Question

i need questions 2-6 done so i can see if my work is right

2 (3 points) Using Horner's scheme, represent (evaluate the binary unsigned whole number 10011101 in decimal CAUTION: You will earn NO CREDITS if you don't use Homer's scheme to more efficiently evaluate polynomial (In particular, you will get a 0 if you resort to evaluating the polynomial directly;"brute force") You will earn NO CREDITS if you simply show the final result and not clearly show working (ie, intermediate steps). 3 (3 points) Using Horner's scheme, represent (evaluate the base-6 unsigned whole number 14032 in decimal CAUTION: You will earn NO CREDITS if you don't use Horner's scheme to more efficiently evaluate polynomial (In particular, you will get a 0 if you resort to evaluating the polynomial directly "brute force") You will earn NO CREDITS if you simply show the final result and not clearly show working (1.e. intermediate steps). 4 (3 points) Using the repeated division method, represent the decimal unsigned whole number 347 in binary. CAUTION: You will earn NO CREDITS if you simply show the final result and not clearly show working (ie, intermediate steps). 5 (3 points) Re-write the result obtained in the preceding question (decimal unsigned whole number 347 in binary) in hex CAUTION You will earn NO CREDITS if you simply show the final result and not clearly show working (e. how you group the bits). 6 (3 points) Using the repeated division method, represent the decimal unsigned whole number 347 in base-3 CAUTION You will earn NO CREDITS if you simply show the final result and not clearly show working (ie, intermediate steps).

Answer 1

2.

10011101 in binary can be written as:

1*2^7 + 0*2^6 + 0*2^5 + 1*2^4  + 1*2^3 + 1*2^2 +0*2^1 + 1*2^0

i.e. (1*2^6 + 0*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 +0)2^1 + 1

On further simplification,

i.e. (((((((1)*2^1 + 0)*2^1 + 0)*2^1 + 1)*2^1 + 1)*2^1 + 1)*2^1 +0)*2^1 + 1

i.e. (((((((1)*2 + 0)*2 + 0)*2 + 1)*2 + 1)*2 + 1)*2 +0)*2 + 1

i.e. 157 in decimal

3.

14032 in base 6 can be written as:

1*6^4 + 4*6^3 + 0*6^2 + 3*6^1 + 2*6^0

i.e. (1*6^3 + 4*6^2 + 0*6^1 + 3)*6^1 + 2

i.e. ((((1)*6^1 + 4)*6^1 + 0)*6^1 + 3)*6^1 + 2

i.e. ((((1)*6 + 4)*6 + 0)*6 + 3)*6 + 2

i.e. 2180 in decimal

4.

Long Division:

Format is 2 |Dividend | Remainder

2|347|1

2|173|1

2|86 |0

2|43 |1

2|21 |1

2|10 |0

2|5 |1

2|2 |0

2|1 |1

Write the remainders from bottom to top: 101011011

347 is 101011011 in binary.

5.

101011011 can be written as 0001 0101 1011

0001 0101 1011 can be written as 15B in hex (0001 is 1 in hex, 0101 is 5, and 1011 is B)

6.

Long Division:

Format is 5 |Dividend | Remainder

5|347|2

5|69 |4

5|13 |3

5|2 |2

Write the remainders from bottom to top: 2342

347 is 2342 in base 5.

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