Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 231 17. ( − 1)( −2)( − 3)( −4) −15 = ( 2 −5 + 4)( 2 −5 + 6)− 15 = ( + 4)( + 6) − 15 = ( 2 +10 + 24)− 15 = 2 + 10 + 9 = ( + 9)( + 1) = ( 2 −5 + 9)( 2 −5 + 1) Jadi, nilai dari + + + = 1+ (−5)+ 9+ 1 = 6. 18. 211 ≡ 4 ( 17) 212 ≡ 16 ( 17) ≡ (−1) ( 17) 213 ≡ 64 ( 17) ≡ 13 ( 17) ≡ (−4)( 17) 2116 ≡ 1 ( 17) 21390 ≡ (2116) 24 ∙ 216 ≡ (1) 24 ∙ 216 ( 17) ≡ 1(−1)(−4)( 17) ≡ 4 ( 17) 19. Untuk menyelesaikan soal di atas kita mulai dari = 0 karena pada ( +) = + () yang diketahui adalah (0) = 2, sehingga kita peroleh: y = 0 → f(x) = x + f(0) → f(x) = x + 2 → f(2019) = 2019 + 2 = 2021 20. 9876543210 12 ≡ (9 ∙ 109 + 8 ∙ 108 + 7 ∙ 107 +6 ∙ 106 + 5 ∙ 105 +4 ∙ 104 + 3 ∙ 103 + 2 ∙ 102 + 10) 12 ≡ [(0 + (8 ∙ 108 12)+ (7 ∙ 107 12)+ 0+ (5 ∙ 105 12) + (4 ∙ 104 12)+ 0 +(2 ∙ 102 12) +10 12)] 12 ≡ (0 + 8+ 4 + 0+ 8 +4 + 0 +8 + 10) 12 ≡ 42 12 ≡ 6 12 21. 2 70 ≡ (2 7 ) 10 13 ≡ 12810 13 ≡ (9 ∙ 13 + 11) 10 13 ≡ 112∙5 13 ≡ 4 5 13 ≡ (78 ∙ 13+ 10) 13 ≡ 10 13 3 70 ≡ (3) 3∙23+1 13 ≡ 3 3∙23 ∙ 3 13 ≡ (27) 23 ∙ 3 13 ≡ (2 ∙ 13 +1) 23 ∙ 3 13 ≡ 1 23 ∙ 3 13 ≡ 3 13 Sehingga, (2 70 + 3 70) 13 ≡ 10 13 + 3 13 ≡ 13 13 ≡ 0 13 Jadi, terbukti 13 membagi habis 2 70 +3 70 .
Solusi Kelas 8 SMP 232 Kumpulan Sedekah Soal Siswa KPM 22. 2 6 ≡ 1 ( 7) 2 7 ≡ (2 6 ) 1 ∙ 2 1 ( 7) ≡ (1) 1 ∙ 2 ( 7) ≡ 2 ( 7) 3 6 ≡ 1 ( 7) 3 8 ≡ (3 6 ) 1 ∙ 3 2 ( 7) ≡ (1) 1 ∙ 9 ( 7) ≡ 2 ( 7) 6 9 ≡ 1 ( 7) 6 9 ≡ (6 6 ) 1 ∙ 6 3 ( 7) ≡ (1) 1 ∙ 216 ( 7) ≡ 6 ( 7) 9 6 ≡ 1 ( 7) 9 10 ≡ (9 6 ) 1 ∙ 9 4 ( 7) ≡ (1) 1 ∙ 6561 ( 7) ≡ 2 ( 7) Jadi, 2 + 2 + 6 + 2 = 12. 23. −1 ≡ 1 7 19−1 ≡ 1 19 7 18 ≡ 1 19 7 55 ≡ (7 18) 3 ∙ 7 1 19 ≡ (1) 3 ∙ 7 ( 19) ≡ 7 ( 19) 7 55 ≡ 7 ( 19) → the remainder: 7÷ 3 = 21 So, the remainder of 7 55 19 divided by 3 is 1. 24. = 2 + 2 2 = 2 2 + 2 2 = 2 + 2 + 2 + 2 − 2 + 2 = ( +) 2 +( − ) 2 10 = 10 2 + 10 2 = ( 2 + 2 )(3 2 + 1 2 ) = 3 2 2 + 2 + 3 2 2 + 2 = 3 2 2 + 6 + 2 + 2 − 6 +3 2 2 = (3 +) 2 + ( −3) 2 Proven
Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 233 25. Misalkan adalah perpotongan dan Karena [] = [] Maka, [] + [] = [] + [] [] = [] [] + [] = [] + [] [] = [] adalah alas dari ∆ dan ∆, maka kedua segitiga memiliki tinggi yang sama dan ∥ Karena ∥ , maka ∆~∆ = = 3 5 Sejak ∆ dan ∆ memiliki tinggi yang sama, maka [] = 3 5 [] = 3 5 × 10 = 6 26. + = 180° − ∠ − ∠ − ∠ = 180° − 60° − 25° − 20° = 75° ∠ = 180° − ( + ) = 180° − 75° = 105° 27. Nickolas Lucano X 0 1 5 X 4 5 X 3 5 X 2 5 X + 9 10 20 X − 11 10 20 X Maka, + 9 10 20 X = − 11 10 20 X − = = = 11 9 20 20 20 2 20 20 200 X X X X Jadi, kapasitas botol Nickolas adalah 200 ml. 28. 3x (x – 3) – (2x (x + 1) + x – 2) = 3x2 – 9x – (2x2 + 2x + x – 2) = 3x2 – 9x – 2x2 – 2x – x + 2 = 3x2 – 2x2 – 9x – 2x – x + 2 = x2 – 12x + 2
Solusi Kelas 8 SMP 234 Kumpulan Sedekah Soal Siswa KPM 29. Volume tabung = 2 r t 2 3 3 60 ( )(2 ) 60 2 30 r r r r = = = Volume bola = 4 3 3 r = = 4 3 30 40 cm 3 30. P = 26 cm, R = 12, r = 2. d 2 = p2 – (R – r)2 = 262 – (12 – 2)2 = 676 – (10)2 = 576 d = 576 = 24 cm 31. (3 7 4 8 + ( )) + = + = = 7 12 7 19 3 8 8 4 38 4 4 4 4 32. Panjang rusuk alas = 5 cm Luas alas = 5 cm × 5 cm = 25 cm2 Volume = 25 × 8 × 1 16 = 200 × 1 16 = 12,5 cm3 33. K = 2 (p + l ) 54 cm = 2 ((3 cm + l) + l ) 54 cm = 2 (3 cm + 2l) 27 cm = 3 cm + 2l 24 cm = 2l 12 cm = l 12 cm + 3 cm = p 15 cm = p L = p × l L = 15 cm × 12 cm = 180 cm²
Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 235 14.3 Solusi Soal Sulit Kelas 8 SMP 34. Keliling segitiga = jumlah ketiga sisinya. KΔ = a + (a + b) + (a + 2b) =3a + 3b = 72 a + b = 72 : 3 = 24 …… (1) Karena segitiga yang dimaksud adalah segitiga siku-siku maka berlaku teorema phytagoras (kuadrat panjang sisi miring sama dengan jumlah kuadrat panjang dua sisi yang lain. (a + 2b)2 = a2 + (a + b)2 a 2 + 4ab + 4b2 = 2a2+2ab + b2 a 2 -2ab-3b2 = 0 (a-b)2 -4b2 = 0 (a-b)2 = (2b)2 a-b = 2b a = 3b …. (2), disubstitusikan ke persamaan (1) 24 = 3b + b 24 = 4b b = 6 a = 3 x 6 = 18 LΔ = 1 2 x a x (a + b) = 1 2 x 18 x 24 = 216 Sebagai alternatif penyelesaian dapat juga diselesaikan dengan menggunakan perbandingan sisi-sisi segitiga siku-siku. Perbandingan segitiga siku-siku membentuk barisan aritmetika sehingga berlaku : : = 3 ∶ 4 ∶ 5. a = 3x, b = 4x, c = 5x KΔ = 3x + 4x + 5x
Solusi Kelas 8 SMP 236 Kumpulan Sedekah Soal Siswa KPM 72 = 12x x = 6 LΔ = ½ (4x)(3x) = 1 2 (4 x 6)(3 x 6) = 216 35. 11 ( ) 4 8 28 80 ... Sampai sebelas suku 3 9 27 81 S = + + + + ( ) ( ) ( ) ( ) 2 3 4 11 11 11 11 1 1 1 1 1 1 1 1 ... 3 9 27 81 1 1 1 1 1 11 ... 3 9 27 81 1 1 1 1 1 1 11 ... 3 3 3 3 3 1 1 3 3 11 1 1 3 1 3 11 4 45 3 4 − − − = + + − + + + − + = + − + − + = + − + − + + − = + + − = + − = 36. = + + + + 2 2 2 2 ... 1 2 3 2 3 4 3 4 5 100 101 102 X + + + + = + + + + 1 2 2 8 5 3 27 8 4 1000000 299 101 ... 1 2 3 2 3 4 3 4 5 100 101 102 Y + = + + + + 5 23 59 1000000 299 101 ... 1 2 3 2 3 4 3 4 5 100 101 102 + + = + + + + 7 25 61 1000002 299 101 ... 1 2 3 2 3 4 3 4 5 100 101 102 X Y + = + + + + 7 25 61 1000002 299 101 ... 6 24 60 100 101 102 = + + + + 1 1 1 1 1 1 1 ... 1 6 24 60 100 101 102 = + + + + + + + + + 1 1 1 1 1 1 1 ... 1 ... 6 24 60 100 101 102
Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 237 = + + + + + 1 1 1 1 100 ... 1 2 3 2 3 4 3 4 5 100 101 102 (There are 100 number 1) Since we know that ( )( ) ( ) ( )( ) = − + + + + + 1 1 1 1 n n n n n n n 1 2 1 1 2 2 So, + + + + + + + + + 1 1 1 1 1 1 1 ... 1 ... 1 2 3 2 3 4 3 4 5 100 101 102 Become + + + + + − + − + + − 1 1 1 1 1 1 1 1 1 1 1 1 ... 1 ... 2 1 2 2 3 2 2 3 3 4 2 100 101 101 102 = + − + − + + − 1 1 1 1 1 1 1 100 ... 2 1 2 2 3 2 3 3 4 100 101 101 102 = + − 1 1 1 100 2 2 10302 = + 2575 100 10302 From this we know that = 100, = 2575, = 10302 So 2 + + = 1002 + 2575 + 10302 = 10000 + 2575 + 10302 = 22877 37. We can add some additional point to help solve this problem as shown in the picture below L and M on AF and AG respectively, and DL//DM//BC = = 1 3 = 1 ⇒ = 1 3
Solusi Kelas 8 SMP 238 Kumpulan Sedekah Soal Siswa KPM = = 1 3 = 2 ⇒ = 1 6 : = 1: 7 = = 1 3 = 2 ⇒ = 2 3 = = 2 3 = 1 ⇒ = 2 3 : = 2: 5 [] = [] = ( − ) [] = ( 2 5 − 1 7 ) [] = 9 35 [] Since [ACD] is 1 3 of [ABC] so we have [] = 9 35 . 1 3 [] = 3 35 [] N and O on AF and AG respectively, and EN//EO//BC = = 2 3 = 1 ⇒ = 2 3 = = 2 3 = 2 ⇒ = 1 3 : = 1: 4 = = 2 3 = 2 ⇒ = 4 3 = = 4 3 = 1 ⇒ = 4 3 : = 4: 7 [] = [] = ( − ) [] = ( 4 7 − 1 4 ) [] = 9 28 [] Since [ACE] is 2 3 of [ABC], so we have [] = 9 28 . 2 3 [] = 3 14 [] So, we get the shaded region area [HIJK] is = []− [] = 3 14 [] − 3 35 [] = ( 15 70 − 6 70) [] = 9 70 [] Then the ratio of shaded region with area of triangle ABC is 9:70
Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 239 38. Perhatikan bahwa: + + + + = + + + + + + ( ) ( ) ( ) = + + + + = + + + 2 2 2 1 2 2 3 3 4 ... 62 63 1 2 2 3 3 4 4 5 ... 61 62 62 63 8 32 64 ... 4028 8 1 8 2 ... 8 31 Maka + + + + = + + + + 1 2 2 3 3 4 ... 62 63 2 2 2 2 1 2 3 ... 31 8 Diketahui bahwa: = − = − = − = − 2 2 2 2 1 2 2 2 2 3 3 3 3 4 4 4 ... 30 31 31 31 ( ) ( ) + + + + = − + − + − + + − = + + + + − + + + + = + + + + − 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 3 3 4 ... 62 63 2 2 3 3 4 4 ... 31 31 1 2 3 ... 31 1 2 3 ... 31 1 2 3 ... 31 496 Maka ( ) + + + + − − + + + + − 1 2 2 3 3 4 ... 62 63 2 2 2 2 256 1 2 3 ... 31 496 8 = + + + + − − + + + + − ( ) ( ) 2 2 2 2 2 2 2 2 1 2 3 ... 31 256 1 2 3 ... 31 496 = − + = 256 496 240, B = 240 = 24 x 31 x 51 Banyak faktor 240 adalah (4 + 1) x (1 + 1) x (1 + 1) = 20 Jadi nilai dari 101 x B = 101 x 20 = 2020. 39. Because 2 2 2 2 prime number 10 100 m m + = + = Let m = 10y ➔ 10 m = y so, 2 y + = 2 prime number 0 mod3 ( ) • Case 1 If y = 1 (mod 3) and y = 3k + 1 (3k + 1)2 + 2 = 9k2 + 6k + 3 ≡ 0 (mod 3) y 2 + 2 not a prime number • Case 2 If y = 2 (mod 3) and y = 3k + 2 (3k + 2)2 + 2 = 9k2 + 12k + 6 ≡ 0 (mod 3) y 2 + 2 not a prime number • Case 3
Solusi Kelas 8 SMP 240 Kumpulan Sedekah Soal Siswa KPM If y = 0 (mod 3) and y = 3k (3k)2 + 2 = 9k2 + 2 ≡ 2 (mod 3) y 2 + 2 a prime number when y ≡ 0 (mod 3) and y is a prime number y is exactly 3 y = 3 ➔ m = 10.y = 10.3 = 30 See that (y – 1)3 = y3 – 3y2 + 3y – 1 3y2 – 3y = y3 – (y – 1)3 – 1 3y(y – 1) = y3 – (y – 1)3 – 1 ( ) ( ( )) 1 1 3 1 1 3 3 y y y y − = − − − ( ) ( ( )) 1 1 3 1 1 3 3 y y y y − = − − − ( ) ( ) ( ) ( ) ( ) ( ) 1 , 1 , 1 30 31 30,31 30,31 30,31 30 31 m m LCM m m GCD m m LCM GCD LCM + = + + = = 2 6 12 20 ... 30,31 1 2 2 3 3 4 ... 30 31 + + + + + = + + + + LCM( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 3 3 3 3 3 3 1 1 1 2 2 1 3 3 1 1 2 3 3 2 3 3 1 1 3 4 4 3 3 3 .... 1 1 30 31 31 30 + 3 3 1 1 1 2 2 3 3 4 ... 30 31 31 1 30 3 3 1 29790 10 3 = − − = − − = − − = − − + + + + = − − = − 9930 10 9920 n = − = = 180 9920 180 10100 2020 5 5 5 n X + + = = = = Jadi, nilai dari X = 2020
Solusi Kelas 8 SMP Kumpulan Sedekah Soal Siswa KPM 241 40. Perhatikan gambar berikut BG = 2√2, AG = 2√3, BO = x Misalkan titik O di AG sehingga BO ⊥ AG Dari Pythagoras didapat: ( ) ( ) 2 2 2 2 ... 1 ... 2 AO AB BO GO BG BO = − = − Persamaan (1) + (2) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 8 AO GO AB BO BG BO AG x x x x + = − + − = − + − = − + − Kuadratkan kedua ruas ( ) ( ) 2 2 2 2 2 2 2 4 2 4 2 2 4 2 2 4 2 4 4 2 2 2 2 2 4 8 12 4 8 2 32 12 12 12 2 2 12 32 2 2 12 32 12 32 12 32 12 32 32 8 12 3 x x x x x x x x x x x x x x x x x x x x = − + − = − + − + − + = − + − + = − + = − + = − + = = = 3 3 8 2 97 97 97 8 8 3 a x = = = 96 y z mod97 karena y dan a relatif prima dan dari teorema Euler didapat: ( ) ( ) ( ) ( ) 96 1mod 1 97 1 96 1mod96 a y a a a y = − = − = Dari persamaan di atas diketahui bahwa z = 1 Jadi, a + z = 97 + 1 = 98 E F A B D O C H G
Solusi Kelas 8 SMP 242 Kumpulan Sedekah Soal Siswa KPM 41. Perhatikan gambar berikut. , sehingga 90 sehingga 90 karena maka ABD BCF DBA FCB b CBF b ABC ABD BCF AB BC = = = − = = Dan akan didapat = BCA 45 Karena AB = BC dan = ABC 90 Maka a + b = 180o –BCA = 180o – 45o = 135o A D 3 B 9 6 E 3 C 3 F