Submitted by Heirio on Mon, 08/07/2017 - 23:08
Forums:
I don't understand part c at all. For the last part of it, I didn't use binary, but series formulae. I got to the same answer, so all is fine there. But with the proving of 2^n (and by extension the part ii b), I have no clue why they doing what they're doing. I can see that the 2*2*2*...*2=2^n but I don't see where the number of options for the weight comes into it. Maybe it's because I'm tired, but could someone please explain? This is probably the hardest assignment I've come across and I've used the hints a bit too much.
On a side note, I feel like I
On a side note, I feel like I'm using the hints too much, largely because I often get very stuck or don't understand the question at all.
Should I just try and power through the questions? I'm worried that if I do that, it'll take so long that I won't have enough time to finish these by October. I'm thinking of looking at he Silkos book to help, just in case my overuse of hints is too detrimental. Is that okay?
Hints
IMO, it is better to use the hints to help you get an answer than to never get an answer to the question. I would suggest having a go without using the hints, then sleep on the problem and try again the next day. If on this second day you are still stuck, then look at the hints.
q4 i part c
When you are putting weights only on one side each weight can either be in the pan or not.
Assume you have weights {A,B,C}. Each weight can either be in the pan or not. The options are:
*** (no weights)
A**
AB*
A*C
ABC
*B*
*BC
**C
There are eight options here (some of them might give the same weight, e.g. if A=B+C, so we have at most eight different weights).
A can either be in the pan or not. This gives 2 options for A.
For each of these options for A, B can either be in the pan or not. The number of different combinations is now $2 \times 2=4$ (for each of the two options for A, there are two options for B).
For each of these combinations of A and B, there are 2 options for C (in the pan or not). The total number of combinations is now $2 \times 2 \times 2=8$.
By extension, if we had $n$ weights we could make at most $2 \times 2 \times \cdots \times 2=2^n$ different weights (including 0).
Let me know if this helps! (or not)
I'll have a go, but judging
I'll have a go, but judging by your response I most likely misread the question.
That itself looks to be something I need to work on.
Another solution to warm up
I'm posting a lot, so I think I'll put this one here. It's not urgent, but I have another solution for the warm up part iii which isn't mentioned. Multiply by x^2 to make a cubic, take it all to one side and factorise the x out. Then you have a quadratic inside. Factorise this and thus you can safely sketch the cubic, from which you can say the values of x for which the inequality holds true.
Yup
This is another way of solving the equation (we have not been exhaustive in our solutions!). Multiplying both sides by $x$ is a bad idea, but multiplying both sides by $x^2$ is fine, as (as long as $x$ is real) we have $x^2 \ge 0$.
Nice solution method, thank you!