SEAMO PAPER E 2019 [PROBLEM And SOLUTION] - BorneoMath

9. How many ways are there to spell “SEAMO” starting from the “S” in the middle, moving only ↑, ↓, ← and → ?
An example is shaded in the figure.

(A) 12
(B) 24
(C) 36
(D) 48
(E) None of the abov

Banyak kata “SEAMO” pada bagian berwarna merah ada 7. Karna simetris dengan bentuk lainnya maka banyak kata “SEAMO” yang terbentuk seluruhnya ada 7× 8 = 56

10. Each corner of a rectangular prism is cut off. In the figure, 2 out of 8 cuts are shown. How many edges does the new figure have? Assume that the planes cutting the prism do not intersect anywhere in or on the prism.

(A) 24
(B) 36
(C) 42
(D) 48
(E) None of the above

11. A palindrome is a positive integer that reads the same both forwards and backwards. For example, 12521 and 270072 are palindromes. A palindrome between 1000 and 10000 is randomly chosen. Find the probability that it is a multiple of 7.

(A) \(\frac{1}{4}\)
(B) \(\frac{1}{5}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{8}\)
(E) None of the above

Misalkan bilangan palindrome adalah \(\overline{abba}\) , banyak bilangan palindrome \(4\) digit adalah \(9 × 10 × 1 × 1 = 90\) bilangan
Bilangan palindrome \(\overline{abba}\) kelipatan 7 maka memenuhi
\(1000𝑎 + 100𝑏 + 10𝑏 + 𝑎 ≡ 0\;𝑚𝑜𝑑\; 7\)
\(1001𝑎 + 110𝑏 ≡ 0\;𝑚𝑜𝑑\; 7\)
\(110𝑏 ≡ 0\;𝑚𝑜𝑑\; 7\)
\(5𝑏 ≡ 0\;𝑚𝑜𝑑\; 7\)
Nilai \(b\) yang memenuhi adalah \(0\) dan \(7\), banyak bilangan palindrome yang habis dibagi \(7\)
adalah \(9 × 2 × 1 × 1 = 18\) bilangan.
Jadi peluangnya adalah
\(\frac{18}{90}=\frac{1}{5}\)

12. Evaluate

\(\sqrt{1+2019^2-\frac{2019^2}{2020^2}}-\frac{1}{2020}\)

(A) 2021
(B) 2020
(C) 2019
(D) 2018
(E) None of the above

13. For how many positive integers \(n\) does
\(1 + 2 + 3 + ⋯+ n\) evenly divide \(12n\)?
(A) 5
(B) 6
(C) 7
(D) 8
(E) None of the above

\(\frac{12𝑛}{1 + 2 + 3 + ⋯ + 𝑛}= 𝑘, 𝑘 ∈ 𝐵^+\)

\(⇒\frac{12𝑛}{\frac{(1 + 𝑛)𝑛}{2}}= 𝑘\)

\(⇒\frac{24}{𝑛 + 1}= 𝑘\)

Karena \(k\) bilangan bulat maka nilai dari \(𝑛 + 1\) merupakan factor dari \(24\) kecuali 1, nilai \(𝑛 +1\) yang memenuhi adalah \(\{2, 3, 4, 6, 8, 12, 24\}\). Jadi banyak nilai \(𝑛\) yang memenuhi adalah \(7\) bilangan.

14. Let ABCD be a square of side 10 cm. E and F are variable points on BC and CD , respectively, with the constraint BE = CF. Find the smallest area of ΔAEF in cm².

(A) 37.5
(B) 50
(C) 62.5
(D) 75
(E) None of the above

Misalkan \(𝐵𝐸 = 𝐶𝐹 = 𝑥\)

\([AEF] = [ABCD] – [ABE] – [ECF] – [ADF]\)
\(= 100 −\frac{1}{2}(10)(𝑥) −\frac{1}{2}(10 − 𝑥)𝑥 −\frac{1}{2}(10 − 𝑥)10\)
\(= 100 − 5𝑥 − 5𝑥 +\frac{1}{2}𝑥^2 − 50 + 5𝑥\)
\(=\frac{1}{2}𝑥^2 − 5𝑥 + 50\)
\(=\frac{1}{2}(𝑥^2 − 10𝑥 + 100)\)
\(=\frac{1}{2}((𝑥 − 5)^2 + 75)\)
Nilai minimum dicapai ketika \(𝑥 − 5 = 0, 𝑥 = 5\)
Luas minimum \([AEF] =\frac{75}{2}= 37,5\)

15. Let \( m =\overline{2x192y20}\) be an 8-digit integer. Given that m is a multiple of 41, how many possible values are there of (x,y)?
(A) 0
(B) 1
(C) 2
(D) 3
(E) None of the above

\( m =\overline{2x192y20}= 20000000 + 𝑥000000 + 100000 + 90000 + 2000 + 𝑦00 + 20 ≡ 0\; 𝑚𝑜𝑑\; 41\)
Karena \( 100\; 𝑚𝑜𝑑\; 41 = 18\) , maka
\( 𝑚 = 20(100)^3 + 𝑥(100)^3 + 10(100)^2 + 9(100)^2 + 20(100) + 𝑦(100) + 20 ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 20(18)^3 + 𝑥(18)^3 + 10(18)^2 + 9(18)^2 + 20(18) + 𝑦(18) + 20 ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 20(10) + 𝑥(10) + 10(37) + 9(37) + 20(18) + 𝑦(18) + 20 ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 36 + 𝑥(10) + 1 + 5 + 32 + 𝑦(18) + 20 ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 94 + 𝑥(10) + 𝑦(18) ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 47 + 5𝑥 + 9𝑦 ≡ 0\; 𝑚𝑜𝑑\; 41\)
\( = 6 + 5𝑥 + 9𝑦 ≡ 0 \; 𝑚𝑜𝑑\; 41\)
\( = 5𝑥 + 9𝑦 ≡ −6\; 𝑚𝑜𝑑\; 41\)
\( = 5𝑥 + 9𝑦 ≡ 35\; 𝑚𝑜𝑑\; 41\)
\( = 40𝑥 + 72𝑦 ≡ 35.8 \; 𝑚𝑜𝑑\; 41\)
\( = −𝑥 − 10𝑦 ≡ 34\; 𝑚𝑜𝑑\; 41\)
\( = 𝑥 + 10𝑦 ≡ 7\; 𝑚𝑜𝑑\; 41\)
Jadi nilai \( x+10y=\{7, 48, 89\}\) , pasangan \((x, y)\) yang memenuhi \( (7, 0), (8, 4)\) dan \( (9, 8)\) ada \(3\) pasangan

16. In a badminton tournament, a player who loses a match is eliminated while the winner moves on to the next round. There is no draw. There are 100 players. 28 of them are selected to go directly to Round 2 without playing in Round 1. The remaining 72 players are paired off to play in Round 1. The tournament continues until one player remains unbeaten. What is the total number of matches played?
(A) 50
(B) 64
(C) 99
(D) 100
(E) None of the above

Round 1 : 72 pemain masing-masing bertanding ada 36 pertandingan
Round 2 : Tersisa 36+28 = 64 pemain, ada 32 pertandingan
Round 3 : Tersisa 32 pemain, ada 16 pertandingan
Round : Tersisa 16 pemain , ada 8 pertandingan
QF : Tersisa 8, ada 4 pertandingan
SF : Tersisa 4 pemain, ada 2 pertandingan
F : Tersisa 2 pemain, ada 1 pertandingan
Jadi banyak pertandingan seluruhnya adalah 36 + 32 + 16 + 8 + 4 + 2 + 1 = 99 matches

17. An odometer is an instrument on the dashboard displaying the cumulative distance travelled by a vehicle in its
lifetime. Leonard drove his car on a trip for an integral number of hours at an average speed of 55 km/h. At the start of the trip, the odometer displayed \(\overline{abc}\) km. At the end of the
trip, it displayed \(\overline{cba}\) km.
Given \(a ≥ 1 , a + b + c ≤ 7\) , find the value of \(a^3 + b^3 + c^3\).
(A) 149
(B) 217
(C) 244
(D) 342
(E) None of the above

\(55n=\overline{cba}-\overline{abc}\)
\(⇒55𝑛 = 100𝑐 + 10𝑏 + 𝑎 − (100𝑎 + 10𝑏 + 𝑐)\)
\(⇒55𝑛 = 99𝑐 − 99𝑎\)
\(⇒5𝑛 = 9𝑐 − 9𝑎 = 9(𝑐 − 𝑎)\)
diperoleh \(𝑐 − 𝑎 = 5\), pasangan \((𝑐, 𝑎)\) yang memenuhi adalah \((6, 1), (7,2), (8,3), (9,4)\), karena
\(𝑎 ≥ 1 , 𝑎 + 𝑏 + 𝑐 ≤ 7\) maka nilai \(a, b\) dan \(c\) yang memenuhi adalah \(1, 0\) dan \(6\). Jadi nilai
dari \(𝑎^3 + 𝑏^3 + 𝑐^3 = 1 + 0 + 216 = 217\)

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