SMO 2015 Results
SMO 2015 Open Questions
SMO 2015 Open Round 2 cut off (~12-13)
SMO 2015 (Open Section) Answers
1. 120
2. 3
3. 1
4. 30
5. 4028
6. 216
7. 1512
8. 3
9. 6
10. 1024
11. 21
12. 925327
13. 10050
14. 38918
15. 3
16. 0
17. 160
18. 88330
19. 5
20. 15
21. 4383
22. 5
23. 102
24. 2
25. 33
Initial answers are provided by Clarence Chew.
SMO (Open) Previous Years' Cut-off Points (Inclusive of Round 2 scores)
2013 Gold (26), Silver (13), Bronze (10), HM (9), Round 2 (14), Top 30 (28)
2012 Gold (21), Silver (8), Bronze (6), HM (5), Round 2 (8)
2011 Gold (18), Silver (9), Bronze (6), HM (5), Round 2 (9)
2010 Gold (38), Silver (9), Bronze (6), HM (5)
2009 Gold (27-29), Silver (9), Bronze (6), HM (5)
2008 Gold (27), Silver (11), Bronze (10), HM (6)
2007 Gold (19-20), Silver (9), Bronze (6), HM (5)
33 comments:
11. 21
14. 38918
18. 88330
21. 4383
25. 33
Why isnt Q22 6?
Question 2 the roots are 1, -0.5+0.5rt5, -0.5-0.5rt5 so sum of squares should give 4.
Q21 should be 7715 by PIE?
All answers should be correct except the 5 answers above.
Question 2 answer is 3 as the terms are distinct.
Answer for q11 is 21
Let u=√(x-4) and v=√(5-x) so u²+v²=1
y=√(38)u+√(403)v
u=(y-√(403)v)/√(38)
Substitute the above equation in u²+v²=1
(y-√(403)v)²+38v²=38
441v²-2y√(403)v +y²-38=0
The above equation has real roots for v so D≥0
4×403y²-4×441y²+4×441×38≥0
441×38-38y²≥0
441≥y²
-21≤y≤21
Maximum value of y is 21
U can use Cauchy Schwarz too
For Q11,
38x-152=38(x-4); 2015-403x=403(5-x).
Hence, by Cauchy-Schwarz inequality,
Let a1=38, a2=402, b1=x-4, b2=5-x,
we have: (38+403)(x-4+5-x) greater than or equal to y^2
hence, 441 >= y^2, so max y=21
Can answer be a 6 digits number for Q12?
For q10 the answer isnt exactly 1024, its 1023.95... but they didnt ask for the ceiling function, does this mean the qn will be voided?
(along with q12 the op 6-digit answer)
During the test SMS has said to fill in the first 5 digits...
For question 10, you can either choose to ignore it, or amend the question to i = 0 such that the answer is exactly 1024.
may i know what is the workings for Qn 18?
What's the cut off point for this year? 11 correct?
Btw, why is the answer for Open q21 4383? I got something like 7000+ due to principle of inclusion and exclusion.
Does anyone know how to calculate the coefficient of x^50 for Q12?
18) f(2n+1)/f(2n) = (1+3f(n))/3f(n). Since f(2n)<6f(n), assume f(2n+1) = 1+3(fn); f(2n) = 3f(n).
It then follows that f(1) = 1, f(3) = 4, f(7) = 13, f(15) = 40, f(31) = 121, f(62) = 363, f(125) = 1090, f(251) = 3271, f(503) = 9814, f(1007) = 29443, f(2015) = 88330, if I am not mistaken.
21) Question asks for number of integers which are divisible by exactly one of the integers in {2, 3, 5, 7}. Using the principle of whatever, it should look like this:
⌊10000/2⌋+⌊10000/3⌋+⌊10000/5⌋+⌊10000/7⌋-2(⌊10000/6⌋+⌊10000/10⌋+⌊10000/14⌋+⌊10000/15⌋+⌊10000/21⌋+⌊10000/35⌋)+3(⌊10000/30⌋+⌊10000/42⌋+⌊10000/70⌋+⌊10000/105⌋)-4(⌊10000/210⌋) = 4383.
I think the answer for question 1 should be 598123000 right? Is there a mistake with the question?
n^5 - 5n^3 + 4n = (n-2)(n-1)(n)(n+1)(n+2), which is always divisible by 120 for all integers n.
Maybe you misintepreted the question?
what is the predicted cutoffs for gold/silver/bronze?
This year was quite easy, so the round 2 cut off point may be about 16. For those who got 15 or below, don't worry, there is always next time!
what is the predicted cutoff for bronze
A more rigorous proof for qn 18
From (2)&(3), we know that f(2n+1)/f(2n) = 1+ 1/3f(n) or which means f(2n+1)>f(2n)
And we have 3f(n)[f(2n+1)-f(2n)]=f(2n)<6f(n) --> f(2n+1)-f(2n) <2
Since f is over positive integers, f(2n+1) -f(2n) = 1 --> f(2n+1) = f(2n)+1
Substitute this back to (2), we'll get f(2n)=3f(n)
Does anyone know how to solve qn 15,19,20 and 22? Thanks in advance!
does anybody know the cutoff for HM or bronze? a rough estimtate is much appreciated thanks!
15) I'm a bit too lazy to post a rigorous proof, but note that cos x is minimised when x is closest to pi/2, and sin x is maximised when x is closest to pi/2.
By deductive reasoning, M is obtained when x = y = 5pi/24, z = pi/12. m is obtained when x = pi/3, y = z = pi/12.
20) 1 + 2 + ... + 512 = 1023 < 2015 ==> largest term is 512.
1 + 2 + 4 + 8 + 16 + 2(32 + 64 + 128 + 256 + 512) = 2015, which takes 15 steps.
22) If x1 = x2 = ... = x5 = 0.4; x6 = 1, the inequality does not hold. Hence, n cannot be 6.
I'll leave it to someone else to post a more rigorous proof as my method is very messy.
does anybody know the special round cutoff this year
Are answers there correct?
How come both my friends and I got 11 correct then he got in but I did not?
Can anyone tell me do I need to shade bubbles like 00120,00003? pls...!!
Isn't the 11th answer (403)^1/2 .
No. Maximum value of y is 21 at x = 1802/441
I got 11 last year and got into round 2 so the cop is wrong lol
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