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#### jamonwindeyer

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #15 on: February 01, 2016, 11:18:46 pm »
+3
Thankyou so much both Jake and Happy Physics Land! Your responses have given me a new confidence that binomial isn't as intimidating as I think it is
Jake, if you don't mind, could you please show me the solution to question 5b from the 2001 3u paper?

Hey Phillorsm! Awesome question, and glad to hear that binomial is growing you! I'll be honest and say it is still is one my least favourite parts of teaching the course, though . That being said, I mustered the courage to tag in for Jake and write a quick solution for you!  this question works best by writing out a whole bunch of terms, so try doing that and following on with this quick solution. If you need a bit more detail, please let me know!

#### jamonwindeyer

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #16 on: February 01, 2016, 11:39:01 pm »
+3
Hi, I'm a bit confused with this induction question. I was wondering if you could explain to me the key steps. Thank you!

The sum of consecutive odd positive integers is divisible by 4.

Hi there IkeaandOfficeworks! I bought a wardrobe from you a few weeks back

So this is an awesome question, thanks so much for posting it.

However, I'm interested, because I don't think you can prove this with induction, because it isn't necessarily true!

The sum of consecutive odd positive integers is divisible by 4. So that means 1+3 should be divisible by 4. And it is. So should 1+3+5? And already we hit an exception the rule.

I am thinking that perhaps there is something missing from your question, some other condition.  Or equally likely, I'm missing something because it is 11:30

So, please get back to us! However, in the mean time, I'll leave you with a general method for induction questions.

1 - Express the question algebraically. Put it into a formula you can manipulate and change to suit your needs, usually in terms of n (EG - Divisibility induction questions like this would have some expression equal to 4M, where M is some integer, some involve inequalities, etc). This is actually sometimes the hardest part!
2- Prove the result for the lowest value you need to (usually 1, sometimes higher)
3 - Now, assume the result is true for n=k. Then, use this result to prove for  n=k+1. Almost always, this involves rearranging to allow a substitution. The algebra is normally fairly easy, but it is hard to know where to go. The solution? Practice makes perfect!
4 - Conclude that since the result is true for n=1, and n=k+1, it is true for n=k+1=2, n=2+1=3, etc etc! This ending is really important, lots of people forget it, and its worth a mark by itself!

I hope this helps for your induction questions, feel free to keep throwing them at us! Sorry I couldn't be of more help with this one

#### Happy Physics Land

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #17 on: February 01, 2016, 11:54:50 pm »
+3
Hey Phillorsm! Awesome question, and glad to hear that binomial is growing you! I'll be honest and say it is still is one my least favourite parts of teaching the course, though . That being said, I mustered the courage to tag in for Jake and write a quick solution for you!  this question works best by writing out a whole bunch of terms, so try doing that and following on with this quick solution. If you need a bit more detail, please let me know!

(Image removed from quote.)

Hey Ikeaand Im sorry everything is out of stock (Haha lm funny!):

There are two major issues with this question that you have proposed, and I think this is a slightly more challenging question than the usual mathematical induction questions which simply throw you an equation and tell you to prove the identity.

In this question, however:
1. It mentioned CONSECUTIVE ODD positive integers, however didnt specify how many there are. Initially I thought this would be an extremely challenging problem because the question sounds exactly like to prove "no matter how many consecutive positive odd integers there are, the sum of all these terms will all be divisible by 4". BUT, when you set out a few simple examples, you will discover that this is not the case. Lets start off with 1+3, they are consecutive and they add to 4, which is divisible by 4. Sweet. Now let's consider 1+3+5, they are consecutive however they add to 9, which is not divisible by 4. Furthermore, if we consider 1+3+5+7+9, we will get a sum of 25 which is also not divisible by 4. This creates a discrepancy to the way I initially defined the question. So I gave it a thought and I discovered that if we only do two consecutive odd numbers, e.g. 1+3 = 4, 3+5 = 8, 5+7 = 12, 7+9 = 16, these sums are all divisible by 4. So I think what the question really means is "to prove that the sum of ADJACENT odd positive integers is divisible by 4".
2. Now, another problem is that we are not provided with an equation to prove, which means that unfortunately we will need to construct our own equations, which, is easy enough. If we let n = the lower of the two adjacent odd numbers, and n+2 = the greater of the two adjacent odd numbers,

Here is my worked solution:

And at the end of your mathematical induction proof, make sure that you put a concluding statement "Since the it is true for n=1, assumed true for n=k and proved true for n=k+1, then by the principle of mathematical induction, the sum of consecutive odd positive integers is divisible by 4".

It is possible, however, that I have understood the question differently from your teacher, and if this is the case, please inform me and I will upload another solution as soon as possible!

Best Regards
Happy Physics Land
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#### Happy Physics Land

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #18 on: February 01, 2016, 11:58:13 pm »
0
Hi there IkeaandOfficeworks! I bought a wardrobe from you a few weeks back

So this is an awesome question, thanks so much for posting it.

However, I'm interested, because I don't think you can prove this with induction, because it isn't necessarily true!

The sum of consecutive odd positive integers is divisible by 4. So that means 1+3 should be divisible by 4. And it is. So should 1+3+5? And already we hit an exception the rule.

I am thinking that perhaps there is something missing from your question, some other condition.  Or equally likely, I'm missing something because it is 11:30

So, please get back to us! However, in the mean time, I'll leave you with a general method for induction questions.

1 - Express the question algebraically. Put it into a formula you can manipulate and change to suit your needs, usually in terms of n (EG - Divisibility induction questions like this would have some expression equal to 4M, where M is some integer, some involve inequalities, etc). This is actually sometimes the hardest part!
2- Prove the result for the lowest value you need to (usually 1, sometimes higher)
3 - Now, assume the result is true for n=k. Then, use this result to prove for  n=k+1. Almost always, this involves rearranging to allow a substitution. The algebra is normally fairly easy, but it is hard to know where to go. The solution? Practice makes perfect!
4 - Conclude that since the result is true for n=1, and n=k+1, it is true for n=k+1=2, n=2+1=3, etc etc! This ending is really important, lots of people forget it, and its worth a mark by itself!

I hope this helps for your induction questions, feel free to keep throwing them at us! Sorry I couldn't be of more help with this one

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
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#### jamonwindeyer

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #19 on: February 02, 2016, 12:19:39 am »
+1
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

Hey HappyPhysicsLand! I was providing a general method, so it entirely depends on the question. But yes, in this case, it would be n=k+2. If it were consecutive multiples of 5, it would be n=k+5, and so on. Apply to the situation at hand.

Also, I think it is highly likely that two consecutive odd integers is what the question specifies. Any even multiple of odd numbers would satisfy the condition, but two is what was most likely asked, which you've covered excellently.

Also, very impressed with your use of set notation in the solution. Little tip, you can use an N (with the same style typography as your Z) to denote all positive integers (and usually including 0), or more correctly, the set of all natural numbers . Virtually no difference to what you've done, and certainly not a necessity, just FYI!

#### jamonwindeyer

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #20 on: February 02, 2016, 12:24:02 am »
0
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

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

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #21 on: February 02, 2016, 10:16:39 am »
+1
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.

Hello Jamon:

Thank your for your recommendations and also for giving approval to my induction proof !  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!
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#### jakesilove

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #22 on: February 02, 2016, 11:03:51 am »
+3
Hello Jamon:

Thank your for your recommendations and also for giving approval to my induction proof !  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!

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
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#### cajama

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #23 on: February 06, 2016, 07:26:52 pm »
+2
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

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #24 on: February 06, 2016, 07:52:10 pm »
+3
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.

#### Happy Physics Land

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #25 on: February 06, 2016, 07:53:02 pm »
+5
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)

Hey Cajama:

Firstly I just wanna to assure you here that they are not typos, Im fairly certain that the questions are correct despite the answer does seem quite ridiculous. Good job on answering part a), remembering to divide by 2! which shows that you have recognised that there is a repetition.

Now for part B, I would like to invite you to draw just four underline dashes, each representing a number that can be placed in that spot. (i.e. __ __ __ __)

On the first underscore dash, there are only 4 numbers that can be placed there， which are 4, 4, 5, 6 (you cannot select 3 as the first digit because the number has to be greater than 4000). On the second underscore, there are also only 4 numbers that can be placed there (because a number, we dont know which one, has already been selected, leaving us with 4 options to choose from.) On the third underscore, we will only have 3 options to choose from because we have already selected two of them from all the numbers that we are provided with. And finally on the last underscore we will only have 2 options to choose from because we have already selected 3 numbers from all the numbers that the question provides us with.

Hence, the total number of different arrangements will seem to be 4 x 4 x 3 x 2 = 96 ways. BUT, WE HAVE NOT YET ACCOUNTED FOR THE REPETITION OF 4 IN THE NUMBERS THAT WE ARE PROVIDED WITH. Hence similar to what you have done in part a), it is necessary for you to divide 96 by 2! and you will obtain an answer of 48.

Ok, now we are up to part c). This one will be extremely similar to part b). I would recommend you to follow what I did in part b) and attempt part c) yourself to check whether you have truly understood what I have done. I will still post the solution below, but have an attempt on yourself.

So, similar to part b), draw four underscores __ __ __ __, each representing a digit.

On the first digit, we can only have three options of numbers, because from all the numbers we are given, only three of them will suit the criteria of "a number that is less than 5000". These three numbers, evidently, are 3, 4, 4. The second digit will have 4 options for us to choose from, because there is a total of 5 numbers which we are provided with and we have already selected one. The third digit is the same, we will have 3 options to choose from because we have already chosen 2 numbers from the 5 for the first two digits. And for the final digit, we will have 2 options to choose from because we have already chosen 3 numbers for the first 3 digits.

Hence the total amount of numbers that we can make below 5000 will be (3 x 4 x 3 x 2) / 2! = 36 (2! is to account for the repetition of 4 in the numbers that we are provided with)

Anyways great questions, hope you have understood my solution. Don't hesitate to post more questions if you need further assistance!!!

Best Regards
Happy Physics Land
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#### Happy Physics Land

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #26 on: February 06, 2016, 07:57:26 pm »
+3
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.

Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.
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#### jakesilove

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #27 on: February 06, 2016, 09:58:44 pm »
+1
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)

Great answers all! Glad this forum is developing so quickly, and there are so many people willing to jump in and help others out!!!

Jake
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#### nerdgasm

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##### Re: 98 in 3U Maths: Ask me Anything!
« Reply #28 on: February 07, 2016, 01:34:50 am »
+4
Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.

You're definitely right. the suggested answers work just as well if the question is considering 4-digit arrangements. I think that your interpretation is more likely to be correct  and that there was no typo. Good job on your explanation too, I thought it was really clear!

*(The following is probably not part of the 3U maths course)*:
I think that the reason why we were able to get the same numerical answers despite our different interpretations is as follows:
Let us define a finite 'pool' of n members (the members are not necessarily distinct). I claim that the number of different n-member arrangements is the same as the number of (n-1)-member arrangements.

My idea is as follows: I propose a mapping from the set of n-member arrangements to the set of (n-1)-member arrangements. This mapping basically chops the last member off the n-member arrangement to get an (n-1)-member arrangement.

To show that this mapping is injective (or one-to-one), I use proof by contradiction: assume we have two distinct n-member arrangements that map to the same (n-1)-member arrangement. Then, it follows the two n-member arrangements must agree in their 1st, 2nd ... (n-2)th and (n-1)th positions. But because the pool is finite, it follows that there is only one member in the pool left to make the nth position. Therefore, the two n-member arrangements must in fact be the same.

To show that this mapping is surjective (or onto), note that for any (n-1)-member arrangement, there must be exactly one member from the original pool that was not used. Stick this member onto the end to form a n-member arrangement, which will map to the (n-1)-member arrangement.

Therefore, as this mapping is both injective and surjective, it follows that there are exactly the same number of n- and (n-1)-member arrangements from a finite pool of n members.

I think this explanation also explains our similarity in answers to parts b) and c) as well: in part b), since the starting digit (to be excluded) is 3 in both of our interpretations, we are essentially trying to place 4, 4, 5 and 6 into a 3-member arrangement or a 4-member arrangement in order to work out how many to exclude, which can be done in exactly the same number of ways. And similarly in part c), we are trying to place 4, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 3, and we are trying to place 3, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 4, both of which can be done in exactly the same number of ways.

#### foodmood16

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