National math Olympiad questions 2010
ā§§. āύāĻŋāĻā§āϰ āϏāĻāĻā§āϝāĻžāĻā§āϞā§āϰ āĻŽāϧā§āϝ āĻĨā§āĻā§ āĻŽā§āϞāĻŋāĻ āϏāĻāĻā§āϝāĻž āĻāĻŦāĻ āϝā§āĻāĻŋāĻ āĻŦāĻž āĻā§āϤā§āϰāĻŋāĻŽ āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻĒā§āĻĨāĻ āĻāϰāĨ¤
111, 239, 379, 455
Determine which of the following are (is) prime and which are (is) composite 111, 239, 379, 455
⧍. āĻā§āύ āϏā§āĻĨāĻžāύā§āϰ 6 āĻŽāĻžāϏā§āϰ āĻŽāĻžāϏāĻŋāĻ āĻāĻĄāĻŧ āĻŦā§āώā§āĻāĻŋāĻĒāĻžāϤā§āϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖ 28.5 āĻŽāĻŋāϞāĻŋāĻŽāĻŋāĻāĻžāϰāĨ¤ āϝāĻĻāĻŋ āϏ⧠āϏāĻŽāϝāĻŧā§ āĻāϰ⧠36 āĻŽāĻŋāϞāĻŋāĻŽāĻŋāĻāĻžāϰ āĻŦā§āĻļāĻŋ āĻŦā§āώā§āĻāĻŋ āĻšāϤ⧠āϤāĻžāĻšāϞ⧠āĻŽāĻžāϏāĻŋāĻ āĻāĻĄāĻŧ āĻŦā§āώā§āĻāĻŋāĻĒāĻžāϤā§āϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻāϤ āĻšāϤā§? āĻŽāĻžāϏāĻŋāĻ āĻāĻĄāĻŧ āĻāϤ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§āϤ⧠?
The average monthly rainfall for 6 months was 28.5 millimeter. If it had rained 36 millimeter more within this six months what would the average have been? By how much would average have been increase?
ā§Š. ABC āĻāĻāĻāĻŋ āϏāĻŽāĻā§āĻŖā§ āϤā§āϰāĻŋāĻā§āĻāĨ¤ āĻāĻ āϤā§āϰāĻŋāĻā§āĻā§āϰ āĻāύā§āϝ AC à AC = AB à AB + BC à BC, āϝāĻĻāĻŋ \[\frac{AB}{AC} = \frac35, \frac{BC}{AC} = \frac45 \] āĻāĻŦāĻ AB, BC, AC āĻāϰ āĻ. āϏāĻž. āĻā§. 1 āĻšāϝāĻŧ āϤāĻžāĻšāϞ⧠AC āĻāϰ āĻŽāĻžāύ āĻāϤ?
ABC is a right angled triangle. For this triangle, AC Ã AC = AB Ã AB + BCÃ BC . If \[\frac{AB}{AC} = \frac35, \frac{BC}{AC} = \frac45 \] and the ged of AB, BC, AC is 1 then find AC.
ā§Ē. āĻā§āϰāĻŋ āĻ āĻŋāĻ āĻāϰāϞ⧠āϏ⧠āĻāĻāĻāĻŋ 10.1 āĻŽāĻŋāĻāĻžāϰ āϞāĻŽā§āĻŦāĻž āĻāĻŦāĻ 4.2 āĻŽāĻŋāĻāĻžāϰ āĻāĻāĻĄāĻŧāĻž āĻāĻāĻāĻŋ āϏāĻŦāĻāĻŋ āĻŦāĻžāĻāĻžāύ āĻāϰāĻŦā§āĨ¤ āĻāĻŋāύā§āϤā§, āĻāĻŽā§āϰ āĻāĻāĻŽāύ āĻ ā§āĻāĻžāύā§āϰ āĻāύā§āϝ āϤāĻžāĻā§ āĻāĻžāϰāĻĻāĻŋāĻā§ āĻŦā§āĻĄāĻŧāĻž āĻĻāĻŋāϤ⧠āĻšāĻŦā§āĨ¤ āĻā§āϰ⧠āĻ āĻŋāĻ āĻāϰāϞ⧠āϏ⧠11.2 āĻŽāĻŋāĻāĻžāϰ āϞāĻŽā§āĻŦāĻž āĻ 5.0 āĻŽāĻŋāĻāĻžāϰ āĻāĻāĻĄāĻŧāĻž āĻŦā§āĻĄāĻŧāĻž āĻĻā§āĻŦā§āĨ¤ āĻŦā§āĻĄāĻŧāĻž āĻ āĻŦāĻžāĻāĻžāύā§āϰ āĻŽāϧā§āϝāĻŦāϰā§āϤ⧠āĻ āĻāĻļā§āϰ āĻā§āώā§āϤā§āϰāĻĢāϞ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰā§āĨ¤
Jerry decided to grow a garden so he could make salad. He wants to make it 10.1 m long and 4.2 m wide. However, in order to avoid Tom from entering his garden he must make a fence surrounding the garden. He decides to make the fence 11.2 m long and 5.0 m wide. What is the area between the fence and the garden?
ā§Ģ. āĻā§āϰāĻŋ āĻ āĻŋāĻ āĻāϰā§āĻā§ āϝā§, āϏ⧠1, 2, 3, 4, 5, 6, 7, 8, 9, 10 āϏāĻāĻā§āϝāĻžāĻā§āϞā§āϰ āĻĒā§āϰāϤā§āϝā§āĻāĻāĻŋ āĻŽāĻžāϤā§āϰ āĻāĻāĻŦāĻžāϰ āĻāϰ⧠āĻāĻāĻāĻŋ āĻŦā§āϤā§āϤā§āϰ āĻĒāĻžāϰāĻĒāĻžāĻļā§ āĻāĻŽāύāĻāĻžāĻŦā§ āĻŦāϏāĻžāĻŦā§ āϝāĻžāϤ⧠āĻĒāĻžāĻļāĻžāĻĒāĻžāĻļāĻŋ āĻĻā§āĻāĻāĻŋ āϏāĻāĻā§āϝāĻžāϰ āĻŦāĻŋāϝāĻŧā§āĻāĻĢāϞ 4 āĻāϰ āĻā§āϝāĻŧā§ āĻāĻŽ āύāĻž āĻšāϝāĻŧāĨ¤ āĻā§āϰāĻŋ āĻā§ āϏā§āĻāĻž āĻĒāĻžāϰāĻŦā§āĨ¤ āĻĒāĻžāϰāϞā§, āϏā§āĻ āĻŦā§āϤā§āϤāĻāĻŋāϤ⧠āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻŦāϏāĻŋāϝāĻŧā§ āĻĻā§āĻāĻžāĻāĨ¤
Jerry decided to find out whether it is possible or not to write all numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 each of them once, around some circle so that the difference of every two neighboring numbers would be not less than 4? If possible, draw the circle with the numbers.
ā§Ŧ. āĻŦā§āϞā§āϝāĻžāĻāĻŦā§āϰā§āĻĄā§ āϞā§āĻāĻž āĻāĻŋāϞ 1, 3, 4, 6, 8, 9, 11, 12, 16āĨ¤ āĻā§āĻĨāĻž āĻĨā§āĻā§ āĻšāĻžāĻāĻŋāϰ āĻšāϞ⧠āĻāĻŽ āĻāϰ āĻā§āϰāĻŋāĨ¤ āĻĒā§āϰāĻĨāĻŽā§ āĻāĻŽ āĻāĻāĻāĻŋ āύāĻŽā§āĻŦāϰ āĻŽā§āĻā§ āĻĢā§āϞāϞā§āĨ¤ āϤāĻāύ āĻā§āϰāĻŋāĻ āĻāĻāĻāĻž āĻŽā§āĻāϞā§āĨ¤ āĻāĻŦāĻžāϰ āĻāĻŽ, āĻāĻŦāĻžāϰ āĻā§āϰāĻŋāĨ¤ āĻāĻāĻžāĻŦā§ āĻĒā§āϰāϤā§āϝā§āĻā§ 4āĻāĻŋ āĻāϰ⧠āύāĻŽā§āĻŦāϰ āĻŽā§āĻāϞā§āĨ¤ āĻĻā§āĻāĻž āĻā§āϞ āĻāĻŽā§āϰ āĻŽā§āĻā§ āĻĢā§āϞāĻž āύāĻŽā§āĻŦāϰāĻā§āϞā§āϰ āϝā§āĻāĻĢāϞ āĻā§āϰāĻŋāϰ āĻŽā§āĻā§ āĻĢā§āϞāĻž āύāĻŽā§āĻŦāϰāĻā§āϞā§āϰ āϝā§āĻāĻĢāϞā§āϰ āϤāĻŋāύāĻā§āύāĨ¤ āĻĒā§āϰāĻļā§āύ āĻšāϞ⧠āϏāĻŦ āĻļā§āώ⧠āĻā§āύ āϏāĻāĻā§āϝāĻžāĻāĻŋ āĻŦā§āϞā§āϝāĻžāĻāĻŦā§āϰā§āĻĄā§ āϰāϝāĻŧā§ āĻā§āϞ?
On the blackboard the numbers 1, 3, 4, 6, 8, 9, 11, 12, 16 are written. Tom and Jerry in turn, one after another, are crossing out four numbers each. Tom starts first, so Jerry is the second. It was found that the sum of all 4 numbers crossed out by Tom appeared to be strictly 3 times as big as the sums of the numbers, which were crossed out by Jerry. What number finally remained on the blackboard?
National math olympiad questions 2010 pdf
ā§. āĻĒāĻžāĻāĻ āĻāύ āϞā§āĻ āĻšāϝāĻŧ āύā§āϞ āĻ āĻĨāĻŦāĻž āϞāĻžāϞ āĻā§āĻĒāĻŋ āĻĒāĻĄāĻŧā§āĻā§āĨ¤ āϝāĻĻāĻŋ āĻāĻĻā§āϰ āĻā§āĻ āĻāĻĨāĻž āĻŦāϞ⧠āϤāĻžāĻšāϞ⧠āύā§āϞ āĻā§āĻĒāĻŋāϰ āϞā§āĻā§āϰāĻž āϏāĻŦāϏāĻŽāϝāĻŧ āϏāϤā§āϝ āĻāĻĨāĻž āĻŦāϞ⧠āĻāĻŦāĻ āϞāĻžāϞ āĻā§āĻĒāĻŋāϰ āϞā§āĻā§āϰāĻž āĻā§āĻŦāϞ āĻŽāĻŋāĻĨā§āϝāĻž āĻāĻĨāĻž āĻŦāϞā§āĨ¤ āϤāĻžāĻĻā§āϰ āĻĒā§āϰāϤā§āϝā§āĻā§ āĻ āύā§āϝ āĻāĻžāϰāĻāύā§āϰ āĻā§āĻĒāĻŋ āĻĻā§āĻāϤ⧠āĻĒāĻžāϝāĻŧāĨ¤
A āĻŦāϞāϞ â āĻāĻŽāĻŋ 4 āĻāĻŋ āύā§āϞ āĻā§āĻĒāĻŋ āĻĻā§āĻāĻāĻŋāĨ¤â
B āĻŦāϞāϞ â āĻāĻŽāĻŋ 3 āĻāĻŋ āύā§āϞ āĻā§āĻĒāĻŋ āĻ 1 āĻāĻŋ āϞāĻžāϞ āĻā§āĻĒāĻŋ āĻĻā§āĻāĻāĻŋāĨ¤â
C āĻŦāϞāϞ â āĻāĻŽāĻŋ āĻĻā§āĻāĻāĻŋ 1āĻāĻŋ āύā§āϞ āĻāϰ 3 āĻāĻŋ āϞāĻžāϞ āĻā§āĻĒāĻŋ āĨ¤â
D āĻŦāϞāϞ â āĻāĻŽāĻŋ 4 āĻāĻŋ āϞāĻžāϞ āĻā§āĻĒāĻŋ āĻĻā§āĻāĻāĻŋāĨ¤â
E āĻāĻŋāĻā§āĻ āĻŦāϞāϞ āύāĻžāĨ¤
āĻāϰāĻž āĻā§ āĻā§āύ āϰāĻ āĻāϰ āĻā§āĻĒāĻŋ āĻĒāĻĄāĻŧā§ āĻāĻā§āĨ¤
Five men are wearing either blue or red hats. A man wearing a blue hat always tells the truth. A man wearing a red hat always lies. They look at each other’s hat.
A Says âI see 4 blue hats.â
B Says âI see 3 blue hats and 1 red hat.â
C Says âI see 1 blue hat and 3 red hats.â
D says âI see 4 red hats.â
E Does not say anything.
Find the color of each person’s hat.
ā§Ž. āĻāĻāĻāĻŋ āϏāύāĻžāϤāύ⧠āĻāĻĄāĻŧāĻŋāϤ⧠āĻāĻŖā§āĻāĻžāϰ āĻ āĻŽāĻŋāύāĻŋāĻā§āϰ āĻāĻžāĻāĻāĻž āĻāĻā§āĨ¤ 3 āĻāĻž āĻ 4 āĻāĻžāϰ āĻŽāϧā§āϝ⧠āϝāĻĻāĻŋ āĻŽāĻŋāύāĻŋāĻā§āϰ āĻāĻžāĻāĻāĻž āĻāĻŖā§āĻāĻžāϰ āĻāĻžāĻāĻāĻžāϰ āĻ āĻŋāĻ āĻāĻĒāϰ⧠āĻĨāĻžāĻā§ āϤāĻāύ āϤāĻžāĻšāϞ⧠āϏāĻŽāϝāĻŧ āĻāϤā§?
Consider a traditional clock, with hands that go around the face. If the two hands point to the same spot on the clock, between the 3 and the 4, what time is it?
⧝. āĻāĻŽāύ āĻā§āώā§āĻĻā§āϰāϤāĻŽ āϏāĻāĻā§āϝāĻž āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰ āϝā§āύ āϏā§āĻāĻŋāĻā§ 4, 5, 6 āĻ āĻĨāĻŦāĻž 9 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻ āĻāϰāϞ⧠1 āĻ āĻŦāĻļāĻŋāώā§āĻ āĻĨāĻžāĻā§ āĻāĻŦāĻ āϏāĻāĻā§āϝāĻžāĻāĻŋ 13 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϝāĻŧāĨ¤
Find the smallest number, divisible by 13, such that the remainder is 1 when divided by 4, 5, 6 or 9.

ā§§ā§Ļ. āύāύā§āĻā§ āĻāϰ āĻĢāύā§āĻā§āϰ āĻŽāĻžāĻā§ āĻĒā§āϰāϤāĻŋāϝā§āĻāĻŋāϤāĻž āĻšāĻā§āĻā§, āĻāĻĻā§āϰāĻā§ āĻāϤāĻā§āϞ⧠āϏāĻāĻā§āϝāĻž āĻĻāĻŋāϝāĻŧā§ āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§, āĻāϰāĻž āϏā§āĻāĻžāύ āĻĨā§āĻā§ āĻĻā§āĻāĻŋ āĻāϰ⧠āϏāĻāĻā§āϝāĻž āύā§āĻŦā§ āϝā§āύ āĻ āϏāĻāĻā§āϝāĻž āĻĻā§āĻāĻŋāϰ āϝā§āĻāĻĢāϞ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϝāĻŧāĨ¤ āύāύā§āĻā§āĻā§ 1 āĻĨā§āĻā§ 40 āĻāϰ āĻŽāϧā§āϝ āĻĨā§āĻā§ āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§, āĻĢāύā§āĻā§āĻā§ 1 āĻĨā§āĻā§ 100 āĻāϰ āĻŽāϧā§āϝ⧠āĻŦāĻŋāĻā§āĻĄāĻŧ āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§āĨ¤ āϝ⧠āϝāϤ āĻŦā§āĻļāĻŋ āĻāĻžāĻŦā§ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āϏāĻāĻā§āϝāĻž āĻŦāĻžāύāĻžāϤ⧠āĻĒāĻžāϰāĻŦā§ āϏ⧠āĻāĻŋāϤāĻŦā§āĨ¤ āĻā§ āĻāĻŋāϤāĻŦā§ āĻāĻŦāĻ āĻā§āύ?
Nonte has been given all the numbers from 1 to 40 & Fonte has been given all the odd numbers from 1 to 100. They have to take any two of given numbers so that the summation of those 2 numbers is divisible of 3. He, who will make the numbers divisible by 3 in most ways, will be announced winner. Who will be winner and why?
Junior Category
ā§§. āϤāĻŋāύ āĻ āĻāĻā§āϰ āĻāĻāĻāĻŋ āϏāĻāĻā§āϝāĻž, 1*3 ; āϏāĻāĻā§āϝāĻžāĻāĻŋ 11 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϞ⧠* āĻāĻŋāĻšā§āύāĻŋāϤ āϏā§āĻĨāĻžāύā§āϰ āĻ āĻāĻāĻāĻŋ āĻāϤ⧠āϏā§āĻāĻž āĻĒā§āϰāĻŽāĻžāĻŖ āϏāĻš āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ The three digit number 1 * 3 is divisible by 11. Find, with proof, the missing digit (represented by the asterisk).
⧍. āĻāĻāĻ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻŦāĻŋāĻļāĻŋāώā§āĻ āĻāĻāĻāĻŋ āĻāϝāĻŧāϤāĻā§āώā§āϤā§āϰ āĻāĻŦāĻ āĻāĻāĻāĻŋ āĻŦāϰā§āĻāĻā§āώā§āϤā§āϰā§āϰ āĻŽāϧā§āϝ⧠āĻā§āύāĻāĻŋāϰ āĻĒāϰāĻŋāϏā§āĻŽāĻž āĻŦāĻĄāĻŧ āϏā§āĻāĻŋ āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖ āϏāĻš āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ A rectangle and a square have the same area, find, with proof, which one has a greater perimeter.
ā§Š. āĻāĻŽ āĻŦāϏ⧠āĻāĻāĻĻāĻŋāύ āϏāĻāĻā§āϝāĻž āύāĻŋāϝāĻŧā§ āĻā§āϞāĻāĻŋāϞāĨ¤ āĻāĻŽ 1 āĻĨā§āĻā§ 22 āĻĒāϰā§āϝāύā§āϤ āϏāĻŦ āϏāĻāĻā§āϝāĻžāĻā§ āĻāĻāĻŦāĻžāϰ āĻāϰ⧠āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻāϰ⧠āĻŽā§āĻ 11 āĻāĻŋ āĻāĻā§āύāĻžāĻāĻļ āϞāĻŋāĻā§āĻā§āĨ¤ āĻāĻā§āύāĻžāĻāĻļāĻā§āϞā§āϰ āϞāĻŦ āĻ āĻšāϰ⧠āϝ⧠āĻā§āύ āϏāĻāĻā§āϝāĻž āϏ⧠āĻŦāϏāĻžāϤ⧠āĻĒāĻžāϰā§āĨ¤ āĻāĻā§āϞā§āϰ āĻŽāĻžāĻā§ āϏāϰā§āĻŦā§āĻā§āĻ āĻāϤāĻā§āϞ⧠āĻĒā§āϰā§āĻŖāϏāĻāĻā§āϝāĻž āĻšāϤ⧠āĻĒāĻžāϰ⧠? One day Tom was playing with numbers. He wrote 11 fractions using all natural numbers from 1 to 22 exactly once – either as numerator or as denominator. How many of these fractions, at most, are integers?
ā§Ē. āĻāĻŽāύ āĻā§āώā§āĻĻā§āϰāϤāĻŽ āϏāĻāĻā§āϝāĻž āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰ āϝā§āύ āϏā§āĻāĻŋāĻā§ 4, 6 āĻ āĻĨāĻŦāĻž 9 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻ āĻāϰāϞ⧠1 āĻ āĻŦāĻļāĻŋāώā§āĻ āĻĨāĻžāĻā§ āĻāĻŦāĻ āϏāĻāĻā§āϝāĻžāĻāĻŋ 13 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϝāĻŧāĨ¤ Find the smallest number, divisible by 13, such that the remainder is 1 when divided by 4, 6 or 9.
ā§Ģ. (m, n) āĻāϰ āϧāύāĻžāϤā§āĻŽāĻ āĻĒā§āϰā§āĻŖ āϏāĻāĻā§āϝāĻžāϰ āĻā§āύ āĻā§āύ āĻŽāĻžāύā§āϰ āĻāύā§āϝ \[M^3 + 1331 = n^3 \] āϏāĻŽā§āĻāϰāĻŖāĻāĻŋ āĻļā§āĻĻā§āϧāĨ¤ Find all pairs of positive integers (m, n) which satisfy \[M^3 + 1331 = n^3 \]
ā§Ŧ. āύāύā§āĻā§ āĻāϰ āĻĢāύā§āĻā§āϰ āĻŽāĻžāĻā§ āĻĒā§āϰāϤāĻŋāϝā§āĻāĻŋāϤāĻž āĻšāĻā§āĻā§, āĻāĻĻā§āϰāĻā§ āĻāϤāĻā§āϞ⧠āϏāĻāĻā§āϝāĻž āĻĻāĻŋāϝāĻŧā§ āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§, āĻāϰāĻž āϏā§āĻāĻžāύ āĻĨā§āĻā§ āĻĻā§āĻāĻŋ āĻāϰ⧠āϏāĻāĻā§āϝāĻž āύā§āĻŦā§ āϝā§āύ āĻ āϏāĻāĻā§āϝāĻž āĻĻā§āĻāĻŋāϰ āϝā§āĻāĻĢāϞ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϝāĻŧāĨ¤ āύāύā§āĻā§āĻā§ 1 āĻĨā§āĻā§ 40 āĻāϰ āĻŽāϧā§āϝ āĻĨā§āĻā§ āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§, āĻĢāύā§āĻā§āĻā§ 1 āĻĨā§āĻā§ 100 āĻāϰ āĻŽāϧā§āϝ⧠āĻŦāĻŋāĻā§āĻĄāĻŧ āϏāĻāĻā§āϝāĻžāĻā§āϞ⧠āĻĻā§āĻāϝāĻŧāĻž āĻšāϝāĻŧā§āĻā§āĨ¤ āϝ⧠āϝāϤ āĻŦā§āĻļāĻŋ āĻāĻžāĻŦā§ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āϏāĻāĻā§āϝāĻž āĻŦāĻžāύāĻžāϤ⧠āĻĒāĻžāϰāĻŦā§ āϏ⧠āĻāĻŋāϤāĻŦā§āĨ¤ āĻā§ āĻāĻŋāϤāĻŦā§ āĻāĻŦāĻ āĻā§āύ ?
Nonte has been given all the numbers from 1 to 40 & Fonte has been given all the odd numbers from 1 to 100. They have to take any two of given numbers so that the summation of those 2 numbers is divisible of 3. He, who will make the numbers divisible by 3 in most ways, will be announced winner. Who will be winner and why?
7. āĻāĻāĻāĻŋ āϤāϞ⧠25 āĻāĻŋ āĻŦāĻŋāύā§āĻĻā§ āĻāĻā§, āĻāϰ āĻŽāϧā§āϝ⧠āĻāĻāĻ āϰā§āĻāĻžāϝāĻŧ āϤāĻŋāύāĻāĻŋ āĻŦāĻŋāύā§āĻĻā§āĻ āĻāĻāĻ āϏāϰāϞāϰā§āĻāĻžāϝāĻŧ āύā§āĻāĨ¤ āϏāĻŦ āĻŦāĻŋāύā§āĻĻā§āĻā§ āĻāϞāĻžāĻĻāĻž āĻāϰ⧠āϰāĻžāĻāϤ⧠āĻšāϞ⧠āĻāĻŽāĻĒāĻā§āώ⧠āĻāϝāĻŧāĻāĻŋ āϰā§āĻāĻž āϞāĻžāĻāĻŦā§?
There are 25 points on a plane, no three of which lie on a line. Find the minimum number of lines needed to separate them from one another.
ā§Ž. āĻā§āϰāĻžāĻĒāĻŋāĻāĻŋāϝāĻŧāĻžāĻŽ āĻšāϞ⧠āϏā§āĻ āĻāϤā§āϰā§āĻā§āĻ āϝāĻžāϰ āĻĻā§āĻāĻāĻŋ āĻŦāĻŋāĻĒāϰā§āϤ āĻŦāĻžāĻšā§ āĻĒāϰāϏā§āĻĒāϰ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āĻāĻŋāύā§āϤ⧠āĻ āύā§āϝ āĻĻā§āĻāĻāĻŋ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āύāϝāĻŧāĨ¤ āĻāĻāĻāĻŋ āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āĻā§āϰāĻžāĻĒāĻŋāĻāĻŋāϝāĻŧāĻŽā§āϰ āĻāϰā§āĻŖ āĻāĻāĻŋāĻā§ āĻĻā§āĻāĻāĻŋ āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āϤā§āϰāĻŋāĻā§āĻā§ āĻāĻžāĻ āĻāϰā§āĻā§āĨ¤ āĻā§āϰāĻžāĻĒāĻŋāĻāĻŋāϝāĻŧāĻžāĻŽā§āϰ āĻā§āĻŖāĻā§āϞā§āϰ āĻŽāĻžāύ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰā§āĨ¤
Trapezium is any quadrilateral two opposite sides of which are parallel and another two are not. The diagonal of isosceles trapezium divides it into tow isosceles triangle. Find the angles of the trapezium?
9. āĻāĻŽ āĻ āĻā§āϰā§āϰ āĻāĻžāĻā§ āĻŽā§āĻ 14 āĻāĻŋ āĻāĻžāĻāϞāϏ āĻāĻā§āĨ¤ āĻāϰ āĻŽāϧā§āϝ⧠ā§Ē āĻāĻŋ āύā§āϞ āĻ 6 āĻāĻŋ āϞāĻžāϞāĨ¤ āϤāĻžāϰāĻž āĻāĻā§āϞ⧠āĻāĻ āϞāĻžāĻāύ⧠āĻāĻŽāύāĻāĻžāĻŦā§ āϏāĻžāĻāĻžāϤ⧠āĻāĻžāϝāĻŧ āϝā§, āĻĒā§āϰāϤāĻŋ āĻĻā§āĻāĻāĻŋ āϞāĻžāϞ āĻāĻžāĻāϞāϏā§āϰ āĻŽāĻžāĻāĻāĻžāύ⧠āĻāĻŽāĻĒāĻā§āώ⧠āĻāĻāĻāĻŋ āύā§āϞ āĻāĻžāĻāϞāϏ āĻĨāĻžāĻāĻŦā§āĨ¤ āϏāĻŽā§āĻāĻžāĻŦā§āϝ āĻāϤāĻāĻžāĻŦā§ āĻāĻŽ āĻ āĻā§āϰ⧠āĻāĻ āĻāĻžāĻ āĻāϰāϤ⧠āĻĒāĻžāϰāĻŦā§āĨ¤
Tom and Jerry have 14 tiles in total. Of them 8 are colored blue and 6 are colored red. They want to arrange them in a straight line such that between any two red tiles there is at least one blue tile. How many possible ways are there of arranging them in this line?
ā§§ā§Ļ. ABCD āĻāĻāĻāĻŋ āĻāϤā§āϰā§āĻā§āĻāĨ¤ āĻāϰ AB, BC āĻ CD āĻŦāĻžāĻšā§āϰ āĻŽāϧā§āϝāĻŦāĻŋāύā§āĻĻā§ āϝāĻĨāĻžāĻā§āϰāĻŽā§ P, Q āĻ R | āϝāĻĻāĻŋ PQ = 3, QR= 4 āĻāĻŦāĻ PR = 5 āĻšāϝāĻŧ āϤāĻŦā§ ABCD āĻāϤā§āϰā§āĻā§āĻā§āϰ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻŦā§āϰ āĻāϰā§āĨ¤
ABCD is a quadrilateral. P, Q and R are the midpoints of AB, BC and CD respectively. If PQ = 3, QR = 4 and PR = 5; find the area of ABCD.
National math olympiad questions 2010 pdf download
Secondary Category
ā§§. āĻāĻāĻāĻŋ āĻĒāĻžāϰā§āĻāĻŋāϤ⧠āϤā§āĻŽāĻŋ āĻāĻžāĻĄāĻŧāĻžāĻ āĻāϰ⧠20 āĻāύ āϞā§āĻ āϰāϝāĻŧā§āĻā§āĨ¤ āĻ āĻĒāĻžāϰā§āĻāĻŋāϤ⧠āϤā§āĻŽāĻŋ āϝāϤāĻāύāĻā§ āĻā§āύ, āĻāĻŦāĻžāϰ āĻ āĻŋāĻ āϤāϤāĻāύāĻā§āĻ āĻā§āύāύāĻžāĨ¤ āϤā§āĻŽāĻŋ āĻāϤāĻāύāĻā§ āĻā§āύ?
There are 20 people in a party excluding you. It is known that you know the same number of people as you don’t know. How many of them do you know?
⧍. ABC āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āϤā§āϰāĻŋāĻā§āĻā§āϰ B āĻā§āĻŖāĻāĻŋ āϏāĻŽāĻā§āĻŖ āĻāĻŦāĻ AB = 3āĨ¤ āĻāϰ āϝā§āĻā§āύ āĻāĻāĻāĻŋ āĻļā§āϰā§āώāĻā§ āĻā§āύā§āĻĻā§āϰ āĻāϰ⧠āĻāĻāĻāĻŋ āĻāĻāĻ āĻŦā§āϝāĻžāϏāĻžāϰā§āϧā§āϰ āĻŦā§āϤā§āϤ āĻāĻāĻāĻž āĻšāϞāĨ¤ āϤā§āϰāĻŋāĻā§āĻā§āϰ āϝ⧠āĻ āĻāĻļ āĻŦā§āϤā§āϤā§āϰ āĻ āύā§āϤāϰā§āĻā§āĻā§āϤ āύāϝāĻŧ āϤāĻžāϰ āϏāϰā§āĻŦā§āĻā§āĻ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻŦā§āϰ āĻāϰāĨ¤
 Isosceles triangle ABC is right angled at B and AB = 3. A circle of unit radius is drawn with its centre on any of the vertices of this triangle. Find the maximum value of the area of that part of the triangle that is not shared by the circle.
ā§Š. āĻŦāĻžāϏā§āϤāĻŦ āϏāĻāĻā§āϝāĻžāϝāĻŧ āϏāĻŽāĻžāϧāĻžāύ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĻ \[\frac{|x^2 â 1|}{x â 2}= x\]
 [āϏāĻžāĻšāĻžāϝā§āϝāĻ ax2 + bx + c = 0 āϏāĻŽā§āĻāϰāĻŖā§āϰ āϏāĻŽāĻžāϧāĻžāύ āĻšāϞ x = \[\frac{ â b Âą \sqrt{b^2 – â 4ac} }{2a}\]
Solve for real x: \[\frac{|x^2 â 1|}{x â 2}= x\]
 [Hint: The solutions to the equation ax2 + bx + c = 0 are x = \[\frac{ â b Âą \sqrt{b^2 – â 4ac} }{2a}\]
ā§Ē. āĻāĻāĻāĻŋ āϧāĻžāϰāĻžāĻā§ āύāĻŋāĻā§āϰ āϏāĻāĻā§āĻāĻž āĻ āύā§āϝāĻžāϝāĻŧā§ āĻāĻ āύ āĻāϰāĻž āĻšāϞāĻ
A(1) = 1
A(n) = f(m) āϏāĻāĻā§āϝāĻ f(m) āĻāĻŦāĻ āϤāĻžāϰ āĻĒāϰ f(m) āϏāĻāĻā§āϝāĻ 0; āĻāĻāĻžāύ⧠111 āĻšāĻā§āĻā§ A(n-1) āĻāϰ āĻ āĻāĻ āϏāĻāĻā§āϝāĻžāĨ¤
A(30) āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ āĻāϞā§āϞā§āĻā§āϝ, 11 āĻā§ 9 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻ āĻāϰāϞ⧠āϝ⧠āĻāĻžāĻāĻļā§āώ āĻĨāĻžāĻā§ āϏā§āĻāĻŋāĻ f(m)āĨ¤
A series is formed in the following manner:
A(1) = 1;
A(n) = f(m) numbers of f(m) followed by f(m) numbers of o; m is the number of digits in A(n-1)
Find A(30). Here f(m) is the remainder when m is divided by 9.
ā§Ģ. ABC āϤā§āϰāĻŋāĻā§āĻā§āϰ B āĻā§āĻŖāĻāĻŋ āϏāĻŽāĻā§āĻŖāĨ¤ â BAC āĻāϰ āϏāĻŽāĻĻā§āĻŦāĻŋāĻāĻŖā§āĻĄāĻ BC āĻā§ D āĻŦāĻŋāύā§āĻĻā§āϤ⧠āĻā§āĻĻ āĻāϰā§āĨ¤ G āĻāĻ āϤā§āϰāĻŋāĻā§āĻā§āϰ āĻŽāϧā§āϝāĻŽāĻžāϤā§āϰāϝāĻŧā§āϰ āĻā§āĻĻāĻŦāĻŋāύā§āĻĻā§ā§ˇ GD|| AB āĻšāϞ⧠â C āĻāϰ āĻŽāĻžāύ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤
Triangle ABC is right angled at B. The bisector of â BAC meets BC at D. Let G denote the centroid (common point of the medians) of the triangle ABC. Suppose that GD is parallel to AB. Find â C.
ā§Ŧ. āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖāϏāĻš āĻāĻŽāύ āϏāĻāϞ āĻĒā§āϰā§āĻŖāĻŦāϰā§āĻ āϏāĻāĻā§āϝāĻž āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰ āϝāĻžāϰāĻž āĻāĻžāϰāĻāĻŋ āĻā§āϰāĻŽāĻŋāĻ āϧāύāĻžāϤā§āĻŽāĻ āĻŦā§āĻā§āĻĄāĻŧ āϏāĻāĻā§āϝāĻžāϰ āĻā§āĻŖāĻĢāϞāĨ¤
Find, with proof, all the perfect squares each of which is the product of four consecutive odd natural numbers.
ā§. āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖāϏāĻš āĻāϝāĻŧ āĻ āĻāĻā§āϰ 13**45* āϏāĻāĻā§āϝāĻžāĻāĻŋāϤ⧠āĻĒā§āϰāϤā§āϝā§āĻāĻāĻŋ āϤāĻžāϰāĻāĻžāĻāĻŋāĻšā§āύāĻā§ (*) āĻāĻŋāύā§āύ āĻāĻŋāύā§āύ āĻ āĻāĻ āĻĻā§āĻŦāĻžāϰāĻž āĻĒā§āϰāϤāĻŋāϏā§āĻĨāĻžāĻĒāĻŋāϤ āĻāϰ āϝā§āύ āϏāĻāĻā§āϝāĻžāĻāĻŋ 792 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ āĻšāϝāĻŧāĨ¤
Replace each asterisk with proof in the six digit number 13**45* by different digits such that the resulting number is divisible by 792.
ā§Ž. ABC āϤā§āϰāĻŋāĻā§āĻā§āϰ A āĻā§āĻŖāĻāĻŋ āϏāĻŽāĻā§āĻŖāĨ¤ BC āĻāϰ āĻāĻĒāϰ D āĻāĻāĻāĻŋ āĻŦāĻŋāύā§āĻĻā§āĨ¤ AC āĻāĻŦāĻ AB āĻāϰ āϏāĻžāĻĒā§āĻā§āώ⧠D āĻāϰ āĻĒā§āϰāϤāĻŋāĻĢāϞāύ āϝāĻĨāĻžāĻā§āϰāĻŽā§ E āĻāĻŦāĻ F āĨ¤ āĻĻā§āĻāĻžāĻ āϝā§, [ABC] âĨ [DEF] āĻāĻŦāĻ āϏāĻŽāϤāĻž āĻšāĻŦāĻžāϰ āĻāύā§āϝ D āĻāϰ āϏāĻāϞ āϏāĻŽā§āĻāĻžāĻŦā§āϝ āĻ āĻŦāϏā§āĻĨāĻžāύ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ ([x] āĻĻā§āĻŦāĻžāϰāĻž x āĻāϰ āĻā§āώā§āϤā§āϰāĻĢāϞ āύāĻŋāϰā§āĻĻā§āĻļ āĻāϰāĻž āĻšāĻā§āĻā§)
Triangle ABC is right angled at A. Let D be a point on BC. E and F are reflections of D on AC and AB respectively. Prove that [ABC] âĨ [DEF]. Find all possible positions of D for equality. (Here [x] denotes the area of x)

⧝. āĻāĻŽāύ āϏāĻāϞ āĻŽā§āϞāĻŋāĻ āϏāĻāĻā§āϝāĻž p āĻāĻŦāĻ āϧāύāĻžāϤā§āĻŽāĻ āĻĒā§āϰā§āĻŖ āϏāĻāĻā§āϝāĻž a, b āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰ āϝā§āύ \[ p^a + p^b\] āĻāĻāĻāĻŋ āĻĒā§āϰā§āĻŖ āĻŦāϰā§āĻ āĻšāϝāĻŧāĨ¤
Find all the prime numbers p and positive integers a and b such that \[ p^a + p^b\] is the square of an integer.
ā§§ā§Ļ. 131 āĻāĻŋ āϧāύāĻžāϤā§āĻŽāĻ āĻĒā§āϰā§āĻŖāϏāĻāĻā§āϝāĻžāϰ āĻāĻāĻāĻŋ āϏā§āĻā§ āϝā§āĻā§āύ āϏāĻāĻā§āϝāĻžāϰ āĻŽā§āϞāĻŋāĻ āĻā§āĻĒāĻžāĻĻāĻāĻā§āϞ⧠42 āĻāϰ āĻā§āϝāĻŧā§ āĻā§āĻāĨ¤ āĻĻā§āĻāĻžāĻ āϝā§, āĻāĻ āϏā§āĻ āĻĨā§āĻā§ āĻāĻŽāύ āĻāĻžāϰāĻāĻŋ āϏāĻāĻā§āϝāĻž āύāĻŋāϰā§āĻŦāĻžāĻāύ āĻāϰāĻž āϝāĻžāĻŦā§ āϝā§āύ āϤāĻžāĻĻā§āϰ āĻā§āĻŖāĻĢāϞ āĻāĻāĻāĻŋ āĻĒā§āϰā§āĻŖ āĻŦā§°ā§āĻ āĻšāϝāĻŧāĨ¤
In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
Higher Secondary
ā§§. āϝāĻĻāĻŋ S = \[ 1^1 + 2^2 + 3^3 + ….. + 2010^{2010} \] āĻšāϝāĻŧ āϤāĻžāĻšāϞ⧠S āĻā§ 2 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻ āĻāϰāϞ⧠āĻāĻžāĻāĻļā§āώ āĻāϤ āĻšāĻŦā§ āϤāĻž āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ Let S = \[ 1^1 + 2^2 + 3^3 + ….. + 2010^{2010} \]. What is the remainder when S is divided by 2?
⧍. ABC āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āϤā§āϰāĻŋāĻā§āĻā§āϰ B āĻā§āĻŖāĻāĻŋ āϏāĻŽāĻā§āĻŖ āĻāĻŦāĻ AB = 3āĨ¤ āĻāϰ āϝā§āĻā§āύ āĻāĻāĻāĻŋ āĻļā§āϰā§āώāĻā§ āĻā§āύā§āĻĻā§āϰ āĻāϰ⧠āĻāĻāĻāĻŋ āĻāĻāĻ āĻŦā§āϝāĻžāϏāĻžāϰā§āϧā§āϰ āĻŦā§āϤā§āϤ āĻāĻāĻāĻž āĻšāϞāĨ¤ āϤā§āϰāĻŋāĻā§āĻā§āϰ āϝ⧠āĻ āĻāĻļ āĻŦā§āϤā§āϤā§āϰ āĻ āύā§āϤāϰā§āĻā§āĻā§āϤ āύāϝāĻŧ āϤāĻžāϰ āϏāϰā§āĻŦā§āĻā§āĻ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻŦā§āϰ āĻāϰāĨ¤
Isosceles triangle ABC is right angled at B and AB = 3. A circle of unit radius is drawn with its centre on any of the vertices of this triangle. Find the maximum value of the area of that part of the triangle that is not shared by the circle.
ā§Š. āĻāĻāĻāĻŋ āϧāĻžāϰāĻžāĻā§ āύāĻŋāĻā§āϰ āϏāĻāĻā§āĻāĻž āĻ āύā§āϝāĻžāϝāĻŧā§ āĻāĻ āύ āĻāϰāĻž āĻšāϞāĻ
A(1) = 1
A(n) = f(m) āϏāĻāĻā§āϝāĻ f(m) āĻāĻŦāĻ āϤāĻžāϰ āĻĒāϰ f(m) āϏāĻāĻā§āϝāĻ ā§Ļ;
āĻāĻāĻžāύ⧠111 āĻšāĻā§āĻā§ A(n-1) āĻāϰ āĻ āĻāĻ āϏāĻāĻā§āϝāĻžāĨ¤
A(30) āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤ āĻāϞā§āϞā§āĻā§āϝ 1 āĻā§ 9 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻ āĻāϰāϞ⧠āϝ⧠āĻāĻžāĻāĻļā§āώ āĻĨāĻžāĻā§ āϏā§āĻāĻŋāĻ f(m)āĨ¤
A series is formed in the following manner:
A(1) = 1;
A(n) = f(m) numbers of f(m) followed by f(m) numbers of o; m is the number of digits in A (n-1)
Find A(30). Here f(m) is the remainder when m is divided by 9.
ā§Ē. āĻāĻāĻāĻŋ āĻŦā§āϤā§āϤ  āĻāϰ āĻā§āϤāϰ⧠P āĻāĻāĻāĻŋ āĻŦāĻŋāύā§āĻĻā§ āĨ¤ āĻāĻ āĻŦāĻŋāύā§āĻĻā§ āĻĻāĻŋāϝāĻŧā§ āĻĻā§āĻāĻŋ āĻā§āϝāĻž āĻāĻāĻāĻž āĻšāϞ āϝāĻžāϰāĻž āĻĒāϰāϏā§āĻĒāϰā§āϰ āĻāĻĒāϰ āϞāĻŽā§āĻŦāĨ¤ āĻāĻ āĻā§āϝāĻž āĻĻā§āĻāĻŋ āĻŦā§āϤā§āϤāĻā§ āĻāĻĄāĻŧāĻŋāϰ āĻāĻžāĻāĻžāϰ āĻĻāĻŋāĻā§ a; b; c; d āĻāĻ āĻāĻžāϰāĻāĻŋ āĻāĻžāĻā§ āĻāĻžāĻ āĻāϰ⧠āϝāĻžāϰ āĻŽāϧā§āϝ⧠a āĻ āĻāĻļāĻāĻŋ āĻŦā§āϤā§āϤā§āϰ āĻā§āύā§āĻĻā§āϰāĻā§ āϧāĻžāϰāĻŖ āĻāϰā§āĨ¤ āĻĻā§āĻāĻžāĻ āϝā§, [a] + [c] âĨ [b] + [d], āĻāĻāĻžāύ⧠[x] āĻĻā§āĻŦāĻžāϰāĻž x āĻāϰ āĻā§āώā§āϤā§āϰāĻĢāϞ āύāĻŋāϰā§āĻĻā§āĻļ āĻāϰā§āĨ¤
Given a point P inside a circle Đŗ, two perpendicular chords through P divide à into distinct regions a; b; e; d clockwise such that a contains the centre of T. Prove that [a] + [c] 2 [b] + [d], where [x] = area of x.
ā§Ģ. āĻāĻāĻāĻŋ 2010 āĻŦāĻžāĻšā§ āĻŦāĻŋāĻļāĻŋāώā§āĻ āϏā§āώāĻŽ āĻŦāĻšā§āĻā§āĻā§āϰ āĻļā§āϰā§āώāĻā§āϞ⧠āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻāϰ⧠āĻāϤāĻā§āϞ⧠āϏā§āώāĻŽ āĻŦāĻšā§āĻā§āĻ āĻāĻāĻāĻž āϏāĻŽā§āĻāĻŦ? (2010 āĻŦāĻžāĻšā§ āĻŦāĻŋāĻļāĻŋāώā§āĻ āϏā§āώāĻŽ āĻŦāĻšā§āĻā§āĻā§āϰ āĻļā§āϰā§āώāĻŦāĻŋāύā§āĻĻā§āĻā§āϞā§āϰ āĻā§āϰāĻŽ āĻŽā§āĻā§āϝ āύāϝāĻŧāĨ¤)
How many regular polygons can be constructed from the vertices of a regular polygon with 2010 sides? (Assume that the vertices of the 2010-gon are indistinguishable)
ā§Ŧ. a āĻāĻŦāĻ b āĻĻā§āĻāĻŋ āĻĒā§āϰā§āĻŖ āϏāĻāĻā§āϝāĻž āϝā§āύ 1 ⤠a, b < 2010 āĻāĻŦāĻ a + b; āĻāĻŽāύ āĻāϤāĻā§āϞ⧠āĻā§āϰāĻŽāĻā§āĻĄāĻŧ (a, b) āĻāĻā§ āϝā§āĻāĻžāύ⧠\[ a^2 + b^2 \], 5 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻāĻžāĻā§āϝ? a + b āĻāϰ āĻŽāĻžāύ āĻā§āύ āĻŽāĻžāύā§āϰ āĻāύā§āϝ \[ a^2 + b^2 \] āϏāĻŦāĻā§āϝāĻŧā§ āĻŦāĻĄāĻŧ āĻšāĻŦā§?
a and b are two positive integers both less than 2010; a b. Find the number of ordered pairs (a, b) such that \[ a^2 + b^2 \] is divisible by 5. Find a + b so that \[ a^2 + b^2 \] is maximum.
ā§. ABC āϤā§āϰāĻŋāĻā§āĻā§āϰ AC > AB āĨ¤ āĻāϰ BC āĻŦāĻžāĻšā§āϰ āϞāĻŽā§āĻŦāϏāĻŽāĻĻā§āĻŦāĻŋāĻāĻŖā§āĻĄāĻ āĻāĻŦāĻ CAB āĻā§āĻŖ āĻāϰ āĻ āύā§āϤāϏāĻŽāĻĻā§āĻŦāĻŋāĻāĻŖā§āĻĄāĻ P āĻŦāĻŋāύā§āĻĻā§āϤ⧠āĻā§āĻĻ āĻāϰā§āĨ¤ P āĻĨā§āĻā§ AB āĻāĻŦāĻ AC āĻŦāĻžāĻšā§āϰ āĻāĻĒāϰ⧠āĻ āĻā§āĻāĻŋāϤ āϞāĻŽā§āĻŦāĻĻā§āĻŦāϝāĻŧ āĻŦāĻžāĻšā§āĻĻā§āĻāĻŋāĻā§ āϝāĻĨāĻžāĻā§āϰāĻŽā§ X āĻāĻŦāĻ Y āĻŦāĻŋāύā§āĻĻā§āϤ⧠āĻā§āĻĻ āĻāϰā§āĨ¤ XY āĻāĻŦāĻ BC āĻāϰ āĻā§āĻĻāĻŦāĻŋāύā§āĻĻā§ Z, \[ \frac{BZ}{ZC} \] āĻāϰ āĻŽāĻžāύ āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤
Let ABC be a triangle with AC > AB: Let P be the intersection point of the perpendicular bisector of BC and the internal angle bisector of angle CAB: Let X and Y be the feet of the perpendiculars from P to lines AB and AC; respectively. Let Z be the intersection point of BZ lines XY and BC: Determine the value of \[ \frac{BZ}{ZC} \].
ā§Ž. āĻāĻŽāύ āϏāĻāϞ āĻŽā§āϞāĻŋāĻ āϏāĻāĻā§āϝāĻž p āĻāĻŦāĻ āĻĒā§āϰā§āĻŖ āϏāĻāĻā§āϝāĻž a, b (āĻāĻŖāĻžāϤā§āĻŽāĻāĻ āĻšāϤ⧠āĻĒāĻžāϰā§) āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰ āϝā§āύ \[ p^a + p^b \] āĻāĻāĻāĻŋ āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻžāϰ āĻŦāϰā§āĻ āĻšāϝāĻŧāĨ¤
Find all prime numbers p and integers a and b (not necessarily positive) such that \[ p^a + p^b \] is the square of a rational number.
Math Olympiad questions for Class 10 in bangladesh
⧝. \[ (x + y)^{2010} \]  āĻāϰ āĻŦāĻŋāϏā§āϤā§āϤāĻŋāϤ⧠āĻŦāĻŋāĻā§āĻĄāĻŧ āϏāĻšāĻā§āϰ āϏāĻāĻā§āϝāĻž āύāĻŋāϰā§āĻŖāϝāĻŧ āĻāϰāĨ¤
Find the number of odd coefficients in expansion of \[ (x + y)^{2010} \].
10. \[a_1,a_2,…,a_k,…,a_n \] āϧāύāĻžāϤā§āĻŽāĻ āĻŦāĻžāϏā§āϤāĻŦ āϏāĻāĻā§āϝāĻžāϰ āĻāĻŽāύ āĻāĻāĻāĻŋ āϧāĻžāϰāĻž āϝāĻžāϰ āĻā§āύ āĻĻā§āĻāĻŋ āĻĒāĻĻāĻ āϏāĻŽāĻžāύ āύāϝāĻŧ āĻāĻŦāĻ \[a_1<a_2<…<a_k \] āĻāĻŦāĻ \[ a_k>a_{k+1}>…>a_n \] āĨ¤ āĻāĻāĻāĻŋ āĻāĻžāϏāĻĢāĻĄāĻŧāĻŋāĻ āĻŽā§āϞāĻŦāĻŋāύā§āĻĻā§ O āĻĨā§āĻā§ āĻŦāĻžāϏā§āϤāĻŦ āĻ āĻā§āώ āĻŦāϰāĻžāĻŦāϰ āĻĄāĻžāύāĻĻāĻŋāĻ āĻŦāϰāĻžāĻŦāϰ āϝāĻĨāĻžāĻā§āϰāĻŽā§ \[a_1,a_2,…,a_n \] āĻĻā§āϰāϤā§āĻŦāĻā§āϞ⧠āĻ āϤāĻŋāĻā§āϰāĻŽ āĻāϰ⧠āϞāĻžāĻĢāĻŋāϝāĻŧā§ āϞāĻžāĻĢāĻŋāϝāĻŧā§ āĻāϞā§āĨ¤ āĻĒā§āϰāĻŽāĻžāĻŖ āĻāϰ āϝā§, āϏāĻŦāĻžāϰ āĻĄāĻžāύ⧠āĻĒā§āĻāĻā§ āĻāĻŋāϝāĻŧā§ āϏ⧠a1, 2, âĸâĸâĸ, āĻŦā§ āϧāĻžāϰāĻžāϰ āĻĒāĻĻāĻā§āϞā§āĻā§ āĻā§āύ āĻāĻ āĻā§āϰāĻŽā§ āĻ āύā§āϏāϰāĻŖ āĻāϰ⧠āĻŦāĻžāĻŽāĻĻāĻŋāĻ āĻŦāϰāĻžāĻŦāϰ āϞāĻžāĻĢāĻŋāϝāĻŧā§ āϞāĻžāĻĢāĻŋāϝāĻŧā§ āϏā§āĻāĻžāύ āĻĨā§āĻā§ āĻŽā§āϞāĻŦāĻŋāύā§āĻĻā§āϤ⧠āĻĢāĻŋāϰ⧠āĻāϏāϤ⧠āĻĒāĻžāϰ⧠āϝā§āύ āϝāĻžāĻŦāĻžāϰ āĻĒāĻĨā§ āϏ⧠āϝ⧠āĻŦāĻŋāύā§āĻĻā§āĻā§āϞā§āϤ⧠āύā§āĻŽā§āĻāĻŋāϞ āĻāϏāĻžāϰ āϏāĻŽāϝāĻŧ āϏā§āĻā§āϞā§āϰ āĻā§āύāĻāĻŋāϤā§āĻ āύāĻž āύā§āĻŽā§ āĻĨāĻžāĻā§ (āĻŽā§āϞāĻŦāĻŋāύā§āĻĻā§ āĻāĻŦāĻ āϏāϰā§āĻŦāĻĄāĻžāύā§āϰ āĻŦāĻŋāύā§āĻĻā§ āĻšāϞ āĻŦā§āϝāϤāĻŋāĻā§āϰāĻŽ) āĨ¤
\[a_1,a_2,…,a_k,…,a_n \] is a sequence of distinct positive real numbers such that \[a_1<a_2<…<a_k \] and \[ a_k>a_{k+1}>…>a_n \]. A Grasshopper is to jump along the real axis, startin g at the point O and making n jumps to the right of lengths \[a_1,a_2,…,a_n \] respectively. Prove that, once he reaches the rightmost point, he can come back to point O by making n jumps to the left of lengths \[a_1,a_2,…,a_n \] in some order such that he never lands on a point which he already visited while jumping to the right. (The only exceptions are point O and the rightmost point)

