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

Studdy Solution

STEP 1

Assumptions1. We are performing operations in different bases, not in the usual base10. . The operations are multiplication and division.
3. We need to convert the results back to the original base.

STEP 2

Let's start with the first operation. We need to convert the numbers from base five to base ten to perform the multiplication.
$31_{\text {five }} = \times5^1 +1 \times5^0 =15 +1 =16_{\text {ten }}$ $_{\text {five }} = \times5^0 =_{\text {ten }}$

STEP 3

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

STEP 4

We need to convert the result back to base five. We find the largest power of that is less than or equal to48, which is $^2 =25$ . The coefficient for $^2$ is $48 \div25 =1$ . The remainder is $48 -25 =23$ .

STEP 5

Next, we find the coefficient for $5^1$ . It is $23 \div5 =4$ . The remainder is $23 -20 =3$ .

STEP 6

Finally, the coefficient for $5^0$ is $3 \div1 =3$ . The remainder is $3 -3 =0$ .

STEP 7

So, $48_{\text {ten }} =143_{\text {five }}$ . Therefore, $31_{\text {five }} \cdot3_{\text {five }} =143_{\text {five }}$ .

STEP 8

Next, let's perform the division. We need to convert the numbers from base five to base ten.
$31_{\text {five }} =3 \times5^1 +1 \times5^0 =15 +1 =16_{\text {ten }}$ $3_{\text {five }} =3 \times5^0 =3_{\text {ten }}$

STEP 9

Now, perform the division in base ten.
$16_{\text {ten }} \div3_{\text {ten }} =5_{\text {ten }} \text { remainder }_{\text {ten }}$

STEP 10

We need to convert the quotient and the remainder back to base five. $5_{\text {ten }} =10_{\text {five }}$ and $_{\text {ten }} =_{\text {five }}$ .

STEP 11

So, $31_{\text {five }} \div3_{\text {five }} =10_{\text {five }} \text { remainder }_{\text {five }}$ .

STEP 12

Finally, let's perform the last multiplication. We need to convert the numbers from base six to base ten.
$32_{\text {six }} = \times6^ +2 \times6^0 =18 +2 =20_{\text {ten }}$ $23_{\text {six }} =2 \times6^ + \times6^0 =12 + =15_{\text {ten }}$

STEP 13

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

STEP 14

We need to convert the result back to base six. We find the largest power of6 that is less than or equal to300, which is $6^3 =216$ . The coefficient for $6^3$ is $300 \div216 =$ . The remainder is $300 -216 =84$ .

STEP 15

Next, we find the coefficient for $^2$ . It is $84 \div36 =2$ . The remainder is $84 -72 =12$ .

STEP 16

Next, we find the coefficient for 6^. It is $12 \div6 =2$ . The remainder is $12 -12 =0$ .

STEP 17

Finally, the coefficient for $6^0$ is $0 \div =0$ . The remainder is $0 -0 =0$ .

STEP 18

So, $300_{\text {ten }} =1220_{\text {six }}$ . Therefore, $32_{\text {six }} \cdot23_{\text {six }} =1220_{\text {six }}$ .

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