This is a paper and pencil exercise. Show your calculations.

1. Complete the table. Show your calculation. Each column represents a base and each row should have the same number in it, represented by the various bases.

2 | 8 | 10 | 16 |
---|---|---|---|

10111100 | 0274 | 188 | 0xbc |

1010101 | 0125 | 85 | 0x55 |

11100000 | 0x340 | 224 | 0xE0 |

111111110101 | 07765 | 4085 | 0xFF5 |

*Solution.*

Let's do the top row

10111100_{2}→ 10 111 100 2 7 4 → 0274 10111100_{2}= 188 1 0 1 1 1 1 0 0 2 2 4 10 22 46 94 188 _________________________________________ 1 2 5 11 23 47 94 188 10111100_{2}→ 1011 1100 b c → 0xbc

Now for the second row.

0125 = 0 1 2 5 1 010 101 → 1010101 1010101_{2}= 101 0101 5 5 → 0x55 0x55 = 5*15 + 5 = 85.

Now for row 3

224 % 16 = 0 = 0x0 224/16 = 14 = 0xD 224 = 0xD0. To get binary: 0xD0 → 111000000 To get Octal 111000000 → 111 000 000 7 0 0 → 0700

Finally the last row.

First decimal F F 5 16 240 4080 _________________________________ 15 255 4085

2. Fill in the blanks. Each question mark stands for a single digit.

0x?4 = 124?_{5}

*Solution.*

Let's give names to the missing digits like so 0xp4 = 124q_{5}. Converting, we know 0xp4 = 16p + 4 and that 124q_{5}= 125 + 50 + 20 + q. Equating these we get 16p + 4 = 195 + q or 16p = 191 + q. By our definition of base, we know 0 ≤ p < 16 and 0 ≤ q < 4, and that p and q must be integers. Let's try all possible values for q. q = 0: This gives 16p = 191. No q works q = 1: This gives 16p = 192. q = 12 = 0xc works q = 2: This gives 16p = 193. No q works q = 3: This gives 16p = 194. No q works q = 4: This gives 16p = 195. No q works We get the solution 0xc4 = 1242_{5}

3. How can you use % to obtain the last digit of a number represented by the
symbol `x`? The last two digits? What can you say in general?

*Solution.* If `x` is a number and `b` is the
base, then `x % b` is the last digit. Here is a quick reason why.

Suppose we have a number in base `b`, `ABCD.... Z`. The
last digit is in the base alphabet so `0 ≤ Z ≤ b`. Any
other term has place value of `b` or higher; each contributes
a term in a sum divisible by `b`. Therefore,

ABC....Z % b = Z

4. **A Balanced Ternary Expansion** It is possible to uniquely write any
integer in an expansion of the form

e_{0}+ 3*e_{1}+ 3^{2}e_{2}+ . . . + e_{n}*3^{n},

where each of the `e _{k}`s are 0, 1 or -1. Find such an expansion for each
of the folllowing numbers: 5, 13, 37 and 79.

*Solution.*

5 = 9 - 3 - 1 13 = 9 + 3 + 1 37 = 27 + 9 + 1 79 = 81 - 3 + 1

5. **Be an agent of Change** Describe how to count out any
amount between $0.00 and $1.00$ using standard coins to use as few coins
as possible. You will find using `//`

and `%`

handy.

*Solution.*

Suppose you have `x`

cents. We will proceed greedily. Let

```
q ← x//25
x ← x %25
```

This procedure uses as many quarters as possible. It then reassigns
`x`

to the amount left over after taking the quarters.

Now do this.

```
d ← x//10
x ← x%10
```

We now oink up the dimes and revise the amount. Repeat this for nickels
and pennies and you will use `q`

quarters, `d`

dimes,
`n`

nickels, and `p`

pennies.