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Further Maths AQA GCSE exams

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Original post by NiamhM1801
Ah we did it as Easter holiday homework - I didn't like paper 1 that much (62/70) but paper 2 was honestly the best one I've ever sat!! (102/105!!)
It was hard but an amazing challenge, I adapted to it better than any other paper.
Good luck for tomorrow - hope you smash it :smile:


Haha, I hope you smash it too :wink:
I am guessing you enjoy maths then, same here. Looks like you're aiming for that 100%, based on those marks!! We only did paper 2 for our mock, on which I didn't do as well as I have been doing on past papers recently, but that was in December - 90/105.

Yeah, hope I can get an A^.
Are you taking Maths/Further Maths to A level?
(edited 7 years ago)
Reply 101
Avatar for Jakir
Jakir
14
Original post by Saihan
Anyone want to prove they are a real A^ candidate?
Answer this question:

Prove that if you add any two digit number from the 9 times table to the reverse of itself (that is, swap the tens digit and units digit) the result is always be 99.


What about 9*11 = 99

It doesn't work
Reply 102
Avatar for Saihan
Saihan
1
Original post by Jakir
What about 9*11 = 99

It doesn't work


Umm sorry if the question isn't clear, but what it means is why does 18+81 (it's reverse) or 72+27 equal 99. You're meant to prove it.
Original post by Jakir
What about 9*11 = 99

It doesn't work


Add.
Original post by Ishan_2000
Haha, I hope you smash it too :wink:
I am guessing you enjoy maths then, same here. Looks like you're aiming for that 100%, based on those marks!! We only did paper 2 for our mock, on which I didn't do as well as I have been doing on past papers recently, but that was in December - 90/105.

Yeah, hope I can get an A^.
Are you taking Maths/Further Maths to A level?


Yeah I love it - wow 100% looks very unlikely. I might be good but I'm not that good!!
That's still a fantastic score, I think you're well in with a chance of getting that A^ :biggrin:
And yup, I'm doing them both along with Physics. What about you?
Original post by Saihan
Umm sorry if the question isn't clear, but what it means is why does 18+81 (it's reverse) or 72+27 equal 99. You're meant to prove it.


Original post by TheDragonGuy
Add.


I understand what he meant.
99 is a 2 digit number in the 9 times table
The reverse of 99 is 99
99+99=198, therefore it doesn't work
Original post by NiamhM1801
I understand what he meant.
99 is a 2 digit number in the 9 times table
The reverse of 99 is 99
99+99=198, therefore it doesn't work


Well you know what he meant anyway...
Original post by TheDragonGuy
Well you know what he meant anyway...


Yeah I know
Bearing in mind there's no way I could prove that algebraically I shouldnt really be saying anything haha
Original post by Saihan
Anyone want to prove they are a real A^ candidate?
Answer this question:

Prove that if you add any two digit number from the 9 times table to the reverse of itself (that is, swap the tens digit and units digit) the result is always be 99.


Given n in the range 1<=n<=10 (forgive my lack of an less than / equal to sign (using C# notation))
The reverse of 9n is 9(11-n) (such that 2*9 (18) mirrors 9*9 (81), and 3*9 (27) mirrors 8*9 (72))
9n + 9(11-n)
9n + 99 - 9n
therefore, sum of 9n + 9(11-n) = 99.

This works with all numbers btw, but the 'reverse' is specific with my given range.
(edited 7 years ago)
Original post by Saihan
Anyone want to prove they are a real A^ candidate?
Answer this question:

Prove that if you add any two digit number from the 9 times table to the reverse of itself (that is, swap the tens digit and units digit) the result is always be 99.


How about you prove that ANY number is divisible by 3 if the sum of its digits are divisible by 3.
Original post by Ano123
How about you prove that ANY number is divisible by 3 if the sum of its digits are divisible by 3.

You can split 10^n (∀n∈Z) into ((10^n) - 1) + 1, e.g. 100 is 99 + 1
So, take the number 3216; it can be split (in base 10) to 3000 + 200 + 10 + 6
Which is 3(999 + 1) + 2(99 + 1) + 1(9 + 1) + 6.
See how everything in the bracket has a 1, and a repetition of 9 (which is a multiple of 3, such that 999 is 3*333), it becomes this;
3 + 2 + 1 + 6 = 12; a multiple of 3, hence the number is divisible by 3. I can ignore the 999, 99 etc. as we already know they're divisible by 3.
(edited 7 years ago)
Original post by lin-e

You can split 10^n (∀n∈Z) into ((10^n) - 1) + 1, e.g. 100 is 99 + 1
So, take the number 3216; it can be split (in base 10) to 3000 + 200 + 10 + 6
Which is 3(999 + 1) + 2(99 + 1) + 1(9 + 1) + 6.
See how everything in the bracket has a 1, and a repetition of 9 (which is a multiple of 3, such that 999 is 3*333), it becomes this;
3 + 2 + 1 + 6 = 12; a multiple of 3, hence the number is divisible by 3. I can ignore the 999, 99 etc. as we already know they're divisible by 3.


Not very rigorous though is it mate? Publish it.

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