Singapore Mathematical Olympiad (SMO) adalah kompetisi matematika yang
diadakan di Singapura. SMO dirancang untuk menantang dan menguji kemampuan
matematika siswa sekolah menengah atas dan siswa muda di Singapura. Tujuan
utama SMO adalah mempromosikan minat dan kecintaan terhadap matematika serta
mengembangkan keterampilan pemecahan masalah dan pemikiran kritis.
Short Questions
11. Find the value of $\sqrt{9999^2+19999}$.
12. If the graphs of $y=x^2+2 a x+6 b$ and $y=x^2+2 b x+6 a$ intersect at only one point in the $x y$-plane, what is the $x$-coordinate of the point of intersection?
13. Find the number of multiples of 11 in the sequence $99,100,101,102, \cdots, 20130$.
14. In the figure below, $B A D, B C E, A C F$ and $D E F$ are straight lines. It is given that $B A=B C, A D=A F, E B=E D$. If $\angle B E D=x^{\circ}$, find the value of $x$.
15. If $a=1.69, b=1.73$ and $c=0.48$, find the value of
$$
\dfrac{1}{a^2-a c-a b+b c}+\dfrac{2}{b^2-a b-b c+a c}+\dfrac{1}{c^2-a c-b c+a b}
$$
16. Suppose that $x_1$ and $x_2$ are the two roots of the equation $(x-2)^2=3(x+5)$. What is the value of the expression $x_1 x_2+x_1^2+x_2^2$ ?
17. Let $A B C D$ be a square and $X$ and $Y$ be points such that the lengths of $X Y, A X$ and $A Y$ are 6,8 and 10 respectively. The area of $A B C D$ can be expressed as $\dfrac{m}{n}$ units where $m$ and $n$ are positive integers without common factors. Find the value of $m+n$.
18. Let $x$ and $y$ be real numbers satisfying the inequality
$$
5 x^2+y^2-4 x y+24 \leq 10 x-1
$$
Find the value of $x^2+y^2$.
19. A painting job can be completed by Team A slone in 2.5 hours or by Team B slone in 75 minutes. On one occssion, after Team A had completed a fraction $\dfrac{m}{n}$ of the job, Team B took over immediately. The whole painting job was completed in 1.5 hours. If $m$ and $n$ are positive integers with no common factors, find the value of $m+n$
20. Let $a, b$ and $c$ be real numbers such that $\dfrac{a b}{a+b}=\dfrac{1}{3}, \dfrac{b c}{b+c}=\dfrac{1}{4}$ and $\dfrac{c a}{c+a}=\dfrac{1}{5}$. Find the value of $\dfrac{24 a b c}{a b+b c+c a}$
21. Let $x_1$ and $x_2$ be two real numbers that satisfy $x_1 x_2=2013$. What is the mininum value of $\left(x_1+x_2\right)^{2 ?}$
22. Find the value of $\sqrt{45-\sqrt{2000}}+\sqrt{45+\sqrt{2000}}$.
23. Find the smallest pooitive integer $k$ such that $(k-10)^{4026} \geq 2013^{2013}$.
24. Let $a$ and $b$ be two real numbers. If the equation $ax+(b-3)=(5 a-1)x+3 b$ has more than one solution, what is the value of $100 a+4 b$ ?
25. Let $S=\{1,2,3, \ldots, 48,49\}$. What is the maximum value of $n$ such that it is posible to select $n$ numbers from $S$ and arrange them in a circle in such a way that the product of any two adjacent numbers in the circle is less than 100 ?
26. Given any 4-digit positive integer a not ending in ' 0 ', we can reverse the digits to obtain another 4-digit integer $y$. For example if $x$ is 1234 then $y$ is 4321 . How many posible 4-digit integers $x$ are there if $y-x=3177 ?$
27. Find the lesst positive integer $n$ such that $2^8+2^{11}+2^n$ is a perfect square.
28. How many 4-digit positive multiples of 4 can be formed from the digits $0,1,2,3,4,5,6$ such that each digit appears without repetition?
29.Let $m$ and $n$ be two positive integers that satify
$$
\dfrac{m}{n}=\dfrac{1}{10 \times 12}+\dfrac{1}{12 \times 14}+\dfrac{1}{14 \times 16}+\cdots+\dfrac{1}{2012 \times 2014} .
$$
Find the smallest possible value of $m+n$.
30. Find the units digit of $2013^1+2013^2+2013^3+\cdots+2013^{2013}$.
31. In $\triangle A B C, D C=2 B D, \angle A B C=45^{\circ}$ and $\angle A D C=60^{\circ}$. Find $\angle A C B$ in degrees.
32. If $a$ and $b$ are positive integers such that $a^2+2 a b-3 b^2-41=0$, find the value of $a^2+b^2$.
33. Evaluate the following sum
$$
\left\lfloor\dfrac{1}{1}\right\rfloor+\left\lfloor\dfrac{1}{2}\right\rfloor+\left\lfloor\frac{2}{2}\right\rfloor+\left\lfloor\dfrac{1}{3}\right\rfloor+\left\lfloor\dfrac{2}{3}\right\rfloor+\left\lfloor\dfrac{3}{3}\right\rfloor+\left\lfloor\dfrac{1}{4}\right\rfloor+\left\lfloor\dfrac{2}{4}\right\rfloor+\left\lfloor\dfrac{3}{4}\right\rfloor+\left\lfloor\dfrac{4}{4}\right\rfloor+\left\lfloor\dfrac{1}{5}\right\rfloor+\cdots,
$$
up to the $2013^{\text {th }}$ term.
34. What is the smallest possible integer value of $n$ such that the following statement is always true?
In any group of $2 n-10$ persons, there are always at least 10 persons who have the same birthdays.
(For this question, you may assume that there are exactly 365 different possible birthdays.)
35. What is the smallest positive integer $n$, where $n \neq 11$, such that the highest common factor of $n-11$ and $3 n+20$ is greater than 1 ?
Short Questions
11. Answer 10000
$$ \sqrt{9999^2+19999}=\sqrt{9999^2+2 \times 9999+1}=\sqrt{(9999+1)^2}=10000 $$ 12. Answer 3 Let $(\alpha, \beta)$ be the point of intersection of the two graphs. Then $$ \beta=\alpha^2+2 a \alpha+6 b=\alpha^2+2 b \alpha+6 a $$ It follows that $2(a-b) \alpha=6(a-b)$. Since the two graphs intersect at only cone point, we see that $a-b \neq 0$ (otherwise the two graphs coincide and would have infinitely many points of intersection). Consequently $2 \alpha=6$, and hence $\alpha=3$. 13. Answer: 1822 The the number of multiples of 11 in the sequence $1,2, \ldots, n$ is equal to $\left\lfloor\frac{n}{11}\right\rfloor$. Thus the snswer to this question is $\left\lfloor\frac{20130}{11}\right\rfloor-\left\lfloor\frac{98}{11}\right\rfloor=1830-8=1822$. 14 Answer: 108 Let $\angle A B C=\alpha$ and $\angle B A C=\beta$. Since $B A=B C$, we have $\angle B C A=\angle B A C=\beta$. As $E B=E D$, it follows that $\angle E D B=\angle E B D=\angle A B C=\alpha$ Then $\angle A F D=\angle A D F=$ $\angle E D B=\alpha$ since $A D=A F$. Note that $\angle D A F=180^{\circ}-\beta$. In $\triangle A B C$, we have $\alpha+2 \beta=180^{\circ}$; and in $\triangle A D F$, we have $2 \alpha+180^{\circ}-\beta=180^{\circ}$. From the two equations, we obtain $\alpha=36^{\circ}$. By considering $\triangle B D E$, we obtain $z=180^{\circ}-2 \alpha=108^{\circ}$. 15. Answer: 20 $$ \begin{aligned} & \dfrac{1}{a^2-a c-a b+b c}+\dfrac{2}{b^2-a b-b c+a c}+\dfrac{1}{c^2-a c-b c+a b} \\ & =\dfrac{1}{(a-b)(a-c)}+\dfrac{2}{(b-a)(b-c)}+\dfrac{1}{(c-a)(c-b)} \\ & =\dfrac{c-b-2(c-a)-(a-b)}{(a-b)(b-c)(c-a)} \\ & =\dfrac{-1}{(a-b)(b-c)} \\ & =\dfrac{-1}{(1,69-1,73)(1,73-0,48)} \\ & =\dfrac{-1}{(-0,04)(1,25)} \\ & =\dfrac{-1}{(-0,05)}=20 \end{aligned} $$ 16. Answer: 60 The equation $(x-2)^2=3(x+5)$ is equivalent to $x^2-7 x-11=0$. Thus $x_1+x_2=7$ and $x_1 x_2=-11$. So $$ x_1 x_2+x_1^2+x_2^2=\left(x_1+x_2\right)^2-x_1 x_2=7^2-(-11)=60 . $$ 17. Answer: 1041 Let the length of the sle be $\varepsilon$. Oberve that sin $\alpha 6^2+8^2=10^2 80 \angle A X Y=90^{\circ}$. This allows us to see that $\triangle A B X$ bs similse to $\triangle X C Y$. Thus $\frac{A X}{X Y}=\frac{A B}{X C}$, i.e $\frac{8}{6}=\frac{s}{s-B X}$. Solving this equstion gives $s=4 B X$ and we can then compute that $$ 8^2=A B^2+B X^2=16 B X^2+B X^2 $$ So $B X=\frac{8}{\sqrt{17}}$ and $s^2=16 \times \frac{64}{17}=\frac{1024}{17}$. Thus $m+n=1041$. 18. Answerz: 125 The lnequelity is equivelent to $$ (x-5)^2+(2 x-y)^2 \leq 0 $$ Thus we must have $(x-5)=0$ and $(2 x-y)=0$, hence $x^2+y^2=5^2+10^2=125$. 19. Answer: 6 Suppose Team $B$ spent $t$ minutes an the job. Then $$ \frac{t}{75}+\frac{90-t}{150}=1 $$ Thus $t=60$ minutes and so Team A completed $\frac{30}{150}=\frac{1}{5}$ of the job. So $m+n=6$. 20. Answer: 4 Taking reciprocals, we find that $\frac{1}{a}+\frac{1}{b}=3 \frac{1}{b}+\frac{1}{c}=4$ and $\frac{1}{a}+\frac{1}{c}=5$. Summing the three equations, we get $$ 12=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2 \times \frac{a b+b c+c a}{a b c} $$ Hence $\frac{24 a b c}{a b+b c+c a}=4$ 21. Answer: 8052 $$ \left(x_1+x_2\right)^2=\left(x_1-x_2\right)^2+4 x_1 x_2 \geq 0+4 \times 2013=8052 $$ If $x_1=x_2=\sqrt{2013}$, then $\left(x_1+x_2\right)^2=8052$. 22. Answer: 10 Let $x_1=\sqrt{45-\sqrt{2000}}$ and $x_2=\sqrt{45+\sqrt{2000}}$. Then $x_1^2+x_2^2=90$ end $$ x_1 x_2=\sqrt{(45-\sqrt{2000})(45+\sqrt{2000})}=\sqrt{45^2-2000}=\sqrt{25}=5 . $$ Thes $$ \left(x_1+x_2\right)^2=x_1^2+x_2^2+2 x_1 x_2=100 $$ As both $x_1$ and $x_2$ are positive, we have $x_1+x_2=10$. 23. Answer: 55 $(k-10)^{402}=\left((k-10)^2\right)^{2013} \geq 2013^{2013}$ ks equvelent to to $(k-10)^2 \geq 2013 . A 8 k-10$ is an Integer and $44^2<2013<45^3$, the minimum value of $k-10$ is 45 , snd thus the minimum value of $k$ is 55 . 24. Answer 19 Rearranging the tems of the equation, we obtain $$ (1-4 a) x=2 b+3 $$ Since the equation has more thsn one solution (i.e., infinitely many solutions), we must have $1-4 a=0$ and $2 b+3=0$. Therefore $a=\frac{1}{4}$ sod $b=-\frac{3}{2}$. Consequently, $100 a+4 b=$ 19. 25. Answer: 18 First note that the product of any two different 2-digt numbers is greaer then 100 . Thus If a 2-digit number is chosen, then the two numbers adjacent to it in the circle must be single digit numbers. Note that at most nine single-digit numbers can be chosen fram $S$, and no matter how these nine numbers $1,2, \ldots, 9$ are arranged in the circle, there is at most one 2-digit number in between them. Hence it follows that $n \leq 18$. Now the following exrengement $$ 1,49,2,33,3,24,4,19,5,16,6,14,7,12,8,11,9,10,1 $$ shows that $n \geq 18$. Consequently we conclude that the msxmum value of $n$ is 18 . 26. Answer: 48 Let $x=\overline{a b c d}$ and $y=\overline{d c b a}$ where $a, d \neq 0$. Then $$ \begin{aligned} y-x & =1000 \times d-d+100 \times c-10 \times c+10 \times b-100 \times b+a-1000 \times a \\ & =999(d-a)+90(c-b)=9(111(d-a)+10(c-b)), \end{aligned} $$ So we have $111(d-a)+10(c-b)=353$. Consider the remainder modulo 10, we obtsin $d-a=3$, which implies thes $c-b=2$. Thus the values of $a$ snd $b$ determines the values of $d$ and $c$ reapectively. a can take on any value from 1 to 6 , sad $b$ can take any value from 0 to 7 , giving $6 \times 8=48$ choice
27. Answer: 12 Let $2^8+2^{11}+2^n=m^2$ and 80 $$ 2^n=m^2-2^8(1+8)=(m-48)(m+48) . $$ If we let $2^k=m+48$, then $2^{n-k}=m-48$ and we have $$ 2^k-2^{n-k}=2^{n-k}\left(2^{2 k-n}-1\right)=96=2^5 \times 3 $$ The mesns thet $n-k=5$ and $2 k-n=2$, giving us $n=12$. Note that a positive intege $k$ is a multiple of 4 if and only if the number formed by the last two digits of $k$ (in the mme order) is a multiple of 4 . There are 12 possible multiples of 4 that can be formed from the digits $0,1,2,3,4,5,6$ without repetition, nomely $$ 20,40,60,12,32,52,04,24,64,16,36,56 \text {. } $$ If 0 appears in the last two digits, there are 5 choices for the first digit and 4 choioes for the second digit. But if 0 does not appear, there are 4 choices for the first digit and also 4 choices for the second digit. Total number is $$ 4 \times 5 \times 4+8 \times 4 \times 4=208 $$
29. Answer: 10571
$$ \begin{aligned} \frac{m}{n}=\frac{1}{4} \sum_{k=5}^{1005} \frac{1}{k(k+1)} & =\frac{1}{4} \sum_{k=5}^{1006} \frac{1}{k}-\frac{1}{k+1} \\ & =\frac{1}{4}\left(\frac{1}{5}-\frac{1}{1007}\right) \\ & =\frac{501}{10070} . \end{aligned} $$ Since $\operatorname{gcd}(501,10070)=1$, we have $m+n=10571$. 30. Answer: 3 Note that the units digit of $2013^1+2013^2+2013^3+\cdots+2013^{2013}$ is equal to the units digit of the following number $$ 3^1+3^2+3^3+\cdots+3^{2013} $$ Since $3^2=9,3^3=27,3^4=81$, the units digits of the sequence of $3^1, 3^2, 3^3, 3^4, \cdots, 3^{2013}$ are $$ \underbrace{3,9,7,1,3,9,7,1, \cdots, 3,9,7,1}_{2012 \text { numbers }}, 3 $$ Furthermore the sum $3+9+7+1$ does not contribute to the units digit, so the answer is 3. Construct a point $M$ an $A D$ so that $C M$ is perpendicular to $A D$. Join $B$ and $M$. Since $\angle A D C=60^{\circ}, \angle M C D=30^{\circ}$. As $\sin 30^{\circ}=\frac{1}{2}, 802 M D=D C$. This means that $B D=M D$ and $\triangle M D B$ is isosceles. It follows that $\angle M B D=30^{\circ}$ and $\angle A B M=15^{\circ}$. We further obeerve that $\triangle M B C$ is sso isceceles and thus $M B=M C$. Now $\angle B A M=\angle B M D-\angle A B M=15^{\circ}$, giving us yet another isceoe les triangle $\triangle B A M$. We now heve $M C=M B=M A, s 0 \triangle A M C$ is also isosceles. This allows is to calculate $\angle A C M=45^{\circ}$ and finally $\angle A C B=30^{\circ}+45^{\circ}=75^{\circ}$. 
32. Answer: 221
We have $a^2+2 a b-3 b^2=(a-b)(a+3 b)=41$. Sinoe 41 is a prime
number, and $a-b<a+3 b$, we have $a-b=1$ and $a+3 b=41$.
Solving the simultaneous equations gives $a=11$ and $b=10$.
Hence $a^2+b^2=21$.
33. Answer: 62 We first note that for $1 \leq r<k,\left\lfloor\frac{r}{k}\right\rfloor=0$ and $\left\lfloor\frac{k}{k}\right\rfloor=1$. The total number of terms up to $\left\lfloor\frac{N}{N}\right\rfloor$ is given by $\frac{1}{2} N(N+1)$, and we hsve the inequality $$ \frac{62(03)}{2}=1953<2013<2016=\frac{63(64)}{2} . $$ So the $2013^{\text {th }}$ term is $\left\lfloor\frac{901}{\hbar}\right\rfloor$, and the sum up to this term is just 62 .
34. Answer: 1648
By the pigeonhole principle in any group of $366 \times 9+1=3296$ persons, there must be at lesst 10 persons who share the same birthday. Hence solving $2 n-10 \geq 3286$ gives $n \geq 1648$. Thus the smallest possible $n$ is 1648 since $2 \times 1647-10=3284<365 \times 9$, and it is possible for each of the 365 different birthdays to be shared by at most 9 persons. 35. Answer 64 Let $d>1$ be the highest common factor of $n-11$ and $3 n+20$. Then $d \mid(n-11)$ and $d \mid(3 n+20)$. Thus $d \mid[3 n+20-3(n-11)$, i.e., $d \mid$ 53. Sino 53 is \& prime and $d>1$, it follows that $d=53$. Therefore $n-11=53 k$, where $k$ is a positive integer, so $n=53 k+11$. Note that for any $k, 3 n+20$ is a multiple of 53 since $3 n+20=3(53 k+11)+20=53(3 k+1)$. Hence $n=64$ (when $k=1)$ is the smallest positive integer such that HCF $(n-11,3 n+20)>$ 1. Sumber : |
-Short Questions){kind=link}
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