April fools day just came and went. Nothing much happened to me and nothing much happened to other people who were tricked by me, mainly because what I did was quite unsuccessful. Never mind, I shall try again next year.

Today, Wei Liang, North, Akash, Tianyuan, Lin Han, Haochi, Zhongli and I went to VJC for their annual Maths competition. Last year Wei Liang, North and I went there too and we got Certificates of Participation. Haha. This year it was almost the same, just that we received an additional 30% discount voucher from Texas Instruments.

This year's competition was similar to last year's. First we took the preliminary round. 30 MCQ questions in 60 minutes. 5, 0 and 2 marks are awarded for each correct, wrong and no answer, respectively. So we actually start off with 60 marks and work our way up (or down!). There was really not enough time so I could answer like 12 to 15 questions only, not saying that most of them are correct anyway. North managed to answer more than half the questions, not bad.

Then after that we went on a tour around the school. I had toured before VJC so it was kinda sian. They made this tour for us so as to stall time for them to mark our papers. After that we went to the Performance Theatre for a talk about Texas Instruments (TI) Graphical Calculators, again to stall time for them to mark our papers. The talk was the same as last year's.

Anyway, they asked 2 volunteers on stage to demonstrate some device for tracking motion. There was one DHS plump nerd - as in totally nerd - who walked onto the stage clumsily and as he walked he tucked in the front of his shirt in front of everybody, and unfortunately leaving the back untucked. Then his movements are a bit 'spastic'. After cooperating with another guy they both hi-5 each other very hard (and note that it's the DHS guy who suggested it and clapped very hard), and both of them received a TI scientific calculator each. When I asked North why he didn't wanna volunteer to go, he said, "Never mind, we go for Graphical Calculators, not Scientific ones."

Well, we went back to one of the LTs for the final round. It was exactly the same as last year, they put the names of the schools in cardboard folded into a triangular prism. Anyway, DHS, RI, RGS and HCI got in the final round. And the nerd was actually one of those who got in, oh man. Shame on NUS High School for not being able to get in. I think more than 9 of those who got in (3 per school, so 12 altogether) are scholars.

The final round consisted of 4 sections: Individual, Team, Audience and Buzzer sections. For individual, each team member would answer 3 questions and other 2 team members are not allow to say anything or solve the question. So 9 questions in total. You can actually see the 3 China Scholars from RI having a very cool attitude. They spend only say, around 30 seconds per question when each is given 100 seconds and after writing their answers, all 3 of them will cross their hands and sit back and relax. Haha.

I could solve one of the questions correctly (so far that I can remember). It went something like this:

Given that [x] means the largest integer equal to or smaller than x, find the number of solutions of x for
[x/2] + [x/3] + [x/6] = x
for integer values of x ≤ 2007.

Answer is 334.
Notice that for x/2 + x/3 + x/6 = x, this statement is true for all real values of x, because:
x * (1/2 + 1/3 + 1/6) = x
x * 1 = x
x = x.

However, in order for [x/2] + [x/3] + [x/6] = x, x must be a multiple of 2, 3 and 6, otherwise the left hand side would be smaller than the right hand side.
LCM of 2, 3 and 6 is 6, so therefore number of integers satisfying that equation is 2007 / 6 = 334 values of x (all multiples of 6 from 6 to 2004).


Oh by the way one of the emcees said, "Other team members are not allowed to discuss." Then the DHS nerd shouted in the microphone with his high-pitched voice, "I never say anything!"

Yeah so RI won that round. After that was team round, whereby each team is given 7 questions to solve within 7 minutes and teams are allowed to discuss. Well just look at RI again, all 3 of them didn't even discuss the answers at all, they just did the questions.

After 420 seconds, the scripts are being marked so the audience round came. There are 10 questions in which questions 1, 4, and 7 are open to the audience sitting at the right block, 2, 5 and 8 in the centre block and 3, 6, 9 in the left block, and if no one answers within a certain time limit, the question is opened to all. The first person with the correct answer would win a TI Scientific Calculator each. And for question 10, it was opened to all and the winner could win a TI Graphical Calculator (which is more expensive).

There are quite a number of questions we can do (as in North, Wei Liang and I), but there are 2 questions I want to rant about.

Given that i² = -1, find the value of
i + 2i² + 3i³ + 4i⁴ + ... + 32i³², in terms of a + bi.

This question is actually quite easy, considering that i is my favourite number. So I went to do and found out that
i¹ = i
i² = -1
i³ = -i
i⁴ = 1
i⁵ = i
...

So hence for 'a', we just have to find those of even i powers. So we have:
2i² + 4i⁴ + ... + 32i³²
= -2 + 4 - 6 + 8 - 10 + ... + 32
= (-2 + 4) + (-6 + 8) + (-10 + 12) + ... + (-30 + 32)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2
= 16

For b, it's the same, and doing the same way, we will get the whole answer as 16 - 16i.

Due to unforgivable carelessness, I got an answer of 16 + 15i, which from another perspective, looks wrong because 15 seems a little bit odd. But nevertheless, I went to try and heard the female emcee said to the male emcee, "15i... ..." Argh.

Another question is:

Find the value of
(1 + tan13°)(1 + tan14°)(1 + tan15°)... ... (1 + tan32°)

Wei Liang and I looked at the question for a while, then suddenly both of us said, "hey!" Because earlier today I discussed with Wei Liang one Maths question from past year Maths Olympiad Paper: Find the value of (1 + tan40°)(1 + tan5°). The answer is 2, because using the identity tan(A+B) = (tanA + tanB)/(1 - tanAtanB), the answer come up to be 2. Therefore if A and B add up to 45 degrees, the answer would be 2.

So we started pairing up the factors together such that
(1 + tan13°)(1 + tan14°)(1 + tan15°)... ... (1 + tan32°)
= (1+tan13°)(1+tan32°) * (1+tan14°)(1+tan31°) * (1+tan15°)(1+tan30°) * ... * (1+tan22°)(1+tan23°)
= 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
= 2¹⁰
= 1024.

In fact, this morning, I borrowed Robin's A Maths textbook to study the trigonometry identities because my textbook don't have, and I thought one question would come out on that. In the end, none came out, maybe only this, which we already know so we just worked out.

Unfortunately, that question was for the audience in the left block, but no one managed to solve it so it was opened to everyone. Both Wei Liang and I raised up our hands and the emcees didn't see us, and chose somebody else from some other place. And thing is that foreign girl who answered got correct. Argh.

After that they went through the answers for the team round before moving on to the buzzer round. Anyway, this is how the conversation went:

[Emcee]: DHS, please press the buzzer and speak into the mic.
*beep*
[the same DHS Nerd as mentioned earlier]: Hello? Hello?

[Emcee]: HCI, please press the buzzer and speak into the mic.
*beep*
[HCI]: Mic test.

[Emcee]: RGS, please press the buzzer and speak into the mic.
*beep*
[RGS]: *thump thump* (because one of the girls was knocking the microphone)

[Emcee]: RI, please press the buzzer and speak into the mic.
*beep*
[RI]: Test.

Haha. Okay I want to end off this post already. So let's just put it that RI won with a very high score, followed up HCI and DHS won RGS by 2 points. Oh man. So wasted. I can't stand the DHS guy. RGS could have caught up and tied with DHS.

*Note: I am not prejudiced against DHS or something, and they aren't nerds. There was only 1 nerd that I saw, and it has been a long time since I have seen a nerd.

Before I end the post, let's dump in one more question that I saw from the Director's Challenge a few weeks ago.

Find a number such that when you move the first digit to the last place, the new number is 1.5 times the original number. (eg: if original number of 1234, new number is 2341; if original number is 123456, new number is 234561.)

The answer is zero. Really. Okay just kidding, let's be more mathematical here, shall we?

Let's have a 4 digit number abcd. After moving we will have bcda. So we have:
1.5 * (1000a + 100b + 10c + 1d) = 1000b + 100c + 10d + a
1500a + 150b + 15c + 1.5d = 1000b + 100c + 10d + a
1499a = 850b + 85c + 8.5d
14990a = 85(100b + 10c + d)
2998a = 17(100b + 10c + d)

Now we have a problem. 17 is a prime number, and 2998 is not divisible by 17. We notice that a cannot be zero, otherwise b = c = d = 0. In order for a, b, c and d to be integers, a must be a multiple of 17, which there are no solutions for a < 10. Hence there are no 4-digit numbers satisfying this question.

Let's try a 5 digit number abcde, and after moving we will have bcdea. So we have:
1.5 * (10000a + 1000b + 100c + 10d + e) = 10000b + 1000c + 100d + 10e + a
149990a = 85(1000b + 100c + 10d + e)
29998a = 17(1000b + 100c + 10d + e)

Note that 29998 is not a multiple of 17, so 5-digit numbers are out.

Using this method, we can find out that all 1-, 2- and 3-digit numbers are also out. And so we need to find a number which is divisible by 17.
For 6 digit numbers, we have 1499990 = 85 ... ...
For 7 digit numbers, we have 14999990 = 85 ... ...
We can ignore the 5 from 5 * 17 = 85 because anything ending with zero will be a multiple of 5. So keep trying by doing long division of 14999... by 17 while adding 9s at the back, and we notice that...
14,999,999,999,999,990 is divisible by 17 exactly!
So hence working backward, we have
1.5(1,000,000,000,000,000a + 100,000,000,000,000b + ... + 100n + 10o + 1p) = 1,000...b + ...

Meaning that abcdefghijklmnop * 1.5 = bcdefghijklmnopa, a sixteen digit number.
14,999,999,999,999,990a = 85(1,000,000,000,000,000b + ... + p)
176,470,588,235,294a = 1,000,000,000,000b + ... + p

If a = 1,
b=1, c=7, d=6, e=4, etc.

Hence, the smallest number satisfying the question is: 1,176,470,588,235,294.
Of course, the next number would be when a = 2.

By the way, during the final round, the slide shows a compass, with the directions N, E, S and W. So we were joking that N is North (duh!), S is me, Sherman, and W and Wei Liang. Poor Akash, his name is not in.