \text { remainder }_{\text {five }}$ c. $32_{\text {six }} \cdot23_{\text {six }}=1220_{\text {six }}$ d. $124_{\text {five }} \div3_{\text {five }}=23_{\text {five }} \text { remainder }0_{\text {five }}$

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QuestionCalculate the following in the 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. $124_{\text{five}} \div 3_{\text{five}}$

Studdy Solution

STEP 1

Assumptions1. We are working with base5 and base6 numbers. . The operations to be performed are multiplication and division.
3. The numbers are already given in the appropriate bases.

STEP 2

We start with the first operation $31_{\text {five }} \cdot_{\text {five }}$ .

STEP 3

Convert the base5 numbers to base10 to make the multiplication easier.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 4

Multiply the base10 numbers.
$16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}$

STEP 5

Convert the result back to base5.
$48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}$

STEP 6

Now, we move on to the second operation $31_{\text {five }} \div3_{\text {five }}$ .

STEP 7

We already converted $31_{\text {five }}$ and $3_{\text {five }}$ to base10 in the previous steps.

STEP 8

Divide the base10 numbers.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}$

STEP 9

Convert the quotient and remainder back to base5.
$5_{\text {ten }} = \cdot5^ =_{\text {five }}$ $_{\text {ten }} = \cdot5^ =_{\text {five }}$

STEP 10

Next, we perform the third operation $32_{\text {six }} \cdot23_{\text {six }}$ .

STEP 11

Convert the base6 numbers to base10.
$32_{\text {six }} =3 \cdot6^ + \cdot6^0 =18 + =20_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 = +3 =15_{\text {ten }}$

STEP 12

Multiply the base10 numbers.
$20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}$

STEP 13

Convert the result back to base6.
$300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}$

STEP 14

Finally, we perform the fourth operation $124_{\text {five }} \div3_{\text {five }}$ .

STEP 15

Convert the base5 numbers to base10.
$124_{\text {five }} = \cdot5^2 +2 \cdot5^ +4 \cdot5^0 =25 +10 +4 =39_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 16

Divide the base10 numbers.
$39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}$

STEP 17

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Studdy Solution

STEP 1

Assumptions1. We are working with base5 and base6 numbers. . The operations to be performed are multiplication and division.3. The numbers are already given in the appropriate bases.

STEP 2

We start with the first operation 31five ⋅five 31_{\text {five }} \cdot_{\text {five }}31five ​⋅five ​.

STEP 3

STEP 4

Multiply the base10 numbers.16ten ⋅3ten =48ten 16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}16ten ​⋅3ten ​=48ten ​

STEP 5

Convert the result back to base5.48ten =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 move on to the second operation 31five ÷3five 31_{\text {five }} \div3_{\text {five }}31five ​÷3five ​.

STEP 7

We already converted 31five 31_{\text {five }}31five ​ and 3five 3_{\text {five }}3five ​ to base10 in the previous steps.

STEP 8

Divide the base10 numbers.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

Convert the quotient and remainder back to base5.5ten =⋅5five =5_{\text {ten }} = \cdot5^ =_{\text {five }}5ten ​=⋅5five =​ten =⋅5five =_{\text {ten }} = \cdot5^ =_{\text {five }}ten ​=⋅5five =​

STEP 10

Next, we perform the third operation 32six ⋅23six 32_{\text {six }} \cdot23_{\text {six }}32six ​⋅23six ​.

STEP 11

Convert the base6 numbers to base10.32six =3⋅6+⋅60=18+=20ten 32_{\text {six }} =3 \cdot6^ + \cdot6^0 =18 + =20_{\text {ten }}32six ​=3⋅6+⋅60=18+=20ten ​23six =⋅6+3⋅60=+3=15ten 23_{\text {six }} = \cdot6^ +3 \cdot6^0 = +3 =15_{\text {ten }}23six ​=⋅6+3⋅60=+3=15ten ​

STEP 12

Multiply the base10 numbers.20ten ⋅15ten =300ten 20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}20ten ​⋅15ten ​=300ten ​

STEP 13

Convert the result back to base6.300ten =⋅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

Finally, we perform the fourth operation 124five ÷3five 124_{\text {five }} \div3_{\text {five }}124five ​÷3five ​.

STEP 15

STEP 16

Divide the base10 numbers.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

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Assumptions1. We are working with base5 and base6 numbers. . The operations to be performed are multiplication and division.
3. The numbers are already given in the appropriate bases.

We start with the first operation $31_{\text {five }} \cdot_{\text {five }}$ .

Multiply the base10 numbers.
$16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}$

Convert the result back to base5.
$48_{\text {ten }} =1 \cdot5^3 +4 \cdot5^1 +3 \cdot5^0 =143_{\text {five }}$

Now, we move on to the second operation $31_{\text {five }} \div3_{\text {five }}$ .

We already converted $31_{\text {five }}$ and $3_{\text {five }}$ to base10 in the previous steps.

Divide the base10 numbers.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }1_{\text {ten }}$

Convert the quotient and remainder back to base5.
$5_{\text {ten }} = \cdot5^ =_{\text {five }}$ $_{\text {ten }} = \cdot5^ =_{\text {five }}$

Next, we perform the third operation $32_{\text {six }} \cdot23_{\text {six }}$ .

Convert the base6 numbers to base10.
$32_{\text {six }} =3 \cdot6^ + \cdot6^0 =18 + =20_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 = +3 =15_{\text {ten }}$

Multiply the base10 numbers.
$20_{\text {ten }} \cdot15_{\text {ten }} =300_{\text {ten }}$

Convert the result back to base6.
$300_{\text {ten }} = \cdot6^3 +2 \cdot6^2 +2 \cdot6^0 =1220_{\text {six }}$

Finally, we perform the fourth operation $124_{\text {five }} \div3_{\text {five }}$ .

Divide the base10 numbers.
$39_{\text {ten }} \div3_{\text {ten }} =13_{\text {ten }} \text { remainder }0_{\text {ten }}$