SMO 2006 Open Solution

8/6/2019 SMO 2006 Open Solution

1/9

Singapore Mathematical SocietySingapore Mathematical Olympiad (SMO) 2006

(Open Section, Round 1 Solutions)

1. Ans: 495Each integer can be written as XIX2X3X4X5 where each Xt = 0,1,2, ... ,9 with Xl + X2 +X3 + X4 + X5 =8. The number of non-negative integer solutions to the above equationis 495. So there are 495 such integers.

2. Ans: 20p + q = p2 + 11 and pq = 135(p + q) + 576 = 15(p + q) + 16.

9 36 4So p + q > 0 and is a multiple of 4. Also p + q =1 + p2:2 . So p2 + 2 is a multiple of9. So p = 5,13,14, .... If p = 5, p + q = 4 and pq = 31. If p = 13, p + q = 20, pq = 91,q =7. Thus p + q =20.3. Ans: 20071f ( x y + 1) =( x ) f ( y ) - f ( y ) - X + 2, so we have f ( y x + 1) =( y ) f ( x ) - f ( x ) - y + 2.Subtracting, we have 0 = f ( x ) - f ( y ) + y - X or f ( x ) + y = f ( y ) + x . Let y = O . Thenf ( x ) = f ( O ) + x . Substitute into the given identity and putting x = y = 0, we get

f ( O ) + 1 = f ( O ) f ( O ) - f ( O ) + 2, or (1(0) - 1)2 = O.Thus f ( O ) =1 and 1 0 f ( 2 0 0 6 ) + f ( O ) =20071.4. Ans: 682

5. Ans: 13122

Let ak denote the number of ways that the kth pass reach A. We have al=O. Ateach pass, the person holding the ball has 2 ways to pass the ball to. So total numberof ways the ball can be passed after the r pass is 2k The number of ways that at the(k + l)th pass, A receives the ball is ak + 1. So ak+l = 2k - ak Thus a1 = 0, a2 = 2,a3 = 2, ... , all = 682.

Construct the permutation from the end. There are 3 choices each for s9, 88, ... , 82and 2 choices for Sl and 1 choice for So. So the answer is 2.38 =13122.

44


2/9

------------------ ----

6. Ans: 1AD bisects the angle LA and IC bisects the angle LC. Now LBCD = LBAD = LA/2.LADC =LB. Hence LICD =(LA+LC)/2 and LDIC =1800-LB-(LA+LC)/2 =(LA + LC)/2. Thus ID =CD. The chord CD subtends an angle of 30 at pointA of circle C Hence it subtends an angle of 60 at the center of circle C ThusID =CD =2sin(60 /2) =1.7. Ans: 500Ifp is a prime, then the highest power of p that divides 2006! is

J ( p ) =l2006/ p J + l2006/ p 2 J + l2006/ p 3 J + ....(Note that the terms in the sum are eventually 0.) The number of consecutive O's atthe end of the base 10 representation of 2006! is the highest power of 10 that divides2006!, which is min{J(2), J ( 5 ) } = ( 5 ) =401 + 80+ 16+ 3=500.8. Ans: 855The map {a , b , c} -----7 {24- a, 24- b , 24- c} is a bijection from the set of 3-element subsetsof {I, ... ,23} with s (A) < 36 onto the set of 3-element subsets of {1, ... ,23} withs (A) > 36. The number of 3-element subsets of {I, ... ,23} is C ; ) =1771. Therefore,the number of 3-element subsets of {I, ... ,23} with s (A) < 36 is

1 ." 2 [1771 - number of 3-element subsets of {I, ... ,23} with s (A) = 36] = 855.

9. Ans: 322Note that x 6 + x - 6 =x 2 + X - 2 ) ( X 4 - 1+ x - 4 ) =( x + x-I)2 - 2)( ( x 2 + .X - 2 ) 2 - 3) =( ( x + x-1)2 - 2 ) ( ( ( x + X-I)2 - 2)2 - 3) = J ( x + X-I). Thus letting z = x + X-I, wehave J ( z ) =z 2 - 2 ) ( ( Z 2 - 2)2 - 3). Therefore, J ( 3 ) = (32 - 2)((32 - 2)2 - 3) = 322.

10. Ans: 49Let R be the foot of the perpendicular from 0to BC. Since LBAC =LCOR, we haveLAQO = LOCP. Thus DCOP is similar to DQOC. Therefore OC/OQ = OP/OC sothat OP .OQ =OC2 =72=49.

45

I


3/9

11. Ans: 3Let DP =x, P E =y, and BC =a. As 6F DP is similar to 6F BC, we have DF / BF =[o: Thus (BF -1/2)/BF =x/a. Solving for BF, we have l/BF =2(a - x)/a.

Similarly, from the fact that 6GPE is similar to 6GBC, we obtain l/CG =2(a - y ) / a. Consequently,1 1BF + CG =(4a - 2(x + y))/a =(4a - a)/a =3.

12. Ans: 6Extend BP meeting AC at E. Then ABE is an isosceles triangle with AB =AE andBP = P E. As P and M are the midpoints of BE and BC respectively, we have P AIis parallel to EC and PM =EC/2 =(26 - 14)/2 =6.13. Ans: 35Join DT and DC. Let M be the foot of the perpendicular from D onto AC. Then OT,DNI and BC are all parallel. Since D is the midpoint of OB, NI is the midpoint of TC.Thus DTC is an isosceles triangle with DT = DC and LTDM = LMDC.

As LTOA = 70 and 60TD is isosceles, we have LOTD = 35. Thus LBCD =LMDC =LTDNI =LOTD =35.14. Ans: 2006Since D and E are the midpoints of the sides AB and AC respectively, we haveparallel to BC and 6PDE is similar to 6PCB with PD : PC =PE : PB =CB = 1: 2.

Let P E =x and P D =y. Then P B =2x and PC =2y. By Pythagoras' theoremapplied to 6PBD and 6PCE, we get (2X)2 + y2 = 18292 and (2y)2 + x2 = 12982.Adding these two equations, we have x2 + y2 =(18292 + 12982)/5. Thus BC2(2X)2 + (2y)2 =4(18292 + 12982)/5 =4024036. Therefore BC =)4024036 =2006.15. Ans: 65536The answer is

(17) (17) (17) . .. ' (17) = 216 = 65536.1 + 3 + 5 + I 17

46


4/9

16. Ans: 275400(cos515 + sin515) -;--(cos 15 + sin 15)= 400(cos415 - cos' 15 sin 15 + cos2 15 sin ' 15 - cos 15 sin ' 15 + sin415)= 400(cos" 15 + sin" 15 - cos 15 sin 15 + cos2 15 sin2 15)= 400((cos'' 15 + sin2 15)2 - cos'' 15 sin'' 15 - cos 15 sin 15)= 400(1 - (~sin300)2 - (~sin300))2 2= 400(1 - 1/16 - 1/4)=275

17. Ans: 4Consider the equation y + l/y =a, where a > O . It can be changed into

y2 _ ay + 1 =O .Observe that it has two positive real solutions:

a v a 2 - 4y = > O .2Thus the equation

211x + x2 =2006 + 2006'has four real solutions (i.e., V2506, 1/ V2506).

18. Ans: 30Note that

a5 - a =a(a - 1)(a 1)(a2 + 1).It is clear that 2 I a5 - a and 3 I a5 - a. We can show that 5 I a5 - a by considering thefive cases: a = = i (mod 5), i= 0,1,2,3,4. Thus 30 I a5 - a.

When a = 2, we have a5 - a = 30. Thus the maximum n is 30.19. Ans: 32752There are 215 mappings from A to B. There is only one mapping f : A ----)Bf ( i ) =0 for all iE A; and there are 15 mappings f : A ----)B with f ( i ) =0 for alliE A\{k} and f(k) =1, for k =1,2, ... ,15. Thus the answer is 32752.

47


5/9

20. Ans: 102Note that from an=10an-1 - 1, we have

a - ~ =10 ( a -1- ~)n 9 n 9for all n ~ 2. Thus, 1 n-1 ( 1 ) n-18an - 9 =10 a1 - 9 =10 9for all ti ~1. Therefore

Observe that for n ~ 2,

Thus10100alOl < < a102.

That is why the answer is 102.

21. Ans: 1900Let A be a vertex of P. First we shall count the number of such triangles having A asa vertex. After taking away A and 3 consecutive vertices of P on each side of A, we areleft with 23 vertices from which we can choose two vertices in such a way that, togetherwith A, a desired triangle can be formed. There are C+(23;3-2)) =(220) ways to do so.Hence there are 30 (220) - 7 - 3 = 1900 such triangles.22. Ans: 486Let S = {x E Z I 2000 :S x :S 4000}, A = {x E S I x is divisible by 4}, B = {x E S Ix is divisible by 100},C ={x E S I x is divisible by 400}. The required answer isI A I I B I IC I - (4000 2000 ) (4000 2000 ) (4000 2000 )_- + - -- - - + 1 - -- - - + 1 + -- - -- + 1 _ 486.4 4 100 100 400 400

23. Ans: 5Our reference day is today, 31-5-2006, Wednesday. We shall first count the number ofdays D from 15-5-1879 to 31-5-2006. The number of leap years between 1879 and 2005IS

48


6/9

From 1-1-1880 to 31-12-2005 there are 2005 - 1879 = 126 years, of which 31 are leapyears. Thus D = 95 x 365+31 x 366+(365-31-28-14)+(31 +28+31 +30+31) = 46464.Since 46464 leaves a remainder of 5 when divided by 7, Albert Einstein was born onFriday. The answer is 5.

24. Ans: 29776Let an denote the number of such n-digit integers. Among these an integers, let bndenote the number of those which end with 2. By symmetry, the number of those whichend with 3 (or 4) is also equal to bn. Hence

an = 2an~1 +~end with 0 or 12an~2~end with 02 or 12

3bn'---v---'end with 2,3 or 4+ 2bn~1~end with 32 or 42

(1 )

(2 )

Thus an - 2an~1 = 3bn = 6an~2 + 6bn~1=6an~2 + 2(an~1 - 2an~2)

an =4an~1 + 2an~2We have al = 4, a2 = 4 x 5 - 3 = 17. By iterating we get a7 = 29776.

25. Ans: 1274By using the fact that (,=J + (~)=(n~l) , we have

Thus when n = 12,s-------- = 1274.23 x 38 x 41 x 43 x 47

49


7/9

Singapore Mathematical SocietySingapore Mathematical Olympiad (SMO) 2006

(Open Section, Special Round Solutions)

1. Set up a coordinate system so that A is the origin and AC is the x-axis. Letthe coordinate of C be (2c,0) and the coordinate of B be (2b,2bV3). Then thecoordinates of D, E, Nand Mare (c, 0), (b, bV3), (c, cV3) and (4b, 0), respectively.

N

A cThus

Also BC2 = (2b - 2c)2 + (2bV3)2 = 4c2 - 8bc + 16b2.Therefore MN = BC. In the right-angled triangle EMN, EL = ~MN. ThusEL =~BC =ED. That is LEDL =LELD.

2. Suppose that the representation uses the reciprocals of k distinct positive integers,Xl, ... ,Xk, where Xi 2 (mod 3). Since 1 = """'_]_,we getL_., x,

2k = = k2 k-1 (mod 3),from which we get k 2 (mod 3). Hence k =2,5,8, .... Since

1 1 1 1 1-+-+-+-+-


8/9

3. Except for 2 or 3, each prime is of the form 6u 1. If a prime p =6u + 1 is in thechain, its successor, if any, must be of the form 2p - 1=2(6u) + 1 since 2p + 1 isdivisible by 3. Hence the successors are:

2 .2(6u) + 1,2 (6u) + 1, ... ,2~(6u) + 1, ...This sequence cannot go on forever giving primes. To prove this claim, we first notethat, by Fermat's Little Theorem, there exists k such that 2k = = 1 (mod 6u + 1).Thus

2 k (6 u) + 1 = = 0 (mod 6u + 1).Hence 2 k (6 u) + 1 is not prime. A similarly argument can be given for the case6u + 1.

4. Let A ={I, 2, ... ,4n}. Let F be the family of subsets in A with ti + I-elements.Then IF I = ( n ~ 1 ) -Note that every n+ I-element in S, is also a member in F. Since S, nSj contains atmost n elements in A, and any n+ I-element in S, is different from any ti+ I-elementin s, for all 1 < i< j < k. Thus

kIF I ~ ' " ( 2n ) = ( 2n ).~ n+I n+I~=1Hence k < ( 4n ) .z., ( 2n )- n+I . n+I

4n x (4n - 1) x ... x 3n2n x (2n - 1) x . . . x n .

It can be shown that(4n-i)(3n+i) 4nx3n-'-----'----'-----'-- < = 6(2n-i)(n+i) - 2nxn

for all 0 :::;i:::;n - 1)/2.Ifn is odd, then

4n x (4n - 1) x x 3n2n x (2n - 1) x x n

(n-l)/2 (4n - i)(3n + i)II < 6(n+l)/2.(2n - i)(n + i)i=OIfn is even, then() / (n-2)/2 ( ')(3 .)4n x 4n - 1 x x 3n = 4n - n 2 II 4n - z n + z :::;61/26n/2 = 6(n+l)/2.

2nx(2n-I)x ... xn 2n-n/2 i=O (2n-i)(n+i)52


9/9

5. We shall prove that for any prime p with 1 < p < n, if pO I n!, then pO I bn-1a(a +b)(a + 2b) .. (a + (n - l)b).If p I b , then as

ex)

L n n< -=--

Date post:	08-Apr-2018
Category:	Documents
Upload:	wmdsg
View:	219 times
Download:	0 times

SMO 2006 Open Solution

Documents