Wednesday, 31 May 2017

SMO 2017 (Open Section) Answers

SMO 2017 Open Round 2 List

Cut off for Rd 2 ~13

SMO 2017 Open Questions

You can find past year SMO papers (2005-2013, Round 1 and 2 with full solutions) here.

SMO 2017 (Open Section) Answers

1. 3
2. 2
3. 4032
4. 11
5. 7
6. 7
7. 2017
8. 10
9. 10
10. 32
11. 43 (assuming solution exists)
12. 7
13. 45
14. 3
15. 1008
16. 2017
17. 4036
18. 35
19. 3
20. 45
21. 6
22. 121
23. 3025
24. 12
25. 28

45 comments:

Anonymous said...

q12 is 14?

Anonymous said...

q14 I got 2018

Anonymous said...

q8 i got 12

Anonymous said...

q22 I got 115

Anonymous said...

q19 i got 1

Unknown said...
This comment has been removed by the author.
Anonymous said...

q22 i got 106

Anonymous said...

q17 I got 2018

Anonymous said...

I think q19 should be 1 too.

Anonymous said...

why is q19 3 and not 8

PSJH said...

q13 should be 60.
Rearranging the given equation, we get a^2 = b(b-c)
If you construct a circle with centre at point A and with radius of c, you can directly apply the above equation to show that BC is tangent to the circle at B (by tangent-secant theorem). From there, x = 60 degrees.

Anonymous said...

why isn't q8 14?

Anonymous said...

@PSJH q13 should be 45.
You can use cosine rule and sine rule to show (you should get sinC = 1/sqrt(2)) and C=135 deg is rejected

mei panchi said...

Q22 I got 121, filled the board with one color, then fit 6 new colors into each row, 6*20+1, the best I can do.

Papa said...

What do you think of this year's paper? Is it easier or more difficult that previous years?

Unknown said...

qn 12 I got 17
and an 17 I got 8072? isn't it correct?

Unknown said...
This comment has been removed by the author.
LL said...

Fixing 1st term does not give unique way

Papa said...

23. I got 3027

Anonymous said...
This comment has been removed by the author.
Anonymous said...

Apparently you have to mod 20 the numerator if you want to divide by 2.

Anonymous said...

@PSJH Q13 should be 45 not 60 just draw an accurate diagram

Anonymous said...

Why do you think will be the cut-off for bronze?

Anonymous said...

question 19 i got 8

units digits of r = 2017^2017 = 7^2017 = 7^2016 * 7 = 1*7 = 7

ar = 7*8/2 = (2)8

R777 said...

You would need the last 2 digits of 2017^2017 to confirm the units digit. Thus, by finding the last 2 digits, which are 77, you get 77*78/2 = 3003 and 3 is the units digit.

SkittlesPony said...

For question 19, you just need 2017^2017 mod 20, which comes out to 17 and is easier to generate. Furthermore, the pattern of a_n repeats after a cycle of 20 numbers.

Anonymous said...

yea what is the cut off for bronze and hm?

Anonymous said...

will 9 marks get me anything haha

Anonymous said...

Bronze for 9 marks maybe. You'll probably need 14-15 marks to get silver and get into second round

Anonymous said...

Is 10 marks good enough to get something?

Skittles Pony said...

By the way, I'm curious: how did you manage to supply all the correct answers within 1 hour after the paper ended anyway?

Anonymous said...

We can bring home our paper and we probably had our answers recorded and the answers are provided by some math whiz so they should all be correct

Anonymous said...

q22 i got 134 by adding7*7+7*7+6*6

Papa said...

http://sms.math.nus.edu.sg/Competitions/SMO2017/SMO%20Open%20R2.pdf

The results are out! Anyone knows the cut off point?

Anonymous said...

I got 10 didn't get in

Anonymous said...

I got 12, didn't get in. My friend got 13 and got in.

Unknown said...

I got 3026...

Anonymous said...

I got 14 and got in.

Anonymous said...

I also got 12 and did not get in

Anonymous said...

9 honorable mention
10-12 bronze
13 silver

Papa said...

How did you know all these results?

Unknown said...

Why q4 is 11?

Nuha Nfbs said...

I have know... Sorry

Anonymous said...

Q11, there is no need to assume solution exists.
The question said that all of a, b, c and d are positive.
You can therefore show all the quantities required during manipulations,
not only exist, but are positive.

Anonymous said...

Q4) Note that you are integrating the floor function of x. Since floor(x) = 0 for 0 < x < 1, integrating from 0 to 1 gives 0. Similarly, integrating from 1 to 2 gives 1. Integrating floor(x) from 0 to integer a gives 0 + 1 + ... + (a - 1). Solving, a = 11

11) When Lim Jeck says "assuming solution exists", he means that it is one of those "flawed questions" where the scenario is impossible. For more examples of such questions, see SMO(J) 2014 Q26 or SMO(J) 2015 Q35.

The issue here is that if a, b, c, d are all positive and a + b + c + d = 8, then a / (b + c + d) + ... + d / (a + b + c) cannot be equal to 3/10 because the expression is naturally greater than 4/3.