Thursday 31 May 2018

SMO 2018 (Open Section) Answers

SMO 2018 Open Rd 2 Namelist

Open Round 2 Cut-off: 13

SMO 2018 Open Rd 2 Questions

SMO 2018 Open Questions

SMO 2018 (Open Section) Answers

1. 3
2. 1540
3. 2
4. 1
5. 360
6. 112
7. 370
8. 8071
9. 1927
10. 2
11. 29
12. 2
13. 10
14. 33
15. 6
16. 270
17. 3
18. 26961
19. ? (*amended. Refer to **)
20. 2700
21. 2018
22. 81
23. 2520
24. 181
25. 70 (*amended)
**Checked with Lim Jeck Q19, he thinks the intended answer is 24 but said it is not attainable

48 comments:

Leeisateam said...

First

Anonymous said...

Autist i was here before u

Anonymous said...

Q15 is 6 I think

Anonymous said...

14 is 31?

Anonymous said...

19 is 15 i think

Lee said...

I think so too

Anonymous said...

I got 3 for 13 odk

Anonymous said...

Q25 is 70, not 63, use Verrier's Lemma+similar triangles.
Q11 is 29, the graph is basically x+y=-1 and (1,1), which is the closest point.
Q15 is 6 I think, if I'm not wrong you trigo bash.

Anonymous said...

Yup 15 is 6

Anonymous said...

Q15 you can also menelaus’ theorem or stewart’s theorem

Anonymous said...

14 is not 33. If it was its its just like asking whats 6*11/2.

Anonymous said...

Q14 is 33.Just do chessboard coloring.

Anonymous said...

Thats like saying q14 is just 66/2

Anonymous said...

hi is this year more difficult than past yrs?

Anonymous said...

Yes hopefully r2 is like 10 or 11

Anonymous said...

can anyone estimate how many is bronze? thanks

Anonymous said...

can someone provide a solution for q15?

Aaaaa said...

What's the ddifficulty of this year's paper? harder or easier

Anonymous said...

You see the thing is yes, it is

Anonymous said...

Q19 is not 15, I managed to find a counter example.
Let m=2 and solve for a1, you get a1^3-2a1^2+a1-1=0. Plug it into a cubic equation solver, you get 1.754877666246692 as one of the roots. Solve for 16a1-a3, you get somethung like 23.998 which is clearly more than 15, so 15 can't be the answer.

Anonymous said...

*something

Anonymous said...

how do you do qn13?

Anonymous said...

Is it harder than last year?

Anonymous said...

Last year was significantly easier i think

Anonymous said...

I got 24 for q19

Anonymous said...

pray for 5 as r2 cutoff tbh

Anonymous said...

i got 6 can i get to top 30?

Anonymous said...

lol got 6 too, careless for 2 questions

Anonymous said...

I got 11 correct... Hopefully can get in rd 2...

Anonymous said...

Can anyone estimate the cutoff point for entering second round?

Anonymous said...

maybe cutoff will be around 12 or 13 again

Anonymous said...

Will 9 points qualify for anything?

Anonymous said...

Q19 should be 24. First notice that the terms from a2 onwards are positive. If 01 and the rest of the terms are positive, so no solution. If a1 is negative, you have 16a1-a_{m+1} is negative, so we can ignore this case. Plug small values of m into WolframAlpha, we have 16a1-a_{m+1}=15, 23.9984, - 6.5..., - -1765...,.... This probably means 24 is the answer as 16a1-a_{m+1} decreases as m increases for m>2.

Anonymous said...

Believe 5 is cutoff point for r2.

Aaaaa said...

c'mon that's absurd. 11 or 12 maybe more realistic

Anonymous said...

Q19 I also got 24. Manipulation of the recurrence relation can get 1/(a_{n})=1/(a_{n}-1)+1/(a_{n+1}-1). Plug into 1/a1+...+1/am =1, we can get 1/(a_{1}-1)-1/(a_{m+1}-1)=1 by method of difference. Then we can express a_{m+1} in terms of a1, which is 1/(2-a_{1}). Substitute it into 16a1-am+1 and differentiate. maximum value is reached when 2-a1=1/4. Sub in get 24.
I think this process is correct but a1=1.75 is clearly not correct. This is absurd... but 15 is clearly wrong given the counter-example

Anonymous said...

Q13 should be 47 with m=39 and n=8

Aaaaa said...

I also got 47 for q13

Anonymous said...

whats verrier's lemma??

Anonymous said...

q19 is 24 when a1=7/4

Anonymous said...

why questions 23 is not 2523

Anonymous said...

@above because it can be -30 instead of 26

Anonymous said...

@above I mean 30 instaed of 29

Anonymous said...

Is the round 2 list out yet?

Anonymous said...

all the round 2 lists are on sms website; https://sms.math.nus.edu.sg/competitions/SMO2018/R2Open-2018.pdf

Anonymous said...

Hi for Q13 shouldn't sinB=sqrt(39)/8 so m+n=47?

Anonymous said...

I’m really dumb so I’m just gonna say
I got 4 for question 1, how you got 3?

So first it’s x=0,y=10.999999999
Then there’s x=3,y=7

So x difference is 3 y difference is 3.999999...

So since it’s right angled triangle using Pythagoras the area should be slightly under 5 which means floor will become 4
How you get 3?

Anonymous said...

U just sketch a quick x-y plan and mark out the areas where floor x = 0,1,2 (because 0 <= x <=3) then mark out the corresponding areas where floor y = 10, 9, 8. You'll get 3 1*1 squares so R = 1+1+1 = 3