Asian Science and Maths Olympiad (ASMO) is a competition platform designed to challenge and evaluate studentβs knowledge in Mathematics and Science at their grade level. The questions in the Olympiad will stretch their knowledge and understanding of the concepts. Our syllabus fits nicely into the syllabus that concentrates on non-routine problem-solution to prepare the students for the competition. With the expansion of STEM education worldwide, ASMO certainly answers the need of it. Students will be well prepared with the skills to meet the science and technology challenges.

In Malaysia, ASMO is officially endorsed by Ministry of Education and all participants will obtain curriculum marks. In 2018 alone, Asian Science and Mathematics Olympiad has received 70,000 entries from across the ASEAN countries. We are targeting for the number to increase at 80,000 for 2019.

We are also proud to present that ASMO International is a new effort by ASMO Malaysia which started in 2017 in Pattaya, Thailand. When it was initially launched, the competition was setup via collaboration with ASMOPSS and ASMO Thai was the host for the competition. In 2018, Malaysia has become the host for the competition and it was participated by 10 Asian countries.

The idea of opening up a new competition platform which is ASMO International is to expand the level of competition and to provide more opportunities for primary and secondary school students to experience international engagement. (sc : http://asmo2u.com/about-us)

Berikut ini problems and solution ASMO 2019 grade 8

1. Determine the smallest natural number x which satisfies the inequality $$x^{2006}>2006^{1003}$$.

$$π₯^{2006} > 2006^{1003}$$
$$β π₯^{2(1003)} > 2006^{1003}$$
$$β π₯^2 > 2006$$
Jadi nilai $$x$$ terkecil adalah $$π₯ = 45$$

2. An amount of money is to be divided equally among a group of students. If there was 15 dollars more than this amount, then there would be enough for each student to receive 65 dollars. However, if each student was to receive 60 dollars, then 100 dollars would be left over. Determine the number of students in the group.

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3. Based on the diagram below, determine the height, h in units.

Gunakan perbandingan segitiga sebangun yaitu segitiga $$π΄π΅πΆ β πΈπ΅πΉ$$ dan $$π΄π΅π· β π΄πΈπΉ$$
Pada segitiga $$π΄π΅πΆ β πΈπ΅πΉ$$

$$\frac{π + π}{π}=\frac{5}{β}β π + π =\frac{5π}{β}$$

Pada segitiga $$π΄π΅π· β π΄πΈπΉ$$

$$\frac{π + π}{π}=\frac{3}{β}β π + π =\frac{3π}{β}$$

Samakam kedua persamaan

$$π + π = π + π$$
$$β\frac{5π}{β}=\frac{3π}{β}$$
$$β5π = 3π$$

Diperoleh perbandingan $$π: π = 5 βΆ 3$$, misalkan $$π = 5π₯$$ dan $$π = 3π₯$$, subtitusi ke persamaan $$π + π =\frac{3π}{β}$$, diperoleh
$$β =\frac{3π}{π + π}=\frac{15π₯}{5π₯ + 3π₯}=\frac{15}{8}$$

Jadi panjang $$h$$ adalah $$\frac{15}{8}$$

4. Jess picks two consecutive integers, one of which ends in a 5. She multiplies the integers together and then squares the result. Determine the last two digits of her answer.

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5. Determine the last digit of $$2^{2^{2007}} + 1$$.

$$(2^{2^2007}+ 1)\; mod\; 10$$
Karena $$2^{2007}\; mod\;4 β‘ 0$$, maka bentuk di atas dapat disederhanakan menjadi $$(2^4 + 1)\; mod\; 10 β‘ (16 + 1)\; mod\; 10 = 17\; mod\; 10 = 7$$

6. The tens digit of a two-digit number is four more than the units digit. When this two-digit number is divided by the sum of its digits, the answer is 8 remainder 3. Determine the sum of the digits of the two-digit number.

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7. Consider the expression below,
$$0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20$$.
Determine the number of ways if three of the β+β signs are changed to βββ signs so that the expression is equal to 100.

Misalkan hasil bagian penjumlahan adalah $$A$$ dan hasil bagian pengurangan adalah $$B$$

$$π΄ + π΅ = 210$$
$$π΄ β π΅ = 100$$
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$$2π΅ = 110$$
$$π΅ = 55$$

Karena $$B$$ merupakan penjumlahan $$3$$ angka minus maka banyak kemungkinan Β yang jumlahnya $$-55$$ adalah $$(20. 19, 16), (20, 18, 17)$$, hanya ada $$2$$, jadi banyak cara mengganti tiga tanda $$β+β$$ menjadi tanda $$β-β$$ sehingga jumlahnya $$100$$ ada $$2$$ cara

8. Carl tells Jill that he is thinking of three positive integers, not necessarily all different. He tells her that the product of his three integers is 36. Moreover, he also tells her the sum of his three integers. However, Jill still cannot figure out what the three integers are. Determine the sum of Carl’s three integers.

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9. Determine the sum of all the real numbers x that satisfy the equation

$$(3^x – 27)^2 + (5^x – 625)^2=(3^x + 5^x – 652)^2$$

Misalkan $$3^π₯ β 27 = π΄$$ dan $$5^π₯ β 625 = π΅$$

$$π΄^2 + π΅^2 = (π΄ + π΅)^2$$
$$βπ΄^2 + π΅^2 = π΄^2 + π΅^2 + 2π΄π΅$$
$$β2π΄π΅ = 0$$
$$βπ΄ = 0 β¨ π΅ = 0$$
$$π΄ = 0 β 3π₯ β 27 = 0 β π₯ = 3$$,
$$π΅ = 0 β 5π₯ β 625 = 0 β π₯ = 4$$

Jadi jumlah semua nilai $$x$$ yang memenuhi adalah 3 + 4 = 7[/latex]

10. If $$x>0$$ and $$(x+\frac{1}{x})^2=49$$, determine the value of $$x^3+\frac{1}{x^3}$$

$$(π₯ +\frac{1}{π₯})^2= 49 β (π₯ +\frac{1}{π₯})= 7$$

Bentuk lain

$$(π₯ +\frac{1}{π₯})^2= 49 β π₯^2 +\frac{1}{π₯^2} + 2 = 49 β π₯^2 +\frac{1}{π₯^2}=47$$

$$π₯^3 +\frac{1}{π₯^3} = (π₯ +\frac{1}{π₯})(π₯^2 β 1 +\frac{1}{π₯^2}) = 7(47 β 1) = 7(46) = 322$$