QuestionPerform operations in given bases:
a. $31_{\text {five }} \cdot 3_{\text {five }}$
b. $31_{\text {five }}+3_{\text {five }}$
c. $51_{\text {six }} \cdot 23_{\text {six }}$
d. $242_{\text {five }}+4_{\text {five }}$
e. $10110_{\text {two }}+10_{\text {two }}$
f. $10011_{\text {two }} \cdot 100_{\text {two }}$

Studdy Solution

STEP 1

Assumptions1. The operations are performed in the base specified for each problem. . The base of a number is the number of unique digits (including zero) used to represent numbers in positional numeral systems.

STEP 2

For problem a, we need to multiply $31_{\text {five }}$ and $_{\text {five }}$ . First, convert these numbers to base10.
$31_{\text {five }} = \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}$ $_{\text {five }} = \cdot5^0 =_{\text {ten }}$

STEP 3

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

STEP 4

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

STEP 5

For problem b, we need to add $31_{\text {five }}$ and $3_{\text {five }}$ . First, convert these numbers to base10.
$31_{\text {five }} =3 \cdot5^1 +1 \cdot5^0 =15 +1 =16_{\text {ten }}$ $3_{\text {five }} =3 \cdot5^0 =3_{\text {ten }}$

STEP 6

Add the base10 numbers.
$16_{\text {ten }} +3_{\text {ten }} =19_{\text {ten }}$

STEP 7

Convert the result back to base5.
$19_{\text {ten }} =3 \cdot5^1 +4 \cdot5^0 =34_{\text {five }}$

STEP 8

For problem d, we need to add $242_{\text {five }}$ and $4_{\text {five }}$ . First, convert these numbers to base10.
$242_{\text {five }} =2 \cdot5^2 +4 \cdot5^1 +2 \cdot5^0 =50 +20 +2 =72_{\text {ten }}$ $4_{\text {five }} =4 \cdot5^0 =4_{\text {ten }}$

STEP 9

Add the base numbers.
$72_{\text {ten }} +4_{\text {ten }} =76_{\text {ten }}$

STEP 10

Convert the result back to base5.
$76_{\text {ten }} =3 \cdot5^2 +0 \cdot5^ + \cdot5^0 =301_{\text {five }}$

STEP 11

For problem c, we need to multiply $51_{\text {six }}$ and $23_{\text {six }}$ . First, convert these numbers to base10.
$51_{\text {six }} =5 \cdot6^ + \cdot6^0 =30 + =31_{\text {ten }}$ $23_{\text {six }} = \cdot6^ +3 \cdot6^0 = +3 =15_{\text {ten }}$

STEP 12

Multiply the base10 numbers.
$31_{\text {ten }} \cdot15_{\text {ten }} =465_{\text {ten }}$

STEP 13

Convert the result back to base6.
$465_{\text {ten }} =2 \cdot6^3 +0 \cdot6^2 +3 \cdot6^ +3 \cdot6^0 =2033_{\text {six }}$

STEP 14

For problem e, we need to add $10110_{\text {two }}$ and $10_{\text {two }}$ . First, convert these numbers to base10.
$10110_{\text {two }} = \cdot2^4 +0 \cdot2^3 + \cdot2^2 + \cdot2^ +0 \cdot2^0 =16 +4 +2 =22_{\text {ten }}$ $10_{\text {two }} = \cdot2^ +0 \cdot2^0 =2_{\text {ten }}$

STEP 15

Add the base10 numbers.
$22_{\text {ten }} +2_{\text {ten }} =24_{\text {ten }}$

STEP 16

Convert the result back to base2.
$24_{\text {ten }} = \cdot2^4 + \cdot2^3 +0 \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =11000_{\text {two }}$

STEP 17

For problem f, we need to multiply $10011_{\text {two }}$ and $100_{\text {two }}$ . First, convert these numbers to base10.
$10011_{\text {two }} = \cdot2^4 +0 \cdot2^3 +0 \cdot2^2 + \cdot2^ + \cdot2^0 =16 +2 + =19_{\text {ten }}$ $100_{\text {two }} = \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =4_{\text {ten }}$

STEP 18

Multiply the base10 numbers.
$_{\text {ten }} \cdot4_{\text {ten }} =76_{\text {ten }}$

STEP 19

Convert the result back to base.
$76_{\text {ten }} =1 \cdot^6 + \cdot^5 + \cdot^4 +1 \cdot^3 +1 \cdot^ + \cdot^1 + \cdot^ =100110_{\text {two }}$

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Studdy Solution

STEP 1

Assumptions1. The operations are performed in the base specified for each problem. . The base of a number is the number of unique digits (including zero) used to represent numbers in positional numeral systems.

STEP 2

STEP 3

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

STEP 4

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

STEP 5

STEP 6

Add the base10 numbers.16ten +3ten =19ten 16_{\text {ten }} +3_{\text {ten }} =19_{\text {ten }}16ten ​+3ten ​=19ten ​

STEP 7

Convert the result back to base5.19ten =3⋅51+4⋅50=34five 19_{\text {ten }} =3 \cdot5^1 +4 \cdot5^0 =34_{\text {five }}19ten ​=3⋅51+4⋅50=34five ​

STEP 8

STEP 9

Add the base numbers.72ten +4ten =76ten 72_{\text {ten }} +4_{\text {ten }} =76_{\text {ten }}72ten ​+4ten ​=76ten ​

STEP 10

Convert the result back to base5.76ten =3⋅52+0⋅5+⋅50=301five 76_{\text {ten }} =3 \cdot5^2 +0 \cdot5^ + \cdot5^0 =301_{\text {five }}76ten ​=3⋅52+0⋅5+⋅50=301five ​

STEP 11

STEP 12

Multiply the base10 numbers.31ten ⋅15ten =465ten 31_{\text {ten }} \cdot15_{\text {ten }} =465_{\text {ten }}31ten ​⋅15ten ​=465ten ​

STEP 13

Convert the result back to base6.465ten =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 }}465ten ​=2⋅63+0⋅62+3⋅6+3⋅60=2033six ​

STEP 14

STEP 15

Add the base10 numbers.22ten +2ten =24ten 22_{\text {ten }} +2_{\text {ten }} =24_{\text {ten }}22ten ​+2ten ​=24ten ​

STEP 16

Convert the result back to base2.24ten =⋅24+⋅23+0⋅22+0⋅2+0⋅20=11000two 24_{\text {ten }} = \cdot2^4 + \cdot2^3 +0 \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =11000_{\text {two }}24ten ​=⋅24+⋅23+0⋅22+0⋅2+0⋅20=11000two ​

STEP 17

STEP 18

Multiply the base10 numbers.ten ⋅4ten =76ten _{\text {ten }} \cdot4_{\text {ten }} =76_{\text {ten }}ten ​⋅4ten ​=76ten ​

STEP 19

Convert the result back to base.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 }}76ten ​=1⋅6+⋅5+⋅4+1⋅3+1⋅+⋅1+⋅=100110two ​

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

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

Add the base10 numbers.
$16_{\text {ten }} +3_{\text {ten }} =19_{\text {ten }}$

Convert the result back to base5.
$19_{\text {ten }} =3 \cdot5^1 +4 \cdot5^0 =34_{\text {five }}$

Add the base numbers.
$72_{\text {ten }} +4_{\text {ten }} =76_{\text {ten }}$

Convert the result back to base5.
$76_{\text {ten }} =3 \cdot5^2 +0 \cdot5^ + \cdot5^0 =301_{\text {five }}$

Multiply the base10 numbers.
$31_{\text {ten }} \cdot15_{\text {ten }} =465_{\text {ten }}$

Convert the result back to base6.
$465_{\text {ten }} =2 \cdot6^3 +0 \cdot6^2 +3 \cdot6^ +3 \cdot6^0 =2033_{\text {six }}$

Add the base10 numbers.
$22_{\text {ten }} +2_{\text {ten }} =24_{\text {ten }}$

Convert the result back to base2.
$24_{\text {ten }} = \cdot2^4 + \cdot2^3 +0 \cdot2^2 +0 \cdot2^ +0 \cdot2^0 =11000_{\text {two }}$

Multiply the base10 numbers.
$_{\text {ten }} \cdot4_{\text {ten }} =76_{\text {ten }}$

Convert the result back to base.
$76_{\text {ten }} =1 \cdot^6 + \cdot^5 + \cdot^4 +1 \cdot^3 +1 \cdot^ + \cdot^1 + \cdot^ =100110_{\text {two }}$