Note: These are my approaches to these problems. There are many ways to tackle each.
60 | 61 | 62 | 63 | 64 | 65 |
70 | 71 | 72 | 73 | 74 | 75 |
{2}^{0} = | 1 |
{2}^{1} = | 2 |
{2}^{2} = | 4 |
{2}^{3} = | 8 |
{2}^{4} = | 16 |
{2}^{5} = | 32 |
{2}^{6} = | 64 |
{2}^{7} = | 128 |
{2}^{8} = | 256 |
{2}^{9} = | 512 |
{2}^{10} = | 1024 |
0 = | {0}_{2} = | 0000{0}_{2} |
1 = | {1}_{2} = | 0000{1}_{2} |
2 = | 1{0}_{2} = | 0001{0}_{2} |
3 = | 1{1}_{2} = | 0001{1}_{2} |
4 = | 10{0}_{2} = | 0010{0}_{2} |
5 = | 10{1}_{2} = | 0010{1}_{2} |
6 = | 11{0}_{2} = | 0011{0}_{2} |
7 = | 11{1}_{2} = | 0011{1}_{2} |
8 = | 100{0}_{2} = | 0100{0}_{2} |
9 = | 100{1}_{2} = | 0100{1}_{2} |
10 = | 101{0}_{2} = | 0101{0}_{2} |
11 = | 101{1}_{2} = | 0101{1}_{2} |
12 = | 110{0}_{2} = | 0110{0}_{2} |
13 = | 110{1}_{2} = | 0110{1}_{2} |
14 = | 111{0}_{2} = | 0111{0}_{2} |
15 = | 111{1}_{2} = | 0111{1}_{2} |
16 = | 1000{0}_{2} = | 1000{0}_{2} |
17 = | 1000{1}_{2} = | 1000{1}_{2} |
18 = | 1001{0}_{2} = | 1001{0}_{2} |
19 = | 1001{1}_{2} = | 1001{1}_{2} |
20 = | 1010{0}_{2} = | 1010{0}_{2} |
21 = | 1010{1}_{2} = | 1010{1}_{2} |
22 = | 1011{0}_{2} = | 1011{0}_{2} |
23 = | 1011{1}_{2} = | 1011{1}_{2} |
24 = | 1100{0}_{2} = | 1100{0}_{2} |
25 = | 1100{1}_{2} = | 1100{1}_{2} |
26 = | 1101{0}_{2} = | 1101{0}_{2} |
27 = | 1101{1}_{2} = | 1101{1}_{2} |
28 = | 1110{0}_{2} = | 1110{0}_{2} |
29 = | 1110{1}_{2} = | 1110{1}_{2} |
30 = | 1111{0}_{2} = | 1111{0}_{2} |
31 = | 1111{1}_{2} = | 1111{1}_{2} |
One of the more important patterns to see is how the digits repeat once padded to the left by zeros. The units (or right-most) digit alternates 0,1,0,1,\mathop{\mathop{…}}. The next alternates in pairs, 0,0,1,1,0,0,1,1,\mathop{\mathop{…}}. The next in groups of four, the next in groups of eight, and so forth.
+ | 0 | 1 | 2 | 3 |
0 | 0 | 1 | 2 | 3 |
1 | 1 | 2 | 3 | 0 |
2 | 2 | 3 | 0 | 1 |
3 | 3 | 0 | 1 | 2 |
3 | 7 | 4 | ||
0 | 0/6 | 1/4 | 0/8 | 2 |
8 | 0/3 | 0/7 | 0/4 | 1 |
0 | 1/5 | 3/5 | 2/0 | 5 |
4 | 1 | 0 |
So 374 ⋅ 215 = 80410.
bit of 17 | doubling of 71 | accumulator |
0 | ||
0 | 71 | 0 |
0 | 142 | 0 |
0 | 284 | 0 |
1 | 568 | 568 |
1 | 1136 | 1704 |
So 71 ⋅ 24 = 1704.
Problem | Actual answer | Calculator’s answer |
8 × 3 | 24 | 34 |
9 × 5 | 45 | 55 |
4 × 2 | 8 | 18 |
8 × 4 | 32 | 42 |
a × b | x - 10 | x |
9 × 6 | 64 - 10 = 54 | 64 |
When the decimal logic was done outside a chip (the early 80s), this type of failure could happen from a simple short-circuit. Now it’s highly doubtful; a failure like this would require massive trauma to the chip, leading to general failure.