Intermediate Maths Challenge - UKMT - 5th February 2015 Watch

Shadoo
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(Original post by Infaredz)
Decided to do a list of the answers in the thread as a sort of general sheet for people to check by.

1. A
2. E
3. E
4. C
5. B
6. D
7. E
8. B
9. B
10. C
11. D
12. D
13. A
14. C
15. C
16. D
17. D
18. A
19. A
20. C
21. D
22. B
23. B
24. E
25. D

Enjoy!
got 60 due to careless errors based on your markscheme. Anyone know when solutions are published?
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alpegu66
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http://www.ukmt-resources.org.uk/IMC15.html
I got 77. Everyone post your results!!
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alpegu66
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(Original post by Shadoo)
got 60 due to careless errors based on your markscheme. Anyone know when solutions are published?
http://www.ukmt-resources.org.uk/IMC15.html
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yeolk
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Terrible, got less than 70 bc of careless mistakes.
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yeolk
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Does anyone know what score you need for silver?
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alpegu66
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(Original post by yeolk)
Does anyone know what score you need for silver?
We can't know yet
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marmbite
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I think I got 60 that's better than I've done before I think. Might be enough for silver, maybe? Just guessing by looking at least years boundaries, can't know for sure though. I'll be happy if it is silver though because I've never got past bronze before...
At least our actual GCSEs are not cryptic like the maths challenge.
Still, well done everyone!
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theworkkid
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Do you think the boundaries will be higher/ lower than last years?
Be honest!
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carpetguy
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(Original post by yeolk)
Does anyone know what score you need for silver?
(Original post by theworkkid)
Do you think the boundaries will be higher/ lower than last years?
Be honest!
I got 79, should be gold maybe not kangaroo
I personally think the paper was easier than last year so boundaries higher:

(Original post by UKMT)
The Intermediate Challenge 2014 thresholds were:
Bronze 42-54 Silver 55-71 Gold ≥ 72+
Please note these are for guidance only; the thresholds are recalculated each year based on the national score distribution.

Follow on rounds
Cayley Olympiad 90
Hamilton 96
Maclaurin 103
Grey Kangaroo 67
Pink 78
Please note these are for guidance only; the thresholds are recalculated each year based on the national score distribution.
I'm in yr 10 so missed out massively on Hamilton... I may actually get pink kangaroo!
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carpetguy
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(Original post by marmbite)
I think I got 60 that's better than I've done before I think. Might be enough for silver, maybe? Just guessing by looking at least years boundaries, can't know for sure though. I'll be happy if it is silver though because I've never got past bronze before...
At least our actual GCSEs are not as cryptic as the maths challenge.
Still, well done everyone!
U mad bro? ¯\_(ツ)_/¯

GCSEs are a walk in the park with crutches and massive fluorescent arrows pointing in the right direction. Not to mention a ginormous red flag at the finish line waving at you the whole way. (Compared to the IMC)
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alpegu66
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(Original post by theworkkid)
Do you think the boundaries will be higher/ lower than last years?
Be honest!
Judging by what my friends got I'd think lower
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Bacopa
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Hi guys I did the paper yesterday and found out the answers today, but I still don't understand the answer to question 10. (remainder question).
Could anyone explain it to me?
Thanks
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alpegu66
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(Original post by Bacopa)
Hi guys I did the paper yesterday and found out the answers today, but I still don't understand the answer to question 10. (remainder question).
Could anyone explain it to me?
Thanks
there must be another way but I just did the multiplication and divided by 8, it took me about 12 min >
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Chittesh14
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(Original post by Bacopa)
Hi guys I did the paper yesterday and found out the answers today, but I still don't understand the answer to question 10. (remainder question).
Could anyone explain it to me?
Thanks
Look, basically when you multiply a odd number by itself, unlimited times, the answer will still remain odd.
For example, 3^3 is a odd number too, and so will the rest of the numbers.
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243.

So all; 3^3, 5^5 and 7^7 are odd.
When you multiply an odd number by an odd number, the answer will be odd. So, 3^3 * 5^5 * 7^7 is an odd number.
When you multiply an odd number, by an even number, it will be even. So the answer is going to be odd, and so will the reminder.
As any number that is even and is divided by 8 will have an even remainder.
For example, 12 / 8 = 1 remainder 4. 28 / 8 = 7 remainder 4.

First, we note that 33 × 55 × 77 is a positive odd integer greater than 1 and so is equal to 2n + 1,
for some positive integer n.

It follows that

2^2 × 3^3 × 5^5 × 7^7= 4×(3^3 × 5^5 × 7^7) = 4(2n + 1)
= 8n + 4.

Therefore when 22 × 33 × 55 × 77 is divided by 8, the quotient is n and the remainder is 4.


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Bacopa
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Thanks a lot!
I understand now
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Chittesh14
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(Original post by Bacopa)
Thanks a lot!
I understand now
Np !


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RaamSongara
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I got 97, do you think I will make it to Cayley?
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Chittesh14
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(Original post by RaamSongara)
I got 97, do you think I will make it to Cayley?
If you're in Year 9, yes easily.
That's a fantastic score for a Year 9, real talent.
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RaamSongara
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Thanks been trying quite hard and revising
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RaamSongara
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I presume the boundaries will be lower seeing as quite a lot of people from my school got a low mark on average
what do you think?
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