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# =Maths Competition= (answers + scores) watch

1. Ok, here are the problems to save the annoyance of opening two windows to check. By the way, easy, medium and hard were a bit misleading... since difficulty is relative, easy to different people means different things so I did my best. The easier ones were therefore easier, in my opinion, than medium and hard.

1. Find the sum and product of all the terms in the sequence: 1, 1/2, 1/4, 1/8 ...
THe sum is 2, because the geometric series has first term 1 and common ratio 1/2. Sum to infinity is S*=a/(1-r) = 1/(1-1/2) = 2. As the term goes to infinity, the "final" term is 0/2, and 0 of anything is 0.
2. Given that: a = 3^243; b = 27^81, what is the value of (a^2)/(b^2) ?
27^81 = (3^3)^81 = 3^243. So a = b and a^n/b^n = 1.
3. Find the nth term of the sequence: 1, 1/2, 1/6, 1/24, 1/120
Guesswork is a great thing sometimes... 1/n!.
4. 12!/n! = 11880. What is the value of n?
12!/n! = 11880
Through messy trial and error, n=8. Checking, (12*11*10*9) = 11880.

---meduim---

5. What is the value of: (1/4) + (1/4)^2 + (1/4)^3 + (1/4)^4 ... ?
Geometric series again, a= 1/4, r=1/4; S*=a/(1-r) =(1/4)/(3/4) = 1/3
6. With how many zeroes does the number 3003! end?
5 is the key number... divide 3003 by all 5^n until 3003/5^n < 1, and add up the rounded down answers to get ([3003/5] + [3003/25] + ... + [3003/625]) = 600 + 120 + 24 + 4 = 748.
7. What is the sum of the digits of (2^202)(5^204)?
Group the 202 sets of 2s and 5s together to give (2^202)(5^204) = (10^202)(5^2). This is 10000.... * 25, the digits are 2 and 5 (rest are zeroes), giving 7 as the sum.
8. Find x, if: 9^(2x) = 2*9^x + 3
9^(2x) - 2.9*(x) + 3 = 0
(9^x-3)(9^x+1)=0
9^x = 3; x = 1/2
9^x = -1; no solution. (thanks to Ralfskini for this elegant solution, the longer route is to use the substitution a = 9^x)

===hard===

9. Convert 1847 (base 9) into base 4
Ouch. Lots of dividing and remainders. Frankly, I can't do this. But the answer was 112030(base4)
10. If y/(x-z) = (x+y)/z = x/y, when x, y and z are distinct positive integers, then what is the ratio of x to y?
By simultaneous equations, z=y(x+y)/x and y/(x-z)=x/y. By substitution, and simplification, 2y²+xy-x²=0; (2y-x)(y+x)=0; y=x/2; x/y = 2.
11. Find the area of the largest equilateral triangle that can be inscribed in a rectangle with sides 10 and 11.
Again, solution way too tricky for the likes of me! 220√3-330 was the answer;
It is straightforward to get a quartic in x. Obviously 100 + y2 = 121 + x2. Then either use the third side or use the angles: y/10 = (tan 30 - x/11)/(1 + tan 30 (x/11)). Hence x4 + 22√3 x3 + 84x2 + 2662√3 x - 4477 = 0. This factorises (x2 + 22√3 x - 37)(x2 + 121). Hence x = 20 - 11√3. So area = (20 - 11√3)2(√3)/4.
12. Find the largest value of k, such that 3^(11) is the sum of k consecutive positive integers.
Take numbers as N+1, N+2, ... , N+k. So k(2N+k+1) = 2·311. Since N >= 0, we have k < 2N+k+1, and hence k < (2·311)1/2. Also k divides 2·311. Largest such k is obviously 2·35.
Sorry for the short change in answers, I don't get the last 4 myself!

Well done everyone who submitted answers, I was pretty impressed with them. The results everyone wanted to know are below, but as a disclaimer I don't know that all the answers above are correct... I wouldn't bet my life on it although I am quite sure. Anyway I'll stop talking.

It'sPhil: 60% - but only answered 8 questions, and got 15/17 so pretty good!
shift3: 52% - dropped some marks on early questions and picked some up later on... not quite the conventional way of doing a test, but strangely impressive!
Ralfskini: 100% - incredible... as you may have read, some of his answers were better than my own and other websites.
integral_neo: 84% - despite the fragmented answers scattered throughout my inbox, a very good score! hardly lost any marks.
Fermat: 84% - somehow dropping his only marks in the easy and medium section. Another impressive result.
MC REN: 16% - but this is not fair! I would award a much higher result for the effort of a GCSE student attempting these questions (and only answering a few). 4/4, 100%

BTW these results were all pretty good, those were tough questions I thought and me or my friends wouldn't be able to get very far. I wouldn't liken it to STEP or AEA as I haven't seen those (more like maths challenges actually), but I'd say getting only a few right deserves a lot of credit as they are mostly hard. I got about 2/3 of them from various internet sources and made the rest up, and I know I would have trouble with a lot of them.

Anyway thanks for replying! Well done everyone
2. I messed up the easier questions?!

Can you please pm me my answers, if you still have them?
3. 9. Convert 1847 (base 9) into base 4
Ouch. Lots of dividing and remainders. Frankly, I can't do this. But the answer was 112030(base4)

9. Isn't too hard if you approach it with a good method and lay your work out well:

1847 from base 9 to base 10 (It is easier to deal with change of base if you go to denary and then convert from that to your next base.)

7 * 9^0 = 7
4 * 9^1 = 36
8 * 9^2 = 648
1 * 9^3 = 729

Total = 1420 (Base 10)

Now we want to go to base 4, so work out the highest power of 4 less than 1420. That is 4^5 = 1024

1420 = 4^5 + 396
396 = 4^4 + 140
140 = 2*4^3 + 12
12 = 0*4^2 + 12
12 = 3*4^1 + 0

So we can now assemble the number in base 4 by taking the number of multiples of the descending powers of 4, remembering to add on a 0 for the missing 4^0 units column.

1847 (Base 9) = 1420 (Base 10) = 112030 (Base 4)

4. brilliant well done Michael!
5. (Original post by mik1a)
8. Find x, if: 9^(2x) = 2*9^x + 3
9^(2x) - 2.9*(x) + 3 = 0
(9^x-3)(9^x+1)=0
9^x = 3; x = 1/2
9^x = -1; no solution. (thanks to Ralfskini for this elegant solution, the longer route is to use the substitution a = 9^x)
Actually there are complex solutions, I think infinitely many, but one is x = (pi*i)/ln[9].

EDIT:

It is a solution because 9^((pi*i)/ln[9]) = e^(i*pi) = -1

EDIT 2:

I was right, there are infinitely many solutions because e^(i(pi +2n*pi)) = -1 where n is an integer. So x = 9^(((1+2n)pi*i)/ln[9]), where n is an integer.

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