Solution of SEAMO 2018 Paper D

Solution of SEAMO 2018 Paper D
SEAMO (Southeast Asian Mathematical Olympiads)
SEAMO 2018 PAPER D
SEAMO 2018 Paper D No. 1
Find the last 2 digits in the sum
1 + 2 + 3 + 4 + … + 2017 + 2018
A. 75
B. 71
C. 56
D. 63
E. None of the above
Solution: Show/Hide Remember:
$1+2+3+...+n=\frac{n(n+1)}{2}$
1 + 2 + 3 + 4 + … + 2017 + 2018
= $\frac{2018(2018+1)}{2}$
= $1009\times 2019$
= 2037171
The last 2 digits are 71.
Answer: B

SEAMO 2018 Paper D No. 2
It is now 3:15 PM. After how many minutes will the minute-hand form an angle of 30° with the hour-hand?
A. $5\frac{9}{11}$
B. $5\frac{3}{11}$
C. $6\frac{9}{11}$
D. $6\frac{9}{12}$
E. None of the above
Solution: Show/Hide If the minute hand turns $\frac{360^\circ }{60}=6^\circ $ in 1 min, the hour hand turns $\frac{360^\circ }{12}\div 60=0.5^\circ $.
Difference in speed is $6^\circ -0.5^\circ =5.5^\circ /\min $
At 3:15 PM, minute hand is pointing at 3 and hour hand is $(0.5\times 15^\circ )=7.5^\circ $ ahead.
Distance to catch up = $(7.5+30)^\circ $ = $37.5^\circ $
$(37.5^\circ +5.5^\circ )\min $ = $6\frac{9}{11}$ minutes
Answer: C

SEAMO 2018 Paper D No. 3
When $n^5-n)$ is divide by 5, where $n\in Z^+$, the remainder is
A. 0
B. 1
C. 2
D. 3
E. 4
Solution: Show/Hide $n^5-n$
= $n(n^4-1)$
= $n(n^2-1)(n^2+1)$
= $n(n-1)(n+1)(n^2+1)$
= $n(n-1)(n+1)[(n^2-4)+5]$
= $n(n-1)(n+1)[(n-2)(n+2)+5]$
= $(n-2)(n-1)n(n+1)(n+2)$ + $5n(n-1)(n+1)$
We khow product of $n$ consecutive numbers is always divisible by $n!$
So the number is divisible by 5, the remainder is 0.
Answer: A

SEAMO 2018 Paper D No. 4
Find the value of $g$ in
SEAMO 2018 Paper D No. 4
A. 9
B. 5
C. 7
D. 4
E. None of the above
Solution: Show/Hide Recall $1^{st}$ cyclic number, 142857
Clearly $g=9$
Solution of SEAMO 2018 Paper D No. 4
Answer: A

SEAMO 2018 Paper D No. 5
It is known that $p$ is a prime number. If $p^2+2$ is also a prime number, find $p$.
A. 2
B. 3
C. 5
D. 7
E. None of the above
Solution: Show/Hide $p=2\to p^2+2=2^2+2=8$ is not a prime number.
$p=3\to p^2+2=3^2+2=11$ is a prime number.
$p=5\to p^2+2=5^2+2=27$ is not a prime number.
$p=7\to p^2+2=7^2+2=51$ is not a prime number.
Answer: B

SEAMO 2018 Paper D No. 6
Find the number of digits in $4^{22}\times 5^{40}$.
A. 22
B. 23
C. 40
D. 42
E. None of the above
Solution: Show/Hide $\begin{align}4^{22}\times 5^{40} &= (2^2)^{22}\times 5^{40} \\ &= 2^{44}\times 5^{40} \\ &= 2^4\times 2^{40}\times 5^{40} \\ &= 16\times (2\times 5)^{40} \\ &= 16\times {{10}^{40}} \\ &= 16\times 1\underbrace{00..00}_{40\,digits} \\ &= 16\underbrace{00..00}_{40\,digits} \end{align}$
The number of digits in ${{4}^{22}}\times 5^{40}$ are 42.
Answer: D

SEAMO 2018 Paper D No. 7
There are 7 black, 5 white and 4 orange balls in a bag. Four balls are drawn without replacement. Find the probability that at least 3 balls are white.
A. $\frac{23}{362}$
B. $\frac{1}{8}$
C. $\frac{23}{363}$
D. $\frac{1}{19}$
E. None of the above
Solution: Show/Hide Total outcomes:
$\begin{align}n(S) &= C_4^{16} \\ &= \frac{16!}{4!\times 12!} \\ &= \frac{16\times 15\times 14\times 13\times 12!}{4\times 3\times 2\times 1\times 12!} \\ &= \frac{16\times 15\times 14\times 13}{4\times 3\times 2\times 1} \\ n(S) &= 1820 \end{align}$
E = at least 3 white balls are drawn
E = 3 white balls and 1 ball of black and orange balls are drawn
Number of favorable outcomes:
$\begin{align}n(E) &= C_3^5\times C_1^{11}+C_4^5 \\ &= \frac{5\times 4\times 3}{3\times 2\times 1}\times 11+5 \\ n(E) &= 115 \end{align}$
Probability that at least 3 balls are white:
$P(E)=\frac{n(E)}{n(S)}=\frac{115}{1820}=\frac{23}{364}$
None of the above
Answer: E

SEAMO 2018 Paper D No. 8
Given that the area of $\Delta ABC$ is 1 and $\frac{BD}{CD}=\frac{CE}{EF}=\frac{AF}{FD}=\frac{1}{2}$, what is the area of $\Delta DEF$?
SEAMO 2018 Paper D No. 8
A. $\frac{7}{75}$
B. $\frac{8}{75}$
C. $\frac{8}{27}$
D. $\frac{1}{3}$
E. $\frac{9}{21}$
Solution: Show/Hide SEAMO 2018 Paper D No. 8
$[ABC]=1$
$\Delta DCA$ and $\Delta ABC$ have the same height
$\frac{BD}{CD}=\frac{1}{2}\Rightarrow \frac{DC}{BC}=\frac{2}{3}$
$\begin{align}\frac{[DCA]}{[ABC]} &= \frac{2}{3} \\ \frac{[DCA]}{1} &= \frac{2}{3} \\ [DCA] &= \frac{2}{3} \end{align}$
$\Delta DCF$ and $\Delta DCA$ have the same height
$\frac{AF}{FD}=\frac{1}{2}\Rightarrow \frac{FD}{AD}=\frac{2}{3}$
$\begin{align}\frac{[DCF]}{[DCA]} &= \frac{2}{3} \\ [DCF] &= \frac{2}{3}\times [DCA] \\ [DCF] &= \frac{2}{3}\times \frac{2}{3} \\ [DCF] &= \frac{4}{9} \end{align}$
$\Delta DEF$ and $\Delta DCF$ have the same height
$\frac{CE}{FE}=\frac{1}{2}\Rightarrow \frac{FE}{FC}=\frac{2}{3}$
$\begin{align}\frac{[DEF]}{[DCF]} &= \frac{2}{3} \\ [DEF] &= \frac{2}{3}\times [DCF] \\ [DEF] &= \frac{2}{3}\times \frac{4}{9} \\ [DEF] &= \frac{8}{27} \end{align}$
Answer: C

SEAMO 2018 Paper D No. 9
Find the positive integral solution of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$
A. (2, 4, 12)
B. (2, 6, 6)
C. (3, 3, 6)
D. (3, 4, 4)
E. All of the above
Solution: Show/Hide When $x$ = 2
$\frac{1}{y} < \frac{1}{3}\le \frac{2}{y}\Rightarrow 3 < y\le 6$
$y$ is integer, $y$ = 4, 5 or 6
When $x$ = 3
$\frac{1}{y} < \frac{1}{2}\le \frac{2}{y}\Rightarrow 2 < y\le 4$
$y$ is integer, $y$ = 3 or 4
Possible (x, y, z)
(2, 4, 12),
(2, 6, 6)
(3, 3, 6)
(3, 4, 4)
So, the answer all of the above.
Answer: E

SEAMO 2018 Paper D No. 10
Today is Saturday. What day is it $3^{100}$ days later?
A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. None of the above
Solution: Show/Hide $\begin{align}3^3 &\equiv 27\bmod 7 \\ &\equiv -1\bmod 7 \end{align}$
$\begin{align}3^{100} &\equiv 3^{3\times 33+1}\bmod 7 \\ &\equiv (3^3)^{33}.3^1\bmod 7 \\ \equiv (-1)^{33}.3\bmod 7 \\ &\equiv -3\bmod 7 \\ 3^{100} &\equiv 4\bmod 7 \end{align}$
$3^{100}$ days later is the same as 4 days after Saturday, namely Wednesday
Answer: D

SEAMO 2018 Paper D No. 11
If $a-b=3$ and $a^2+b^2=5$, then the value of $\frac{1}{a}-\frac{1}{b}$ is
A. $\frac{1}{2}$
B. 1
C. $\frac{3}{2}$
D. 2
E. None of the above
Solution: Show/Hide $a^2+b^2=5$
$\begin{align}a-b &= 3 \\ (a-b)^2 &= 3^2 \\ a^2+b^2-2ab &= 9 \\ 5-2ab &= 9 \\ -2ab &= 4 \\ ab &= -2 \end{align}$
$\begin{align}\frac{1}{a}-\frac{1}{b} &= \frac{b-a}{ab} \\ &= \frac{-(a-b)}{ab} \\ &= \frac{-3}{-2} \\ &= \frac{3}{2} \end{align}$
Answer: C

SEAMO 2018 Paper D No. 12
15 countries took part in The World Cup. Each team played exactly 2 matches. The winner scored 3 points and the loser scored 0 points. Each team got 1 point for a draw. How many more points at most did the champion score than the runner up?
A. 48
B. 51
C. 53
D. 56
E. 59
Solution: Show/Hide The champion scored at most
$3\times 14\times 2$ = 84 points
The minimum score of the runner-up
$1\times 14\times 2$ = 28 points
84 – 28 = 56 points
Answer: D

SEAMO 2018 Paper D No. 13
It is known that $m$, $n$ and $p$ are positive integers, where $m < n < p$. If $2^m+2^n+2^p=2336$, find the value of $m$.
A. 4
B. 5
C. 6
D. 7
E. Unable to determine
Solution: Show/Hide $\begin{align}2^m+2^n+2^p &= 2336 \\ &= 2^5\times 73 \\ &= 2^5\times ({{2}^{3}}\times 9+1) \\ &= 2^8\times 9+2^5 \\ &= 2^8\times ({{2}^{3}}+1)+2^5 \\ 2^m+2^n+2^p &= 2^{11}+2^8+2^5 \end{align}$
$m < n < p$
Clearly $m$ = 5
Answer: B

SEAMO 2018 Paper D No. 14
How many odd digits are there in the result of 1111111111 $\times $ 9999999999?
A. 10
B. 11
C. 8
D. 9
E. None of the above
Solution: Show/Hide $1111111111\times 9999999999$
= $1111111111\times (1\underbrace{000...00}_{10}-1)$
= $\underbrace{111...11}_{10}\underbrace{000...00}_{10}-\underbrace{111...11}_{10}$
Clarly there are 10 odd digits.
Answer: A

SEAMO 2018 Paper D No. 15
It is known that $m$ is a positive integer such that $m\times 874$ is 5-digit number ending with 92. Possible value (s) of $m$ is/are
i. 48 ii. 58 iii. 68 iv. 108
A. i only
B. ii only
C. i and iii
D. ii and iv
E. i, ii and iii
Solution: Show/Hide Given $874m=100a+92$
Where $a$ is positive integer
$100\le a\le 999$
$12\le m\le 114$
$\begin{align}a &= \frac{874m-92}{100} \\ &= \frac{437m-46}{50} \\ &= \frac{437m-46}{50} \\ a &= 9m-1-\frac{13m-4}{50} \end{align}$
$13m-4$ is a multiple of 50, unit digit of $m$ is 8.
Let $m=10n+8$; $4\le 10m\le 106$
When $n$ is positive integer
$\begin{align}13m-4 &= 13(10n+8)-4 \\ 13m &= 130n+100 \end{align}$
$n=5\to m=58$
$n=10\to m=108$
Answer: D

SEAMO 2018 Paper D No. 16
In the figure below, ABCD is a rectangle. E and F are mid-points of AD and CE, respectively. If the area of $\Delta BDF$ is 7,5 $\text{cm}^2$, find (in $\text{cm}^2$) the area of rectangle ABCD.
Solution of SEAMO 2018 Paper D No. 16
A. 45
B. 56
C. 58
D. 60
E. None of the above
Solution: Show/Hide Let S be the area of rectangle ABCD
$\begin{align}[CDF] &= \frac{1}{2}\times [CDE] \\ &= \frac{1}{2}\times \frac{1}{4}S \\ [CDF] &= \frac{1}{8}S \end{align}$
$\begin{align}[BCF] &= \frac{1}{2}\times [BCE] \\ &= \frac{1}{2}\times \frac{1}{2}S \\ [BCF] &= \frac{1}{4}S \end{align}$
$\begin{align}[BDF] &= 7.5 \\ [BCD]-[BCF]-[CDF] &= 7.5 \\ \frac{1}{2}S-\frac{1}{4}S-\frac{1}{8}S &= 7.5 \\ \frac{1}{8}S &= 7.5 \\ S &= 7.5\times 8 \\ S &= 60\,\text{cm}^2 \end{align}$
Answer: D

SEAMO 2018 Paper D No. 17
$\frac{2}{3}$ of the journey from Lethbridge to Medicine Hat is highway, while the remaining $\frac{1}{3}$ is country road. Connecting the highway to the country road is the town of St. Paul. A van left Lethbridge for Medicine Hat, travelling at 100 km/h on highway and 60 km/h on country road. A car left Medicine Hat for Lethbridge, travelling at 70 km/h on country road and 110 km/h on highway. The car and the van passed each other 44 km away from St. Paul. Find, in km, the distance from Lethbridge to Medicine Hat.
A. 248
B. 339
C. 441
D. 453
E. None of the above
Solution: Show/Hide Let the distance between the 2 cities be $x$ m.
The car fork $\frac{x}{3\times 70}$ h on the country road.
The distance covered by the van along the highway in $\frac{x}{3\times 70}$ h
= $100\times \frac{x}{3\times 70}$
= $\frac{10}{21}x$ km
$\frac{2}{3}x > \frac{20}{21}x$
The two vehicles passed each other along the highway.
$\begin{align}\left( \frac{2}{3}x-44 \right)\div 100 &= \frac{1}{210}x+\frac{44}{110} \\ \frac{2}{3}x-44 &= \frac{100}{210}x+\frac{4400}{110} \\ \frac{2}{3}x-44 &= \frac{10}{21}x+40 \\ \frac{2}{3}x-\frac{10}{21}x &= 84 \\ \frac{4}{21}x &= 84 \\ x &= 84\times \frac{21}{4} \\ x &= 441 \end{align}$
The distance from Lethbridge to Medicine Hat is 441 km.
Answer: C

SEAMO 2018 Paper D No. 18
Let $n$ be a positive integer such that $n^2+15n+36$ is a perfect square. Find the value of $n$.
A. 13
B. 14
C. 15
D. 16
E. None of the above
Solution: Show/Hide $n^2+15n+36=(n+3)(n+12)$
Both $(n+3)$ and $(n+12)$ must be square numbers
By examination, $n=13$.
Answer: A

SEAMO 2018 Paper D No. 19
The chord AB, common to the circles with centers M and N, has length $2\sqrt{3}$. If forms 1 side of an equilateral triangle in circle M and 1 side of a right hexagon in circle N. Find the ratio of the areas of both circles.
Solution of SEAMO 2018 Paper D No. 19
A. 1 : $\sqrt{3}$
B. 3 : $2\sqrt{3}$
C. 1 : 3
D. $\sqrt{3}$ : 4
E. None of the above
Solution: Show/Hide Solution of SEAMO 2018 Paper D
$\begin{align}AM &= \sqrt{AP^2+MP^2} \\ &= \sqrt{\left( \sqrt{3} \right)^2+1^2} \\ AM &= 2 \end{align}$
$r_1$ = 2
$r_2=AB\to r_2=2\sqrt{3}$
The ratio of the areas of both circles:
= $\pi r_1^2:\pi r_2^2$
= $r_1^2$ : $r_2^2$
= $2^2$ : $(2\sqrt{3})^2$
= $4$ : 12
= 1 : 3
Answer: C

SEAMO 2018 Paper D No. 20
Starting from any letter ‘M’ below, the word MADAM can be spelled by moving up, down, left or right to an adjacent letter. How many ways are there to spell MADAM?
Solution of SEAMO 2018 Paper D
A. 90
B. 120
C. 144
D. 169
E. None of the above
Solution: Show/Hide 12 ways to spell M to D
12 ways to spell D to M
$12\times 12$ = 144 ways
Answer: C

SEAMO 2018 Paper D No. 21
It is known that $Q_{n+1}=\frac{1}{1+\frac{1}{Q_n}}$ for $n$ = 1, 2, 3, 4, …, 2017, 2018. Evaluate $Q_1Q_2+Q_2Q_3+Q_3Q_4+...+Q_{2017}Q_{2018}$ where $Q_1=1$.
Solution: Show/Hide $Q_1=1$
$Q_2=\frac{1}{1+\frac{1}{Q_1}}=\frac{1}{1+\frac{1}{1}}=\frac{1}{2}$
$Q_3=\frac{1}{1+\frac{1}{Q_2}}=\frac{1}{1+\frac{1}{\frac{1}{2}}}=\frac{1}{3}$


$Q_n=\frac{1}{n}$
$Q_1Q_2+Q_2Q_3+Q_3Q_4+...+Q_{2017}Q_{2018}$
= $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{2017\times 2018}$
= $\left( \frac{1}{1}-\frac{1}{2} \right)$ + $\left( \frac{1}{2}-\frac{1}{3} \right)$ + $\left( \frac{1}{3}-\frac{1}{4} \right)$ + … + $\left( \frac{1}{2017}-\frac{1}{2018} \right)$
= $1-\frac{1}{2}$ + $\frac{1}{2}-\frac{1}{3}$ + $\frac{1}{3}-\frac{1}{4}$ + … + $\frac{1}{2017}-\frac{1}{2018}$
= $1-\frac{1}{2018}$
= $\frac{2017}{2018}$
Answer: 2017/2018

SEAMO 2018 Paper D No. 22
It is known that $n$ is a 10-digit number of the form $\overline{2017x2018y}$, where $x$ and $y$ can be any integer 0 to 9. How many such numbers $n$ are there that are divisible by 33?
Solution: Show/Hide Since $33=3\times 11$
Check divisibility by 11:
$(2+1+x+0+8)$ – $(0+7+2+1+y)$ = $11n$
$(11+x)-(10+y)=11n$
$1+x-y=11n$
$x-y=11n-1$
$x$ and $y$ can be any integer 0 to 9
$-9\le x-y\le 9$
$-9\le 11n-1\le 9$
$-8\le 11n\le 10$
$-\frac{8}{11}\le n\le \frac{10}{11}$
$n=0$
$\begin{align}x-y &= 11n-1 \\ x-y &= 11\times 0-1 \\ x-y &= -1 \\ y-x &= 1 \end{align}$
(1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8), (8,9)

Check divisibility by 3:
$2+0+1+7+x+2+0+1+8+y$
= $21+x+y$
(1, 2), (4,5), (7, 8) are possible.
There are 3 such numbers.
Answer: 3

SEAMO 2018 Paper D No. 23
It is known that $a$, $b$ and $c$ are constants and
When $x=1$, $x^3+ax^2+bx+c=1$
When $x=2$, $x^3+ax^2+bx+c=2$
When $x=8$, $x^3+ax^2+bx+c=P$
When $x=-5$, $x^3+ax^2+bx+c=Q$
Find $(P-Q)$.
Solution: Show/Hide When $x=1$
$1+a+b+c=1$
$a+b+c=0\,....\,(1)$
When $x=2$
$2^3+a.2^2+b.2+c=2$
$4a+2b+c=-6$
Solve (1) and (2),
$\frac{\begin{align}4a+2b+c &= -6 \\ a+b+c &= 0 \end{align}}{3a+b=-6}-$
When $x=8$
$8^3+a.8^2+b.8+c=P$
$512+64a+8b+c=P$
When $x=-5$
$(-5)^3+a(-5)^2+b(-5)+c=Q$
$-125+25a-5b+c=Q$
P – Q = ($512+64a+8b+c$) – ($-125+25a-5b+c$)
P – Q = 637 + 39a + 13b
P – Q = 637 + 13(3a + b)
P – Q = 637 + 13(-6)
P – Q = 559
Answer: 559

SEAMO 2018 Paper D No. 24
Given that $(x-3)$ is a positive integer and a factor of $3x^2-2x+9$, find the sum of all possible values of $x$.
Solution: Show/Hide Since $\frac{3x^2-2x+9}{(x-3)}=3x+7+\frac{30}{x-3}$
$(x-3)$ is a factor of 30.
$(x-3)$ = 1, 2, 3, 5, 6, 10, 15, 30
$x$ = 4, 5, 6, 8, 9, 13, 18, 33
The sumb of all possible values of $x$:
= 4 + 5 + 6 + 8 + 9 + 13 + 18 + 33
= 96
Answer: 96

SEAMO 2018 Paper D No. 25
It is known that circles with centers $O_1$ and $O_2$ are tangent at point M. AB and CD are external tangents to both the circles.
Solution of SEAMO 2018 Paper D
Given that $O_2E\bot O_1A$, $\angle AO_1C=120^\circ $ and $O_2B$ = 1 cm, find the area of the shaded region. (Leave your answer in terms of $\pi $.
Solution: Show/Hide Let the 2 radii be R and r.
$\angle AO_1C=120^\circ $
$\begin{align}\angle AO_1O_2 &= \angle EO_1O_2 \\ &= \frac{1}{2}\times \angle AO_1C \\ &= \frac{1}{2}\times 120^\circ \\ \angle AO_1O_2 &= 60^\circ \end{align}$
Consider triangle ABC:
$\begin{align}\cos \angle AO_1O_2 &= \cos 60^\circ \\ \frac{EO_1}{O_1O_2} &= \frac{1}{2} \\ \frac{R-r}{R+r} &= \frac{1}{2} \\ 2R-2r &= R+r \\ R &= 3r \end{align}$
Since $r=O_2B\to r=1\,\text{cm}$
$O_1A=R=3r=3\times 1=3\,cm$
$O_1O_2=R+r=3+1=4\,cm$
$O_1E=R-r=3-1=2\,cm$
$\begin{align}O_2E &= \sqrt{(O_1O_2)^2-(O_1E)^2} \\ &= \sqrt{4^2-2^2} \\ &= \sqrt{12} \\ O_2E &= 2\sqrt{3}\,cm \end{align}$
Area of trapezium $O_2BAO_1$;
$\begin{align}\left[ O_2BAO_1 \right] &= \frac{(O_1A+O_2B)}{2}\times AB \\ &= \frac{(O_1A+O_2B)}{2}\times AB \\ &= \frac{3+1}{2}\times 2\sqrt{3} \\ \left[ O_2BAO_1 \right] &= 4\sqrt{3} \end{align}$
Sector $AO_1M=\frac{60^\circ .\pi .3^2}{360^\circ }=\frac{3}{2}\pi \,\text{cm}^2$
Sector $BO_2M=\frac{120^\circ .\pi .1^2}{360^\circ }=\frac{1}{3}\pi \,\text{cm}^2$
Shaded region = $[O_2BAO_1]$ - sector $AO_1M$ - sector $BO_2M$
= $4\sqrt{3}-\frac{3}{2}\pi -\frac{1}{3}\pi $
= $\left( 4\sqrt{3}-\frac{11}{6}\pi \right)\,\text{cm}^2$
Answer: $\left( 4\sqrt{3}-\frac{11}{6}\pi \right)\,\text{cm}^2$

Post a Comment for "Solution of SEAMO 2018 Paper D"