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QuestionCalculate the following operations in the given bases: a. 31five ⋅3five 31_{\text {five }} \cdot 3_{\text {five }} b. 31five +3five 31_{\text {five }} + 3_{\text {five }} c. 51six ⋅23six 51_{\text {six }} \cdot 23_{\text {six }} d. 242five ÷4242_{\text {five }} \div 4 e. 10110two +10two 10110_{\text {two }} + 10_{\text {two }} f. 10011two ⋅100two 10011_{\text {two }} \cdot 100_{\text {two }}

Studdy Solution

STEP 1

Assumptions1. The operations are performed in the given base. . The base of a number is the number of unique digits, including zero, used to represent numbers in a positional numeral system.
3. The base of a number can be changed to another base using appropriate conversion methods.

STEP 2

Let's start with the first operation 31five ⋅five 31_{\text {five }} \cdot_{\text {five }}. First, we need to convert these numbers from base five to base ten to perform the multiplication.
31five =⋅51+1⋅50=15+1=16ten 31_{\text {five }} = \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}five =⋅50=ten _{\text {five }} = \cdot5^0 =_{\text {ten }}

STEP 3

Now, perform the multiplication in base ten.
16ten ⋅3ten =48ten 16_{\text {ten }} \cdot3_{\text {ten }} =48_{\text {ten }}

STEP 4

Finally, convert the result back to base five.
48ten =3⋅2+2⋅1+3⋅0=323five 48_{\text {ten }} =3 \cdot^2 +2 \cdot^1 +3 \cdot^0 =323_{\text {five }}So, 31five ⋅3five =323five 31_{\text {five }} \cdot3_{\text {five }} =323_{\text {five }}.

STEP 5

Next, let's perform the operation 242five ÷4five 242_{\text {five }} \div4_{\text {five }}. First, convert these numbers from base five to base ten.
242five =2⋅52+4⋅51+2⋅50=50+20+2=72ten 242_{\text {five }} =2 \cdot5^2 +4 \cdot5^1 +2 \cdot5^0 =50 +20 +2 =72_{\text {ten }}4five =4⋅50=4ten 4_{\text {five }} =4 \cdot5^0 =4_{\text {ten }}

STEP 6

Now, perform the division in base ten.
72ten ÷4ten =18ten 72_{\text {ten }} \div4_{\text {ten }} =18_{\text {ten }}

STEP 7

Finally, convert the result back to base five.
18ten =3⋅51+3⋅50=33five 18_{\text {ten }} =3 \cdot5^1 +3 \cdot5^0 =33_{\text {five }}So, 242five ÷4five =33five 242_{\text {five }} \div4_{\text {five }} =33_{\text {five }}.

STEP 8

Next, perform the operation 31five +3five 31_{\text {five }}+3_{\text {five }}. Convert these numbers from base five to base ten.
31five =3⋅51+1⋅50=15+1=16ten 31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}3five =3⋅50=3ten 3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}

STEP 9

Now, perform the addition in base ten.
16ten +3ten =19ten 16_{\text {ten }} +3_{\text {ten }} =19_{\text {ten }}

STEP 10

Finally, convert the result back to base five.
19ten =3⋅5+4⋅50=34five 19_{\text {ten }} =3 \cdot5^ +4 \cdot5^0 =34_{\text {five }}So, 31five +3five =34five 31_{\text {five }}+3_{\text {five }} =34_{\text {five }}.

STEP 11

Next, perform the operation 10110two +10two 10110_{\text {two }}+10_{\text {two }}. Convert these numbers from base two to base ten.
10110two =⋅4+0⋅3+⋅+⋅+0⋅0=16+4+=22ten 10110_{\text {two }} = \cdot^4 +0 \cdot^3 + \cdot^ + \cdot^ +0 \cdot^0 =16 +4 + =22_{\text {ten }}10two =⋅+0⋅0=ten 10_{\text {two }} = \cdot^ +0 \cdot^0 =_{\text {ten }}

STEP 12

Now, perform the addition in base ten.
22ten +2ten =24ten 22_{\text {ten }} +2_{\text {ten }} =24_{\text {ten }}

STEP 13

Finally, convert the result back to base two.
24ten =⋅2+⋅23+0⋅22+0⋅2+0⋅20=11000two 24_{\text {ten }} = \cdot2^ + \cdot2^3 +0 \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =11000_{\text {two }}So, 10110two +10two =11000two 10110_{\text {two }}+10_{\text {two }} =11000_{\text {two }}.

STEP 14

Next, perform the operation 51six ⋅23six 51_{\text {six }} \cdot23_{\text {six }}. Convert these numbers from base six to base ten.
51six =⋅6+⋅60=30+=31ten 51_{\text {six }} = \cdot6^ + \cdot6^0 =30 + =31_{\text {ten }}23six =2⋅6+3⋅60=12+3=ten 23_{\text {six }} =2 \cdot6^ +3 \cdot6^0 =12 +3 =_{\text {ten }}

STEP 15

Now, perform the multiplication in base ten.
31ten ⋅15ten =465ten 31_{\text {ten }} \cdot15_{\text {ten }} =465_{\text {ten }}

STEP 16

Finally, convert the result back to base six.
465ten =2⋅63+0⋅62+3⋅6+3⋅60=2033six 465_{\text {ten }} =2 \cdot6^3 +0 \cdot6^2 +3 \cdot6^ +3 \cdot6^0 =2033_{\text {six }}So, 51six ⋅23six =2033six 51_{\text {six }} \cdot23_{\text {six }} =2033_{\text {six }}.

STEP 17

Finally, perform the operation 10011two ⋅100two 10011_{\text {two }} \cdot100_{\text {two }}. Convert these numbers from base two to base ten.
10011two =⋅24+0⋅23+0⋅22+⋅2+⋅20=16+2+=19ten 10011_{\text {two }} = \cdot2^4 +0 \cdot2^3 +0 \cdot2^2 + \cdot2^ + \cdot2^0 =16 +2 + =19_{\text {ten }}100two =⋅22+0⋅2+0⋅20=4ten 100_{\text {two }} = \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =4_{\text {ten }}

STEP 18

Now, perform the multiplication in base ten.
ten ⋅4ten =76ten _{\text {ten }} \cdot4_{\text {ten }} =76_{\text {ten }}

STEP 19

Finally, convert the result back to base two.
76ten =1⋅6+⋅5+⋅4+1⋅3+1⋅+⋅1+⋅=100110two 76_{\text {ten }} =1 \cdot^6 + \cdot^5 + \cdot^4 +1 \cdot^3 +1 \cdot^ + \cdot^1 + \cdot^ =100110_{\text {two }}So, 10011two ⋅100two =100110two 10011_{\text {two }} \cdot100_{\text {two }} =100110_{\text {two }}.

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