Pre RMO Solutions

Pre RMO Hints and Solutions 1. Score of 89, 88 and 85 are impossible to score as to score 88 or 89, one needs to get more than 29 questions correct and atleast one question wrong, whereas to score 85 one needs to get more than 28 questions correct with two questions wrong which is impossible as there are only 30 questions. 1

2. 𝑎2 + 𝑏 2 + 𝑐 2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎 = 2 ((𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2 ) . Thus for it to be equal to zero, all the numbers must be equal which is not possible. If two numbers are equal, then the expression can be equal to 1 (𝑇𝑎𝑘𝑒 𝑎 = 𝑏 = 1, 𝑐 = 2) or is greater than 3 (Take 𝑎 = 𝑏 = 1, 𝑐 = 3). If all three are distinct, then their differences are atleast 1, 1 and 2 (Take 𝑎 = 1, 𝑏 = 2, 𝑐 = 3) giving the least value of the expression in this case as 3. Thus the expression cannot be equal to 2 either. 3. As 3𝑥 + 4𝑦 = 2013, we get 𝑦 = positive, we get 𝑥 =

(2013−4𝑦) 3

3(671−𝑥) , 4

therefore 𝑦 must be a multiple of 3. As 𝑥 is

> 0, or 503.25 > 𝑦. Thus 𝑦 takes the values 3,6,9, … ,501,

i.e. total 167 values. As for every such value of 𝑦, there is one positive integer value of 𝑥, we get 167 such pairs. 4. Let 𝑛 = 10𝑥 + 𝑦, thus we get two equations for two variable 𝑥 and 𝑦 as 10𝑥 + 𝑦 = 2(10𝑦 + 𝑥) + 7 and 10𝑥 + 𝑦 = 7(𝑥 + 𝑦) + 6. Solving simultaneously, we get 𝑥 = 8 and 𝑦 = 3. 2 𝑝

2 𝑝

2 𝑝

5. If we let 𝑥1 = 𝑝, then 𝑥2 = , 𝑥3 = 𝑝, 𝑥4 = , … , 𝑥99 = 𝑝, 𝑥100 = , i.e. 𝑥𝑖 = 𝑝 for odd value 2

of 𝑖 and 𝑥𝑖 = 𝑝 for even value of 𝑖. Therefore if 𝑥𝑎 𝑥𝑏 𝑥𝑐 𝑥𝑑 = 4, we must have two of 2

𝑥𝑎 , 𝑥𝑏 , 𝑥𝑐 , 𝑥𝑑 as 𝑝 and the other two as 𝑝 since 𝑝 is rational. Therefore two of 𝑎, 𝑏, 𝑐, 𝑑 are odd and the other two are even. Therefore the product 𝑎𝑏𝑐𝑑 must be divisible by 4. Thus 𝑎𝑏𝑐𝑑 ≠ 945 𝑜𝑟 270 𝑜𝑟 210. Also for {𝑎, 𝑏, 𝑐, 𝑑} = {2,4,5,7}, we get 𝑎𝑏𝑐𝑑 = 280. 6. Observe that 35 = 243 = 61 ∙ 4 − 1, thus 310 = (35 )2 = 61 ∙ 𝑘 + 1 for some integer 𝑘. Therefore 32010 = (310 )201 = (61 ∙ 𝑘 + 1) ∙ (61 ∙ 𝑘 + 1) ∙ … ∙ (61 ∙ 𝑘 + 1) = 61𝑚 + 1 for some integer 𝑚. i.e. the remainder if 32010 is divided by 61 is 1. Now 32015 = 243 ∙ 32010 = (61 ∙ 3 + 60) ∙ (61 ∙ 𝑚 + 1) = 61 ∙ 𝑛 + 60 for some integer 𝑛. i.e. the remainder when 32015 is divided by 61 is 60. Similarly deduce that the remainder when 112014 is divided by 61 is also 60 by noting 112 = 61 ∙ 2 − 1 and writing 112014 as 112 ∙ 112 ∙ … ∙ 112 (1007) times to get 112014 = 61 ∙ 𝑝 − 1 = 61 ∙ (𝑝 − 1) + 60 for some integer 𝑝. Thus the required remainder is obtained by dividing 60 + 60 = 120 by 61 giving 𝑟 = 59. 7. Take 𝑎 = 123 and 𝑏 = 133 to write 256 as 𝑎 + 𝑏. Therefore 2563 − 1233 − 1333 = (𝑎 + 𝑏)3 − 𝑎3 − 𝑏 3 = 3𝑎𝑏(𝑎 + 𝑏) = 3 ∙ 123 ∙ 133 ∙ 256. 8. Let 𝑥 100 = (𝑥 2 − 1) ∙ 𝑞(𝑥) + 𝑎𝑥 + 𝑏. Get the values of 𝑎 and 𝑏 by putting 𝑥 = 1 and −1. 9. 𝑁 − 1 is the LCM of first ten natural numbers. (Here the number 1 is excluded from dividing N or else the question wouldn’t have made sense).

10. The perfect squares are 22 , 44 , 66 , … , 100100 as well as 11 , 99 , 2525 , 4949 , 8181 . 11. Let the altitudes meet in 𝐻 (They are concurrent). Then the quadrilaterals 𝐴𝐸𝐻𝐹 and 𝐶𝐸𝐻𝐷 are cyclic (Why?). Therefore ∠𝐹𝐸𝐻 = ∠𝐹𝐴𝐻 = 90 − ∠𝐵. Similarly ∠𝐷𝐸𝐻 = ∠𝐷𝐶𝐻 = 90 − ∠𝐵 (Why?). 12. Get the remainder as (𝑝 − 1)𝑥 + (𝑞 − 7) by carrying out long division. For the remainder to be zero we must have 𝑝 = 1, 𝑞 = 7. 13. Drop perpendicular 𝑃𝑆 on side 𝑄𝑅 meeting 𝐶𝐷 at the point 𝑇. Thus 𝑄𝑆 = 𝑆𝑅 = 6 and 𝑃𝑆 = 𝑃𝑇

𝑇𝐶

8. Now ∆𝑃𝑇𝐶~∆𝑃𝑆𝑅. Thus 𝑃𝑆 = 𝑆𝑅 i.e.

8−𝑏 8

=

𝑎/2 6

or 2𝑎 + 3𝑏 = 12. 1 2

14. Draw parallel to side 𝐵𝐶 from the point 𝐸 meeting the side 𝐴𝐵 in 𝐿. Thus 𝐸𝐿 = 𝐵𝐶 = 4 1

(Why?) and 𝐹𝐿 ∥ 𝐴𝐷 as well as 𝐹𝐿 = 2 𝐴𝐷 = 3 (Why?). Since ∆𝐸𝐹𝐿 is right angled, 𝐸𝐹 = 5. 15. Observe that 𝛼 2 = 𝛼 + 1 and 𝛽 2 = 𝛽 + 1, thus 𝛼 𝑛+1 = 𝛼 𝑛 + 𝛼 𝑛−1 and 𝛽 𝑛+1 = 𝛽 𝑛 + 𝛽 𝑛−1 giving 𝑉𝑛+1 = 𝑉𝑛 + 𝑉𝑛−1 . As 𝑉0 = 𝛼 0 + 𝛽 0 = 2 and 𝑉1 = 𝛼 + 𝛽 = 1, we get 𝑉2 = 𝑉1 + 𝑉0 , 𝑉3 = 𝑉2 + 𝑉1 , 𝑉4 = 𝑉3 + 𝑉2 and so on. 16. 𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑎 ∙ 111111 = 3 ∙ 7 ∙ 37 ∙ 11 ∙ 13 ∙ 𝑎 = 𝑏 ∙ 𝑐 ∙ 𝑏𝑐 ∙ (𝑐 + 𝑐 − 𝑏) ∙ (𝑐 + 𝑏 + 𝑏) ∙ (𝑐 − 𝑏). Compare to get 𝑏 = 3, 𝑐 = 7 and 𝑎 = 4. 3

3

17. Assume 𝑥 = √9 + 4√5 + √9 − 4√5, thus 𝑥 3 = 18 + 3𝑥 or (𝑥 − 3)(𝑥 2 + 3𝑥 + 6) = 0 or 𝑥 = 3. 18. For 𝑛 = 41, 82, 123, … the expression is divisible by 41. (But it takes prime values for 𝑛 = 1, 2, 3, … , 39). ! 19. Observe that 𝑛4 + 𝑛2 + 1 = (𝑛4 + 2𝑛2 + 1) − 𝑛2 = (𝑛2 + 1)2 − 𝑛2 = (𝑛2 + 1 + 𝑛)(𝑛2 + 1 − 𝑛), then proceed. 20. Join 𝐵 and 𝐷 and use midpoint theorem. 21. 𝑥 = 1 is the identical root (Obtained by hit and trial method). 22. The quadrilateral 𝑅𝑄𝑂𝑇 is cyclic (Why?). Thus ∠𝑄𝑅𝑂 = ∠𝑄𝑇𝑂 = ∠𝑃𝑇𝑆 = 450 (Why?). As ∠𝑇𝑆𝑃 = 900 , ∠𝑆𝑃𝑇 = 450, Thus 𝑆𝑇 = 𝑆𝑃 = 6. 23. 𝑉 is the midpoint of 𝑋𝑌. Let 𝑃𝑉 = 𝑥. Drop perpendicular from 𝑃 to 𝑌𝑍 to complete a right triangle. Use Pythagorus’s Theorem to get (𝑥 − 5)2 = 152 + (30 − 𝑥)2, i.e. 𝑥 = 22. 24. 4(𝑎2 + 𝑏 2 + 𝑐 2 ) − 12(𝑎 + 𝑏 + 𝑐) + 27 = 0 or (2𝑎 − 3)2 + (2𝑏 − 3)2 + (2𝑐 − 3)2 = 0 3

giving 𝑎 = 𝑏 = 𝑐 = 2.

1 2

1

1 2

1

1

25. (𝑥 2 + 𝑥 2 ) = 𝑥 4 + 𝑥 4 + 2 = 121, thus 𝑥 2 + 𝑥 2 = ±11. (𝑥 + 𝑥) = 𝑥 2 + 𝑥 2 + 2 = 13 𝑜𝑟 − 1

9. Thus 𝑥 + 𝑥 = √13 𝑜𝑟 − √13. 26. Let 𝐴𝐵 = 3𝑥, thus 𝐵𝐶 = 4𝑥 (Why?). Use Pythagorus’s theorem to get 𝑥 = 7. 27. Add all the three equations to get 𝑎2 + 6𝑎 + 𝑏 2 + 2𝑏 + 𝑐 2 + 4𝑐 = −14 or (𝑎 + 3)2 + (𝑏 + 1)2 + (𝑐 + 2)2 = 0, i.e. 𝑎 = −3, 𝑏 = −1, 𝑐 = −2. 28. Assume the radii or the three circles to be 𝑟1 , 𝑟2 , 𝑟3 . Solve the simultaneous equations to get the radii as 1, 2, 3. 1

1

1

1

1 1 1 + + 𝑦 𝑧

29. Assume 2𝑥 = 3𝑦 = 6−𝑧 = 𝑘, therefore 𝑘 𝑥 = 2, 𝑘 𝑦 = 3 and 𝑘 𝑧 = 6. Thus 𝑘 𝑥 1 𝑥

1 𝑦

1 𝑧

1

=2∙3∙6=

1 to get + + = 0. (Here it was assumed that 𝑥, 𝑦, 𝑧 are non-zero or otherwise the 1

1

1

expression 𝑥 + 𝑦 + 𝑧 wouldn’t have been defined.) 30. Put 𝑎 = 𝑡 ∙ 𝑏 to get (𝑡 2 + 1)3 = (𝑡 3 + 1)2 or 3𝑡 4 − 2𝑡 3 + 3𝑡 2 = 0 or 3𝑡 2 − 2𝑡 + 3 = 0. 𝑎

𝑏

1

Thus 𝑏 + 𝑎 = 𝑡 + 𝑡 =

𝑡 2 +1 𝑡

2

= 3.

Author’s Comments on the paper It’ll be almost impossible for any student to solve all or nearly all questions in the time given, however that is not at all expected. What is expected is the student’s ability to think beyond textbooks and the will to lookout for challenging problems which precisely is the goal of Olympiads. A student is not expected to know all the techniques used here to solve a particular question, but with a little innovative thinking and with some patience and time, many questions should be doable within the knowledge acquired at the school level.