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STEP foundation module help pls

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This question is from Assignment 11 of the step foundation module and Ik its a warm up question but I'm stuck on it. I understand that it's has something to do with triangular numbers but i cant solve the question

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Reply 1
Original post by Fndjdidisjb
C:\Users\44786\Pictures\Screenshots\Screenshot 2024-01-24 191402.png
This question is from Assignment 11 of the step foundation module and Ik its a warm up question but I'm stuck on it. I understand that it's has something to do with triangular numbers but i cant solve the question

Also not sure if you can open the link
Original post by Fndjdidisjb
Also not sure if you can open the link

Try uploading to imgur.com or your favourite image hosting site and linking to that.
(edited 2 months ago)
I think you're referring to this.
If you are looking at the T(k+1) = T(k) + k + 1 problem... well... It shouldn't take too long to brute force until you find T(n)>100.

I think finding out 20 terms directly is often worthwhile in the exploration/find-the-pattern phase.
Of course if you want to find, say, T(n)>10^2024, then brute force by hand is impossible - but I'll still find out what the pattern could possibly be.
(edited 2 months ago)
Reply 4
Original post by Fndjdidisjb
C:\Users\44786\Pictures\Screenshots\Screenshot 2024-01-24 191402.png
This question is from Assignment 11 of the step foundation module and Ik its a warm up question but I'm stuck on it. I understand that it's has something to do with triangular numbers but i cant solve the question

It would probably help to say what youre stuck with. As tony says, if necessary write down the first few values so 1,3,6,10 ... and you should recognise the triangular numbers. You should know the basics of triangular numbers, so the gauss trick, the fact that the closed form expression is quadratic as the difference (k+1) is linear, how it relates to an arithmetic series etc. Similarly for (basic) step being able to prove the previous is probably worth practising if youre unsure about them. If youre ok with that then solving the inequality is simply a quadratic, so gcse, but the triangular number expression gives n^2~200 (as well as working through the quadratic formula) which is ~a square number you could simply write down (though step would require some working/justification)
(edited 2 months ago)
Reply 5
@tonyiptony @mqb2766

Assuming you've identified the worksheet correctly here, I'd say it's fairly explicit that you're supposed to do it by calculating the values. I'd actually be hesitant to do anything other than brute force if I wanted to be sure of the marks.

[I actually did the brute force just to see how painful it was, and it was sub 30 second to just list the numbers, so...]
Reply 6
Original post by DFranklin
@tonyiptony @mqb2766

Assuming you've identified the worksheet correctly here, I'd say it's fairly explicit that you're supposed to do it by calculating the values. I'd actually be hesitant to do anything other than brute force if I wanted to be sure of the marks.

[I actually did the brute force just to see how painful it was, and it was sub 30 second to just list the numbers, so...]

Dont really disagree, though 14^2~200 is almost a write down. The previous stuff was just a bit more rambling as the foundation modules are more about getting you up to speed and knowing stuff about the triangular numbers would be "expected".
Reply 7
Thank you so much everyobe. firstly I tried writing the sequence till I got something over 100 and i got the ans. But then i wanted to do the gauss's trick. I tried applying that but still couldn't get to the final ans but i feel like my method is right.
https://imgur.com/hgdns5d this is my working. now how do i solve this quadratic. This is what I'm stuck at. I can do with calc but without calc, i can't.
(edited 2 months ago)
Reply 8
Original post by ghostwalker
Try uploading to imgur.com or your favourite image hosting site and linking to that.

Thnx for this. I've always struggled with sharing photos on the student room
Reply 9
Original post by Fndjdidisjb
C:\Users\44786\Pictures\Screenshots\Screenshot 2024-01-24 191402.png
This question is from Assignment 11 of the step foundation module and Ik its a warm up question but I'm stuck on it. I understand that it's has something to do with triangular numbers but i cant solve the question

What do you need the STEP for, just out of interest?
Original post by Fndjdidisjb
Thank you so much everyobe. firstly I tried writing the sequence till I got something over 100 and i got the ans. But then i wanted to do the gauss's trick. I tried applying that but still couldn't get to the final ans but i feel like my method is right.
https://imgur.com/hgdns5d this is my working. now how do i solve this quadratic. This is what I'm stuck at. I can do with calc but without calc, i can't.

Quadratic formula (I'll skip the "dealing with inequality" part) tells us n>1+8012n>\frac{-1+\sqrt{801}}{2}. Hopefully my mental maths is right.

So we need some sort of estimation of how big sqrt(801) is at least. Well 25^2 is 625, 30^2 is 900, so 4 integer choices left.
A little bit of case checking (or if you happen to remember your squares up to 30) gives us 28<sqrt{801}<29, so n>27/2. Hence take n=14 works (do verify 14 works, and not 13). Everything can be done by hand.

(Alternatively, subbing 1,2,... into the quadratic until one works is also okay. Not pretty, but gets the job done the fastest)

Now I should stress that while you have the idea of using Gauss' trick (whatever that means), it is not obvious why it works from the recurrence relation. You need to justify why the recurrence is somehow related to triangular numbers in the first place. This takes, perhaps, a proof by induction, which takes significantly more time than just calculating the recurrence 13/14 times.
(edited 2 months ago)
Reply 11
Original post by Fndjdidisjb
Thank you so much everyobe. firstly I tried writing the sequence till I got something over 100 and i got the ans. But then i wanted to do the gauss's trick. I tried applying that but still couldn't get to the final ans but i feel like my method is right.
https://imgur.com/hgdns5d this is my working. now how do i solve this quadratic. This is what I'm stuck at. I can do with calc but without calc, i can't.

You can simply note that its the product of consecutive integers which is > 200 so as 14^2 = 196, then 14*15>200 whereas 13*14<200.

If you solve it using the quadratic formula, you have to make a similar argument, for
n^2 + n - 200 = 0
So the roots are approximately equal (sum to -1) and their product is -200 (opposing signs) which is similar to the previous argument about consecutive integers, though going through with the quadratic formula you get
-1/2 +/- sqrt(200.25)
which is again the 14^2 is about 200 argument. The positve root is a bit less than 14 (due to the -1/2) so n=14.

You could also have noted the sign change in the quadratic between n=13 and n=14 which would be a way to solve it using the quadratic, but without actually calculating the root.

Its also worth noting that while its referred to as the gauss trick, really its just the arithmetic series expression so n*(first+last)/2.
(edited 2 months ago)
Reply 12
Original post by vnayak
What do you need the STEP for, just out of interest?

so i applied to imperial computing but im still waiting to hear back for interview. I'm also on a gap year and ive literally forgotten all my maths so thought STEP would help me get back on track
Reply 13
Original post by tonyiptony
Quadratic formula (I'll skip the "dealing with inequality" part) tells us n>1+8012n>\frac{-1+\sqrt{801}}{2}. Hopefully my mental maths is right.

So we need some sort of estimation of how big sqrt(801) is at least. Well 25^2 is 625, 30^2 is 900, so 4 integer choices left.
A little bit of case checking (or if you happen to remember your squares up to 30) gives us 28<sqrt{801}<29, so n>27/2. Hence take n=14 works (do verify 14 works, and not 13). Everything can be done by hand.

(Alternatively, subbing 1,2,... into the quadratic until one works is also okay. Not pretty, but gets the job done the fastest)

Now I should stress that while you have the idea of using Gauss' trick (whatever that means), it is not obvious why it works from the recurrence relation. You need to justify why the recurrence is somehow related to triangular numbers in the first place. This takes, perhaps, a proof by induction, which takes significantly more time than just calculating the recurrence 13/14 times.

Thank you for your detailed explanation. Triangular numbers are relatedto arithmetic series and Gauss' trick is all about the sum of a list of numbers,
Reply 14
Original post by mqb2766
You can simply note that its the product of consecutive integers which is > 200 so as 14^2 = 196, then 14*15>200 whereas 13*14<200.

If you solve it using the quadratic formula, you have to make a similar argument, for
n^2 + n - 200 = 0
So the roots are approximately equal (sum to -1) and their product is -200 (opposing signs) which is similar to the previous argument about consecutive integers, though going through with the quadratic formula you get
-1/2 +/- sqrt(200.25)
which is again the 14^2 is about 200 argument. The positve root is a bit less than 14 (due to the -1/2) so n=14.

You could also have noted the sign change in the quadratic between n=13 and n=14 which would be a way to solve it using the quadratic, but without actually calculating the root.

Its also worth noting that while its referred to as the gauss trick, really its just the arithmetic series expression so n*(first+last)/2.

The first line u typed is really smart. I never thought of it that way. Thanks for helping.
Reply 15
Also for this question from assignment 13. https://imgur.com/a/tsbmDRP . I understand the question and i know that its very basic but I'm kinda confused on part d. I worked out the two solutions so p=q=6 and p=1 and q=10 but I just don't know how to use part (c) in it and how to use k in it.
Reply 16
Original post by Fndjdidisjb
Also for this question from assignment 13. https://imgur.com/a/tsbmDRP . I understand the question and i know that its very basic but I'm kinda confused on part d. I worked out the two solutions so p=q=6 and p=1 and q=10 but I just don't know how to use part (c) in it and how to use k in it.

Youre starting with n=6,p=6,q=6 and are told the set of solutions corresponds to 6,p+5k,q-4k where k is an integer. So what values might k be to generate a valid solution for a new p and q. Youve got one (what value of k did it correspond to), did you try any other values of k, what did you find? At a very simple level you could imagine that if you just considered £5 and £1 nptes say, if you change the number of £5 notes by k (integer), then you need to change the number of £1 notes by -5k to keep the total amount of money the same.

A bit like triangular numbers, the question (with n=6) becomes a diophantine equation in 2 integer variables and its worth sometimes having a read about the background behind a question, so the fact that 4, 5 are co prime and the solution repeats every ... Then even if you dont remember all the theory, youre aware of the basic properties if you meet a similar question.
(edited 2 months ago)
Reply 17
Original post by Fndjdidisjb
so i applied to imperial computing but im still waiting to hear back for interview. I'm also on a gap year and ive literally forgotten all my maths so thought STEP would help me get back on track

Ah fair enough...I've got the offer for the same course so we might meet! What else have you been doing to prepare for it?
Reply 18
Original post by mqb2766
Youre starting with n=6,p=6,q=6 and are told the set of solutions corresponds to 6,p+5k,q-4k where k is an integer. So what values might k be to generate a valid solution for a new p and q. Youve got one (what value of k did it correspond to), did you try any other values of k, what did you find? At a very simple level you could imagine that if you just considered £5 and £1 nptes say, if you change the number of £5 notes by k (integer), then you need to change the number of £1 notes by -5k to keep the total amount of money the same.

A bit like triangular numbers, the question (with n=6) becomes a diophantine equation in 2 integer variables and its worth sometimes having a read about the background behind a question, so the fact that 4, 5 are co prime and the solution repeats every ... Then even if you dont remember all the theory, youre aware of the basic properties if you meet a similar question.

Thanks so much. Good point. I’ll read the background topic and then try doing it.
Reply 19
Original post by Fndjdidisjb
Thanks so much. Good point. I’ll read the background topic and then try doing it.

It was more the other way round so do the question (which is literally write down the solution for the two valid (non-zero) values of k), but then do a bit of reading about the general topic. All the co prime, multiple solutions, ... stuff is done for you in the question, but a bit of insight is always useful.

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