SMO 2011 Junior Special Round Participants List
##Click here for errors in SMO 2011 Solution Book's Junior Rd 1 Q16, 19, 26 and 28. Document is prepared by wmdsg.
1. D
2. C
3. C
4. B
5. B
6. A
7. B
8. E
9. C
10. E
11. 2
12. 5
13. 3
14. 129
15. 0
16. 7981 (3 Sep: answer in solution book is 9241 - refer to ## above)
17. 224
18. 1001
19. 256 (3 Sep: answer in solution book is 16 - refer to ##)
20. 4022
21. 101
22. 19
23. 169
24. 288
25. 14
26. 3 (3 Sep: answer in solution book is 350 - refer to ##)
27. 19
28. 10301(? see ** below and ##)
29. 15
30. 109
31. 4
32. 98
33. 30
34. 4022
35. 10
*Answers for Qn 14 and Qn 19 are provided by sz and wmdsg, respectively. Thanks.
** Lim Li: as pointed out by GeNiUs in comments, there may be something wrong with Qn 28. The answer is unknown at the moment.
National Average Statistics
SMO (Junior) Previous Years' Cut-off Points (Inclusive of Round 2 scores)
2010 Gold (24), Silver (9), Bronze (8), HM (5)
2009 Gold (11), Silver (7), Bronze (6), HM (5)
2008 Gold (23), Silver (14), Bronze (11), HM (10)
2007 Gold (20), Silver (14), Bronze (11), HM (10)
Update:
2011 Gold (12), Silver (9), Bronze (7), HM (6), Cut-off for Round 2: 11, Cut-off for Top 30: about 34
66 comments:
For Q28 could you explain your working, I got 10403
thanks,
an anonymous NUS High Year 1
how is the first answer "d"???
i got "none of the abv"coz when u work it out u get 2048/1024 - 12/1024 or 506/1024??????
~~anonymous Sec 1 student
Hey i am also nushigh yr1!
yup Q1 should be "e"
an anonymous NUS High Year 1
question 1 is lame, i like the pablo question
how do you solve the pablo Qn???
Q1 answer id D i cked with calculator.
typo sry checked not anything else
qn 1 is d nt e.
qn 1 is d, I double checked with my calculator again after the test
Q23 I got 169... Cases for 4 is 6
I'm not sure question 11 answer is 2,because like how 1/3+1/3+1/3=square of 1 (1), 1/9(3 squared)+1/9+1/9 does not equal to 1, so it may not be as simple as just squaring the sum.
For Q23, this is my method.
Let C(n) be the number of possibilities to fit 1X1, 1X2 and 1X4 tiles into a 1Xn block. Then
C(n)=C(n-1)+C(n-2)+C(n-4)
Since C(1)=1, C(2)=2, C(3)=3, C(4)=6, we can continue calculating
C(5)=6+3+1=10
C(6)=10+6+2=18
C(7)=18+10+3=31
C(8)=31+18+6=55
C(9)=55+31+10=96
C(10)=96+55+18=169
I suppose that you got 151 by letting C(0)=0, then C(4)=5, ...C(10)=151, but I think you cannot do that since you have to start from C(1) and not C(0).
Oh well, you are still too pro
Yes Genius you are right. Answer is 169.
Thanks :D
Hi,
For questions 10, 25 and 28, can you provided your solution? I got (B) for qns 10, (7) for qns 24 and (10302) for qns 28. Can you please confirm your answers. Thank you!
Sorry to bother but can someone explain why question 28 is not 10403? Thanks!
Wait... How did you get (C) for question 9??
Solution for Q28.
You can see that the length from smallest to the largest is 1,1,2,2,3,3...100,100,101,101-1.
Sum is [((1+101)/2)x101x2] -1 = 10301
Lim Li
Solution for Q9.
no. of ways to put the knights is 64P2=4032.
When they attack each other, they are in a 3x2 or 2x3 block as shown below:
K00
00K or
00K
K00 or
K0
00
0K or
0K
00
K0
Hence no. of ways both knights attack each other is 7x6x2x2x2=336
Therefore no. of ways both knights do not attack each other is 4032 - 336 = 3696 (C)
Lim Li
I am Hi.
But Lim Li, for question 28 you cannot add from 2. The length on the left will always be even. The first (smallest length) cannot be 1. That is why I think there is something wrong with it
Hi GeNiUs
I think you are a genius. You are right that the length on the left will always be even. And the first length cannot be 1. I think it can be any even number. LJ and LM are in camp and I can't check with them. I also think there is something wrong with this question. Can u solve it?
Lim Li
The inner most two parallel lines' difference should be 1, as stated by the question. That means that the second line from inside is 1, and then the first line can be any even number.
Another possibility is that there really is something wrong with the question and the 100 is labelled wrongly. In that case, the 100 might be the whole length from the ending point to the bottom left-hand corner. Then we add 1+1+2+2+3+...+99+99+100+99=10099
And by the way, I am not a genius, I just like Maths :)
G, question 11, answer is 2. x/a + y/b + z/c = sqrt(2)
Then, square both sides to get (x/a)^2 + (y/b)^2 + (z/c)^2 + 2(xyc + xzb + yza)/abc = 2
also, given in qn, a/x + b/y + c/z = 0 implies (xyc + xzb + yza)/xyz = 0 implies xyc + xzb + yza = 0
Hence, 2(xyc + xzb + yza)/abc = 0 and your answer is 0.
Q28 ans=10400, bcs of 100 on left side, sequence is 2+3+3+4+4+..+101+101+102.
Q11 ans=2
But meesiam,
Like Lim Li pointed out, you cannot be sure that the first (Smallest) length is 2. And by the way, even if it's 2, it should be
2+1+3+2+4+3+5+4+6...+100+99+101+99
since the distance between two adjacent parallel lines must be 1
Oops. it should be 2+3+3+4+..+101+101+101=10399
the left and top are even number, the right and bottom are odd number. but the last top length is same as previous right length.
But since the difference between two parallel lines is 1, it should be 2+1+3+2+4+3+5... +100+99+101+99
why is qn 2 (D)? can somebody tell me the 4 roots of the equation??
The answer for question 2 as posted is C. The three roots are -1, 2011 and -2011 since it can be factorized into
(x+1)^3 times (x+2011)(x-2011)
A spirangle progressively gaining length from the previous segment, tf your ans of 2+1+3+2+4+3+.. had decreasing then increasing progression, which is incorrect.
The ending portion is the same length as the previous segment.
The question is surely wrong.
if you look at the figure, you can see that the last one is 1 which is impossible coz it HAS to be an even number.........
but if you try to do it their way, you will get something like-
101+101+101+100+100+99+99+98+98...2+2+1+1= 10403
it cud also be 2+2+3+3+4+4+....100+100+101+101+101= 10401
what do you mean by cut-off points "inclusive of round 2 scores??"
Can someone plz send me the full solution? My email is zetrio@hotmail.com
For Q24 ans is not 24x4x4=384 because there are some that are impossible combinations.
e.g
1 2 4 3
3 4 1 2
? ? 2 4
? ? 3 1
LL
Got it, thanks!
wmdsg
o....NOW i get it....
Ur genius!!!!!!
but how did u get 288????
Look at some of the solutions at
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=409461
Hello, I am a regular visitor of your blog. I am in Primary 6 this year. I was surfing the internet when I encountered this alphanumeric puzzle. I have tried solving it but I cannot do so.
N U S+
H I G H+
S C H
----------------
R I G H T
-----------------
Each letter represents a distinct value. Find each letter.
Will you please help me solve this? Thanks in advance.
Is + a letter???
@GeNiUs - + is not a letter. It is the operation.
Are the letters purposely aligned to the left or is it supposed to be aligned to the right?
Sorry for asking so much but I need clarifications.
@GeNiUs - Sorry, the letters are aligned to the right.
NUS
+ HIGH
+ SCH
---------
RIGHT
R=1 firstly.
Now we have H=8 or 9, since N+I+S<30
If H=8, then
I can only be 0.
and N+I+S>17, that is, N+S>17
However, this is impossible since N and S represent different numbers. Thus, H=9
Again, I can only be 0.
Until now, we have
NUS
+ 90G9
+ SC9
--------
10G9T
You noticed that the lowest S can be is 4 (since 0 and 1 are taken, and 2, 3 will result in T being 0 or 1), so there must be a carry-over of 2 from the rightmost addition. So U+G+C=7 or 17 (cannot be 27 for obvious reasons)
Now if U+G+C=7, the only ways to make this with three different numbers is 1+2+4, 0+1+6, 0+2+5 or 0+3+4, and all of which are not possible as they contain number(s) that are already taken.
Therefore, U+G+C=17
Then now, there are several ways to form this with the digits we have left.
2+7+8=17
3+6+8=17
4+5+8=17
4+6+7=17
While doing this, we notice that
9+9+S=20+T
S=T+2
so 4+5+8=17 does not work because after we have that, we cannot find two numbers S and T such that S=T+2
So we only have three sets of numbers possible for U, G, C:
{2, 7, 8}
{4, 6, 7}
{3, 6, 8}
Knowing that U+G+C=17 also means that there is one carry-over to the next addition sequence, so
N+S+1=10+G
N+S=9+G ----- (1)
(because there is one carry-over for the 9)
Okay now for the annoying bit.
Consider the set {4+6+7}
Then the numbers left are 2, 3, 5 and 8
so the only possible S=T+2 is when S=5, T=3
And since from (1), N+5=9+G,
N=4+G and is only possible when G=4, N=8
Since U and C are interchangeable, we have two answers
865
+ 9049
+ 579
--------
10493
and
875
+ 9049
+ 569
--------
10493
Consider the set {2, 7, 8}
The numbers left are 3, 4, 5 and 6
There two possibilities for S=T+2, that is, 5 and 3, 6 and 4
If S=5, T=3,
from (1) again,
N=G+4, which is only possible for N=6, G=2
There are two solutions then,
675
+ 9029
+ 589
--------
10293
685
+ 9029
+ 579
--------
10293
When S=6, T=4, then
From (1),
N=G+3
which is only possible for N=5, G=2
which lead us to another two answers
576
+ 9029
+ 689
--------
10294
and
586
+ 9029
+ 679
--------
10294
Okay don't yawn, just a little bit more... ;D
Consider the last set {3, 6, 8}
The numbers left are 2, 4, 5 and 7
there are two possible S=T+2
When S=4, T=2, we have from (1) yet again
N=G+5
which is only possible when N=8, G=3. But 8 is already taken! So this case is not possible.
When S=7, T=5,
N=G+2
which is also not possible for the numbers we have left.
Thus there are 6 unique set of answers for
{C, G, H, I, N, R, S, T, U} :
{7, 4, 9, 0, 8, 1, 5, 3, 6}
{6, 4, 9, 0, 8, 1, 5, 3, 7}
{7, 2, 9, 0, 6, 1, 5, 3, 8}
{8, 2, 9, 0, 6, 1, 5, 3, 7}
{8, 2, 9, 0, 5, 1, 6, 4, 7}
{7, 2, 9, 0, 5, 1, 6, 4, 8}
and the equations are
NUS+HIGH+SCH=RIGHT
865+9049+579=10493
875+9049+569=10493
685+9029+579=10293
675+9029+589=10293
576+9029+689=10294
586+9029+679=10294
Tedious but worth it :D
Thanks for the great problem!
For this kind of alphanumeric puzzles, no matter how easy or hard it is, or how elegant the method is, it requires patience. Lots and lots of them. Good thing I am bored and have nothing to do XD
@meesiam
I am not sure about the definition of a spirangle, but from the picture it can be a rectangular shape and the odd lengths and even length are increasing progressively but not together
@ GeNiUs
ha, if u focus on the central starting part, it looks rectangular, cause it's the start. if u look at the outermost part, it is squarish, which is the definition of spirangle. No worries, the solution be out in sept.
u did well for that alphanumeric puzzle, do u major in maths since u maths-lover?
@GeNiUs: Thanks. I got it.
wen will results cum out?
@meesiam
Well, you can say that Maths is my best subject in school
:D
@Lim Li
Sorry!!! I did not see your post!
Thanks, that is a great solution, much better than my case-by-case evaluation
Solution for Q27.
Note that the amount can be in the form 5k, 5k+1, 5k+2, 5k+3, 5k+4. Smallest for 5k is 5, 5k+3 is 8, 5k+2 is 12. You can easily see that smallest for 5k+1 is 8x2=16, and for 5k+4 is 12x2=24. Hence, largest amount you can't purchase using only coupons is 24-5=19.
Lim Min
Could you please explain your working for Question 11? Thanks.
Could you please explain your working for questions 11, 12 and 13?
Please explain your working for questrion 16, 18, 21 and 19. Thanks.
Please explain your working for: 32 34 24 25 29 30. Thanks.
Question 11:
Let x/a=p, y/b=q, z/c=r
Then
p+q+r= sqrt(2)
(p^2)+(q^2)+(r^2)+2(pq+qr+rp)= 2
Also,
(1/p)+(1/q)+(1/r)= 0
(pq+qr+rp)/(pqr)= 0
pq+qr+rp= 0
so we have
(p^2)+(q^2)+(r^2)+2(0)= 2
(p^2)+(q^2)+(r^2)= 2
as desired
i found this alphanumeric puzzle
ABC+BCA=ACBA
each letter represents a number. i need help
thanks
How to solve question 6?
Why Q 16 and Q 26 no one get it correct?
@Google I'm a little late to the party, but anyway, no one got Q16 and Q26 correct because the answer keys are wrong in the first place
http://sms.math.nus.edu.sg/Publication/Errata_2011.pdf
Also, from the official errata, the answer for Q28 is quoted to be 10403, not 10301.
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