HSC Stuff > HSC Mathematics Extension 1
3U Maths Question Thread
jamonwindeyer:
--- Quote from: Happy Physics Land on February 01, 2016, 11:58:13 pm ---Hey Jamon:
Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?
Best Regards
Happy Physics Land
--- End quote ---
Oh, and I almost forgot, don't forget to conclude your induction proof! You've proved for n=k+2, you need to extrapolate this for further numbers. Just a simple, it is true for n=1, and n=k+2, so it is true for n=1+2=3, n=3+2=5, etc, for any sum of two consecutive odd positive integers, will suffice.
Happy Physics Land:
--- Quote from: jamonwindeyer on February 02, 2016, 12:24:02 am ---Oh, and I almost forgot, don't forget to conclude your induction proof! You've proved for n=k+2, you need to extrapolate this for further numbers. Just a simple, it is true for n=1, and n=k+2, so it is true for n=1+2=3, n=3+2=5, etc, for any sum of two consecutive odd positive integers, will suffice.
--- End quote ---
Hello Jamon:
Thank your for your recommendations and also for giving approval to my induction proof :D ! I think that the use of N notation to represent all the positive integers is quite a clever idea and I really liked the format of the conclusion, very succinct but effective as a concluding statement! But overall thank you for checking my solution!
jakesilove:
--- Quote from: Happy Physics Land on February 02, 2016, 10:16:39 am ---Hello Jamon:
Thank your for your recommendations and also for giving approval to my induction proof :D ! I think that the use of N notation to represent all the positive integers is quite a clever idea and I really liked the format of the conclusion, very succinct but effective as a concluding statement! But overall thank you for checking my solution!
--- End quote ---
Hey all, just thought I'd quickly tag in here.
In regards the the Induction question, Happy Physics Land I think you're right on the money. The question MUST be asking about only two consecutive odd numbers, just because otherwise it isn't true. Which leads me to think that the question itself is phrased badly.
So I think overall a great group effort that eventually got to the right place.
Great job all!
Jake
cajama:
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
nerdgasm:
I think there is a typo; 40000 and 50000 should be what is written in the question (given the answers provided), or else every 5-digit arrangement would be larger than both 4000 and 5000.
Assuming that 40000 and 50000 are respectively meant:
For part b), I find it easier to look at which arrangements are not counted, because there are fewer cases to consider. In this case, only those arrangements starting with a 3 will be not counted. So, let's assume the starting digit of the arrangement is 3, so we have 3 _ _ _ _.
We then have to place 4, 4, 5 and 6 into the remaining four slots. This can be done in 4!/2! = 12 ways, by a similar argument to what you did in part a). All other arrangements work, so the total number of successful arrangements is 60 - 12 = 48.
For part c), we can again use the same 'counting the complement' technique. This time, we exclude those arrangements starting with 5 or 6. By the same working out as in part b), we can deduce that if the starting digit is 5, there are 12 arrangements (as you then have 5 _ _ _ _ with 3, 4, 4, 6 to place), and there are also 12 arrangements if the starting digit is 6 (6 _ _ _ _ , with 3, 4, 4, 5 to place). Hence, the total number of arrangements less than 50000 is 60 - 12 -12 = 36.
It is also possible to count the arrangements that are included for part c), but you do have to be a bit careful: if the starting digit is 3, then there are 12 successful arrangements. However, note that if the starting digit is 4, then you have 4 _ _ _ _ with 3, 4, 5, 6 to place. Because the remaining digits are all distinct, there are 4! ways to place them, and hence there are 24 successful arrangements. Therefore, the number of successful arrangements is 12 + 24 = 36, which matches the result we obtained earlier.
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