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
First
ReplyDeleteAutist i was here before u
DeleteQ15 is 6 I think
ReplyDelete14 is 31?
ReplyDelete19 is 15 i think
ReplyDeleteI think so too
ReplyDeleteI got 3 for 13 odk
ReplyDeleteQ25 is 70, not 63, use Verrier's Lemma+similar triangles.
ReplyDeleteQ11 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.
Yup 15 is 6
ReplyDeleteQ15 you can also menelaus’ theorem or stewart’s theorem
ReplyDelete14 is not 33. If it was its its just like asking whats 6*11/2.
ReplyDeleteQ14 is 33.Just do chessboard coloring.
ReplyDeleteThats like saying q14 is just 66/2
DeleteYou see the thing is yes, it is
Deletehi is this year more difficult than past yrs?
ReplyDeleteYes hopefully r2 is like 10 or 11
Deletecan anyone estimate how many is bronze? thanks
ReplyDeletecan someone provide a solution for q15?
ReplyDeleteWhat's the ddifficulty of this year's paper? harder or easier
ReplyDeleteQ19 is not 15, I managed to find a counter example.
ReplyDeleteLet 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.
*something
ReplyDeletehow do you do qn13?
ReplyDeleteIs it harder than last year?
ReplyDeleteLast year was significantly easier i think
ReplyDeleteI got 24 for q19
ReplyDeletepray for 5 as r2 cutoff tbh
ReplyDeletei got 6 can i get to top 30?
ReplyDeletelol got 6 too, careless for 2 questions
ReplyDeleteI got 11 correct... Hopefully can get in rd 2...
ReplyDeleteCan anyone estimate the cutoff point for entering second round?
ReplyDeletemaybe cutoff will be around 12 or 13 again
ReplyDeleteWill 9 points qualify for anything?
ReplyDeleteQ19 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.
ReplyDeleteBelieve 5 is cutoff point for r2.
ReplyDeletec'mon that's absurd. 11 or 12 maybe more realistic
DeleteQ19 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.
ReplyDeleteI 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
Q13 should be 47 with m=39 and n=8
ReplyDeleteI also got 47 for q13
Deletewhats verrier's lemma??
ReplyDeleteq19 is 24 when a1=7/4
ReplyDeletewhy questions 23 is not 2523
ReplyDelete@above because it can be -30 instead of 26
ReplyDelete@above I mean 30 instaed of 29
ReplyDeleteIs the round 2 list out yet?
ReplyDeleteall the round 2 lists are on sms website; https://sms.math.nus.edu.sg/competitions/SMO2018/R2Open-2018.pdf
ReplyDeleteHi for Q13 shouldn't sinB=sqrt(39)/8 so m+n=47?
ReplyDeleteI’m really dumb so I’m just gonna say
ReplyDeleteI 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?
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
Delete