SOAL DAN PEMBAHASAN INVITATIONAL WORLD YOUTH MATHEMATICS INTERCITY COMPETITION 2001

1. Find all integers n such that 1 + 2 + .. + n is equal to a 3-digit number with identical digits. Answer: By using series formula, we can find 1 + 2 + ⋯+ 𝑛 =

𝑛(𝑛+1) 2

2(1 + 2 + ⋯ + 𝑛) = 𝑛(𝑛 + 1) Suppuse a 3-digit number is AAA, AAA= A(111) 2A(111)= n(n+1) We can get the value of A by subtitute A into 1,2,3,4,5, ... If A= 1 2.1.(111)= n(n+1) 6.37= n(n+1)

impossible, because 6 and 37 is not consecutive number

If A= 2 2.2.(111)= n(n+1) 12.37= n(n+1)

impossible, because 12 and 37 is not consecutive number

If A= 3 2.3.(111)= n(n+1) 18.37= n(n+1)

impossible, because 18 and 37 is not consecutive number

If A= 4 2.4.(111)= n(n+1) 24.37= n(n+1)

impossible, because 24 and 37 is not consecutive number

If A= 5 2.5.(111)= n(n+1) 30.37= n(n+1)

impossible, because 30 and 37 is not consecutive number

If A= 6 2.6.(111)= n(n+1) 36.37= n(n+1)

36 dan 37 is consecutive number. So, the value of n is 36

2. In a convex pentagon ABCDE, ∠A = ∠B = 120°, EA = AB = BC = 2, and CD = DE = 4. Find the area of the pentagon ABCDE.

Answer: If, ∠A = ∠B = 120°, EA = AB = BC = 2, and CD = DE = 4. According to that information, we can draw:

From the above figure, we can say that the area of pentagon equivalent with the area of seven equilateral triangle.

By using phytagorean theorem, 𝑡 = √22 − 12 𝑡 = √3 𝐴∆=

2𝑥√3 2

= √3 𝑐𝑚

The area of pentagon is 𝐴 = 7√3 cm 3. If I place a 6 cm by 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the 2

triangle on the square, I can cover up to 3 of the square. What is the area of the triangle? Answer: Under construction!

4. Find a set of four consecutive positive integers such that the smallest is a multiple of 5, the second is a multiple of 7, the third is a multiple of 9, and the largest is a multiple of 11. Answer: Suppose the four consecutive four number are a, a+1, a+2, and a+3. Amount of first, second, and third number is a+a+1+a+2= 3a+3 a+a+1+a+2= 3(a+1) Amount of first, second, and third number is equal to three times of second number and multiple of 21 (because second number multiple of 7). Using table we can find each of number. 1st number a 55

2nd number a+1 56

3rd number a+2 57

4th number a+3 58

a+(a+1)+(a+2) 3(a+1) 21x8= 168

90

91

92

93

21x13= 273

125

126

127

128

21x18= 378

160

161

162

163

21x23= 483

195

196

197

198

21x28= 588

230

231

232

233

21x33= 693

265

266

267

268

21x38= 798

300

301

302

303

21x43= 903

335

336

337

338

21x48= 1008

370

371

372

373

21x53= 1113

405

406

407

408

21x58= 1218

440

441

442

443

21x63= 1323

475

476

477

478

21x68= 1428

Addition Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 4th number Not suitable in 3rd number Not suitable in 3rd number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 3rd and 4th number Not suitable in 4th number

From the above table, in the row 5 and 13, 1st, 2nd, and 3rd suitable In the row 5, 3(a+1)= 21x23 In the row 13, 3(a+1)= 21x68 The difference is 945= 21x45 1st number a 160

2nd number a+1 161

3rd number a+2 162

4th number a+3 163

a+(a+1)+(a+2) 3(a+1) 21x23= 483

475

476

477

478

21x68= 1428

790

791

792

792

21x113= 2373

1105

1106

1107

1108

21x158= 3318

Addition Not suitable in 4th number Not suitable in 4th number Not suitable in 4th number Not suitable in 4th number

1420

1421

1422

1423

21x203= 4263

1735

1736

1737

1738

21x248= 5208

Not suitable in 4th number correct

The number are 1735, 1736, 1737, and 1738. 5. Between 5 and 6 o’clock, a lady looked at her watch. She mistook the hour hand for the minute hand and vice versa. As a result, she thought the time was approximately 55 minutes earlier. Exactly how many minutes earlier was the mistaken time? Answer: Under construction 6. In triangle ABC, the incircle touches the sides BC, CA and AB at D, E and F respectively. If the radius of the incircle is 4 units and if BD, CE and AF are consecutive integers, find the length of the three sides of ABC. Answer: Under construction 7. Determine all primes p for which there exists at least one pair of integers x and y such that p+1=2x2 and p2+1= 2y2 Answer: Amount from p+1=2x2 and p+1= 2y2 is 𝑝+1=2𝑥 2 𝑝2 +1=2𝑦2 2 𝑝+𝑝 +2=2𝑥 2 +2𝑦 2 (1)

+

Different from p+1=2x2 and p+1= 2y2 is 𝑝+1=2𝑥2 𝑝2 +1=2𝑦 2 2 𝑝−𝑝 =2𝑥2 −2𝑦2 (2)

–

Different from (1) and (2) is 𝑝+𝑝2 +2=2𝑥 2 +2𝑦 2 𝑝−𝑝2 =2𝑥 2 −2𝑦 2 2𝑝2 +2=4𝑦 2

We get 2p2+2= 4y2 or p2=2y2-1 Now, we can subtitute y by 0, ±1, ±2, ±3, ±4, ... For y=1 p2=2.12-1=1 (1 is not prime number) For y=2 p2=2.22-1=7 (7 is prime number) The value of p is 7. 8. Find all real solutions of Answer:

√3𝑥 2 − 18𝑥 + 52 + √2𝑥 2 − 12𝑥 + 162 = √−𝑥 2 + 6𝑥 + 280 √3(𝑥 − 3)2 + 25 + √2(𝑥 − 3)2 + 144 = √−(𝑥 − 3)2 + 289 √3(𝑥 − 3)2 + 25 + √2(𝑥 − 3)2 + 144 = √−(𝑥 − 3)2 + 289 Suppose 𝑎 = (𝑥 − 3)2 √3𝑎 + 25 + √2𝑎 + 144 = √−𝑎 + 289 2

(√3𝑎 + 25 + √2𝑎 + 144) = (√−𝑎 + 289)

2

3𝑎 + 25 + 2𝑎 + 144 + 2√(3𝑎 + 25)(2𝑎 + 144) = −𝑎 + 289 5𝑎 + 169 + 2√6𝑎2 + 482𝑎 + 3600 = −𝑎 + 289 6a − 120 = −2√6𝑎2 + 482𝑎 + 3600 3a − 60 = −√6𝑎2 + 482𝑎 + 3600 2

(3a − 60)2 = (−√6𝑎2 + 482𝑎 + 3600) 9𝑎2 − 360a + 3600 = 𝑎2 + 482𝑎 + 3600 3𝑎2 − 842𝑎 = 0 a(3a-842)= 0 𝑎=0

842

or

𝑎=

(𝑥 − 3) = 0

or

(𝑥 − 3)2 =

𝑥=3

or

x= √

2

3

842 3

842 3

+3

9. Simplify into a single numerical value. Answer:

√12 − √24 + √39 − 2√26 − √12 + √24 + √39 + 2√26 = = √12 + √39 − (√24 + 2√26) − √12 + √39 + √24 + 2√26 Suppose 𝑎 = 12 + √39 and 𝑏 = √24 + 2√26 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 𝑎 − 𝑏 + 𝑎 + 𝑏 − 2√(𝑎 − 𝑏)(𝑎 + 𝑏) 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 2𝑎 − 2√𝑎2 − 𝑏 2 2

2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2√(12 + √39) − (√24 + 2√26)

2

2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2√183 + 24√39 − 128 − 16√39 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2√55 + 8√39 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2√55 + 2√16.39 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2(√16 + √39) 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 24 + 2√39 − 2√16 − 2√39) 2

(√𝑎 − 𝑏 − √𝑎 + 𝑏) = 16 (√𝑎 − 𝑏 − √𝑎 + 𝑏) = ±4 10. Let M = 1010101…01 where the digit 1 appears k times. Find the least value of k so that 1001001001001 divides M ? Answer: Under construction