National math Olympiad questions 2010

 

ā§§. āύāĻŋāĻšā§‡āϰ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋āϰ āĻŽāĻ§ā§āϝ āĻĨ⧇āϕ⧇ āĻŽā§ŒāϞāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž āĻāĻŦāĻ‚ āϝ⧌āĻ—āĻŋāĻ• āĻŦāĻž āĻ•ā§ƒāĻ¤ā§āϰāĻŋāĻŽ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻĒ⧃āĻĨāĻ• āĻ•āϰāĨ¤

111, 239, 379, 455

Determine which of the following are (is) prime and which are (is) composite 111, 239, 379, 455

⧍. āϕ⧋āύ āĻ¸ā§āĻĨāĻžāύ⧇āϰ 6 āĻŽāĻžāϏ⧇āϰ āĻŽāĻžāϏāĻŋāĻ• āĻ—āĻĄāĻŧ āĻŦ⧃āĻˇā§āϟāĻŋāĻĒāĻžāϤ⧇āϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖ 28.5 āĻŽāĻŋāϞāĻŋāĻŽāĻŋāϟāĻžāϰāĨ¤ āϝāĻĻāĻŋ āϏ⧇ āϏāĻŽāϝāĻŧ⧇ āφāϰ⧋ 36 āĻŽāĻŋāϞāĻŋāĻŽāĻŋāϟāĻžāϰ āĻŦ⧇āĻļāĻŋ āĻŦ⧃āĻˇā§āϟāĻŋ āĻšāϤ⧋ āϤāĻžāĻšāϞ⧇ āĻŽāĻžāϏāĻŋāĻ• āĻ—āĻĄāĻŧ āĻŦ⧃āĻˇā§āϟāĻŋāĻĒāĻžāϤ⧇āϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻ•āϤ āĻšāϤ⧋? āĻŽāĻžāϏāĻŋāĻ• āĻ—āĻĄāĻŧ āĻ•āϤ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒ⧇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 āĻāĻ•āϟāĻŋ āϏāĻŽāϕ⧋āĻŖā§€ āĻ¤ā§āϰāĻŋāϭ⧁āϜāĨ¤ āĻāχ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ āϜāĻ¨ā§āϝ AC × AC = AB × AB + BC × BC, āϝāĻĻāĻŋ \[\frac{AB}{AC} = \frac35, \frac{BC}{AC} = \frac45 \] āĻāĻŦāĻ‚ AB, BC, AC āĻāϰ āĻ—. āϏāĻž. āϗ⧁. 1 āĻšāϝāĻŧ āϤāĻžāĻšāϞ⧇ AC āĻāϰ āĻŽāĻžāύ āĻ•āϤ?

ABC is a right angled triangle. For this triangle, AC × AC = AB × AB + BC× BC . If \[\frac{AB}{AC} = \frac35, \frac{BC}{AC} = \frac45 \] and the ged of AB, BC, AC is 1 then find AC.

ā§Ē. āĻœā§‡āϰāĻŋ āĻ āĻŋāĻ• āĻ•āϰāϞ⧋ āϏ⧇ āĻāĻ•āϟāĻŋ 10.1 āĻŽāĻŋāϟāĻžāϰ āϞāĻŽā§āĻŦāĻž āĻāĻŦāĻ‚ 4.2 āĻŽāĻŋāϟāĻžāϰ āϚāĻ“āĻĄāĻŧāĻž āĻāĻ•āϟāĻŋ āϏāĻŦāϜāĻŋ āĻŦāĻžāĻ—āĻžāύ āĻ•āϰāĻŦ⧇āĨ¤ āĻ•āĻŋāĻ¨ā§āϤ⧁, āϟāĻŽā§‡āϰ āφāĻ—āĻŽāύ āϠ⧇āĻ•āĻžāύ⧋āϰ āϜāĻ¨ā§āϝ āϤāĻžāϕ⧇ āϚāĻžāϰāĻĻāĻŋāϕ⧇ āĻŦ⧇āĻĄāĻŧāĻž āĻĻāĻŋāϤ⧇ āĻšāĻŦ⧇āĨ¤ āĻœā§‡āϰ⧀ āĻ āĻŋāĻ• āĻ•āϰāϞ⧋ āϏ⧇ 11.2 āĻŽāĻŋāϟāĻžāϰ āϞāĻŽā§āĻŦāĻž āĻ“ 5.0 āĻŽāĻŋāϟāĻžāϰ āϚāĻ“āĻĄāĻŧāĻž āĻŦ⧇āĻĄāĻŧāĻž āĻĻ⧇āĻŦ⧇āĨ¤ āĻŦ⧇āĻĄāĻŧāĻž āĻ“ āĻŦāĻžāĻ—āĻžāύ⧇āϰ āĻŽāĻ§ā§āϝāĻŦāĻ°ā§āϤ⧀ āĻ…āĻ‚āĻļ⧇āϰ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ⧋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. āĻœā§‡āϰāĻŋ āĻ āĻŋāĻ• āĻ•āϰ⧇āϛ⧇ āϝ⧇, āϏ⧇ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋āϰ āĻĒā§āϰāĻ¤ā§āϝ⧇āĻ•āϟāĻŋ āĻŽāĻžāĻ¤ā§āϰ āĻāĻ•āĻŦāĻžāϰ āĻ•āϰ⧇ āĻāĻ•āϟāĻŋ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻĒāĻžāϰāĻĒāĻžāĻļ⧇ āĻāĻŽāύāĻ­āĻžāĻŦ⧇ āĻŦāϏāĻžāĻŦ⧇ āϝāĻžāϤ⧇ āĻĒāĻžāĻļāĻžāĻĒāĻžāĻļāĻŋ āĻĻ⧁āχāϟāĻŋ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻŦāĻŋāϝāĻŧā§‹āĻ—āĻĢāϞ 4 āĻāϰ āĻšā§‡āϝāĻŧ⧇ āĻ•āĻŽ āύāĻž āĻšāϝāĻŧāĨ¤ āĻœā§‡āϰāĻŋ āϕ⧀ āϏ⧇āϟāĻž āĻĒāĻžāϰāĻŦ⧇āĨ¤ āĻĒāĻžāϰāϞ⧇, āϏ⧇āχ āĻŦ⧃āĻ¤ā§āϤāϟāĻŋāϤ⧇ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻŦāϏāĻŋāϝāĻŧ⧇ āĻĻ⧇āĻ–āĻžāĻ“āĨ¤

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§āĻ˛ā§āϝāĻžāĻ•āĻŦā§‹āĻ°ā§āĻĄā§‡ āϞ⧇āĻ–āĻž āĻ›āĻŋāϞ 1, 3, 4, 6, 8, 9, 11, 12, 16āĨ¤ āϕ⧋āĻĨāĻž āĻĨ⧇āϕ⧇ āĻšāĻžāϜāĻŋāϰ āĻšāϞ⧋ āϟāĻŽ āφāϰ āĻœā§‡āϰāĻŋāĨ¤ āĻĒā§āϰāĻĨāĻŽā§‡ āϟāĻŽ āĻāĻ•āϟāĻŋ āύāĻŽā§āĻŦāϰ āĻŽā§āϛ⧇ āĻĢ⧇āϞāϞ⧋āĨ¤ āϤāĻ–āύ āĻœā§‡āϰāĻŋāĻ“ āĻāĻ•āϟāĻž āĻŽā§āĻ›āϞ⧋āĨ¤ āφāĻŦāĻžāϰ āϟāĻŽ, āφāĻŦāĻžāϰ āĻœā§‡āϰāĻŋāĨ¤ āĻāĻ­āĻžāĻŦ⧇ āĻĒā§āϰāĻ¤ā§āϝ⧇āϕ⧇ 4āϟāĻŋ āĻ•āϰ⧇ āύāĻŽā§āĻŦāϰ āĻŽā§āĻ›āϞ⧋āĨ¤ āĻĻ⧇āĻ–āĻž āϗ⧇āϞ āϟāĻŽā§‡āϰ āĻŽā§āϛ⧇ āĻĢ⧇āϞāĻž āύāĻŽā§āĻŦāϰāϗ⧁āϞ⧋āϰ āϝ⧋āĻ—āĻĢāϞ āĻœā§‡āϰāĻŋāϰ āĻŽā§āϛ⧇ āĻĢ⧇āϞāĻž āύāĻŽā§āĻŦāϰāϗ⧁āϞ⧋āϰ āϝ⧋āĻ—āĻĢāϞ⧇āϰ āϤāĻŋāύāϗ⧁āύāĨ¤ āĻĒā§āϰāĻļā§āύ āĻšāϞ⧋ āϏāĻŦ āĻļ⧇āώ⧇ āϕ⧋āύ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋ āĻŦā§āĻ˛ā§āϝāĻžāĻ•āĻŦā§‹āĻ°ā§āĻĄā§‡ āϰāϝāĻŧ⧇ āϗ⧇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§­. āĻĒāĻžāρāϚ āϜāύ āϞ⧋āĻ• āĻšāϝāĻŧ āύ⧀āϞ āĻ…āĻĨāĻŦāĻž āϞāĻžāϞ āϟ⧁āĻĒāĻŋ āĻĒāĻĄāĻŧ⧇āϛ⧇āĨ¤ āϝāĻĻāĻŋ āĻāĻĻ⧇āϰ āϕ⧇āω āĻ•āĻĨāĻž āĻŦāϞ⧇ āϤāĻžāĻšāϞ⧇ āύ⧀āϞ āϟ⧁āĻĒāĻŋāϰ āϞ⧋āϕ⧇āϰāĻž āϏāĻŦāϏāĻŽāϝāĻŧ āϏāĻ¤ā§āϝ āĻ•āĻĨāĻž āĻŦāϞ⧇ āĻāĻŦāĻ‚ āϞāĻžāϞ āϟ⧁āĻĒāĻŋāϰ āϞ⧋āϕ⧇āϰāĻž āϕ⧇āĻŦāϞ āĻŽāĻŋāĻĨā§āϝāĻž āĻ•āĻĨāĻž āĻŦāϞ⧇āĨ¤ āϤāĻžāĻĻ⧇āϰ āĻĒā§āϰāĻ¤ā§āϝ⧇āϕ⧇ āĻ…āĻ¨ā§āϝ āϚāĻžāϰāϜāύ⧇āϰ āϟ⧁āĻĒāĻŋ āĻĻ⧇āĻ–āϤ⧇ āĻĒāĻžāϝāĻŧāĨ¤

A āĻŦāϞāϞ “ āφāĻŽāĻŋ 4 āϟāĻŋ āύ⧀āϞ āϟ⧁āĻĒāĻŋ āĻĻ⧇āĻ–āĻ›āĻŋāĨ¤â€

B āĻŦāϞāϞ “ āφāĻŽāĻŋ 3 āϟāĻŋ āύ⧀āϞ āϟ⧁āĻĒāĻŋ āĻ“ 1 āϟāĻŋ āϞāĻžāϞ āϟ⧁āĻĒāĻŋ āĻĻ⧇āĻ–āĻ›āĻŋāĨ¤â€

C āĻŦāϞāϞ “ āφāĻŽāĻŋ āĻĻ⧇āĻ–āĻ›āĻŋ 1āϟāĻŋ āύ⧀āϞ āφāϰ 3 āϟāĻŋ āϞāĻžāϞ āϟ⧁āĻĒāĻŋ āĨ¤â€

D āĻŦāϞāϞ “ āφāĻŽāĻŋ 4 āϟāĻŋ āϞāĻžāϞ āϟ⧁āĻĒāĻŋ āĻĻ⧇āĻ–āĻ›āĻŋāĨ¤â€

E āĻ•āĻŋāϛ⧁āχ āĻŦāϞāϞ āύāĻžāĨ¤

āĻ“āϰāĻž āϕ⧇ āϕ⧋āύ āϰāĻ™ āĻāϰ āϟ⧁āĻĒāĻŋ āĻĒāĻĄāĻŧ⧇ āφāϛ⧇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 “I see 4 blue hats.”

B Says “I see 3 blue hats and 1 red hat.”

C Says “I see 1 blue hat and 3 red hats.”

D says “I see 4 red hats.”

E Does not say anything.

Find the color of each person’s hat.

ā§Ž. āĻāĻ•āϟāĻŋ āϏāύāĻžāϤāύ⧀ āϘāĻĄāĻŧāĻŋāϤ⧇ āϘāĻŖā§āϟāĻžāϰ āĻ“ āĻŽāĻŋāύāĻŋāĻŸā§‡āϰ āĻ•āĻžāρāϟāĻž āφāϛ⧇āĨ¤ 3 āϟāĻž āĻ“ 4 āϟāĻžāϰ āĻŽāĻ§ā§āϝ⧇ āϝāĻĻāĻŋ āĻŽāĻŋāύāĻŋāĻŸā§‡āϰ āĻ•āĻžāρāϟāĻž āϘāĻŖā§āϟāĻžāϰ āĻ•āĻžāρāϟāĻžāϰ āĻ āĻŋāĻ• āĻ“āĻĒāϰ⧇ āĻĨāĻžāϕ⧇ āϤāĻ–āύ āϤāĻžāĻšāϞ⧇ āϏāĻŽāϝāĻŧ āĻ•āϤ⧋?

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?

⧝. āĻāĻŽāύ āĻ•ā§āώ⧁āĻĻā§āϰāϤāĻŽ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ āϝ⧇āύ āϏ⧇āϟāĻŋāϕ⧇ 4, 5, 6 āĻ…āĻĨāĻŦāĻž 9 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ 1 āĻ…āĻŦāĻļāĻŋāĻˇā§āϟ āĻĨāĻžāϕ⧇ āĻāĻŦāĻ‚ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋ 13 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āĻšāϝāĻŧāĨ¤

Find the smallest number, divisible by 13, such that the remainder is 1 when divided by 4, 5, 6 or 9.

math Olympiad questions

ā§§ā§Ļ. āύāĻ¨ā§āĻŸā§‡ āφāϰ āĻĢāĻ¨ā§āĻŸā§‡āϰ āĻŽāĻžāĻā§‡ āĻĒā§āϰāϤāĻŋāϝ⧋āĻ—āĻŋāϤāĻž āĻšāĻšā§āϛ⧇, āĻ“āĻĻ⧇āϰāϕ⧇ āĻ•āϤāϗ⧁āϞ⧋ āϏāĻ‚āĻ–ā§āϝāĻž āĻĻāĻŋāϝāĻŧ⧇ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇, āĻ“āϰāĻž āϏ⧇āĻ–āĻžāύ āĻĨ⧇āϕ⧇ āĻĻ⧁āϟāĻŋ āĻ•āϰ⧇ āϏāĻ‚āĻ–ā§āϝāĻž āύ⧇āĻŦ⧇ āϝ⧇āύ āϐ āϏāĻ‚āĻ–ā§āϝāĻž āĻĻ⧁āϟāĻŋāϰ āϝ⧋āĻ—āĻĢāϞ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āĻšāϝāĻŧāĨ¤ āύāĻ¨ā§āĻŸā§‡āϕ⧇ 1 āĻĨ⧇āϕ⧇ 40 āĻāϰ āĻŽāĻ§ā§āϝ āĻĨ⧇āϕ⧇ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇, āĻĢāĻ¨ā§āĻŸā§‡āϕ⧇ 1 āĻĨ⧇āϕ⧇ 100 āĻāϰ āĻŽāĻ§ā§āϝ⧇ āĻŦāĻŋāĻœā§‹āĻĄāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇āĨ¤ āϝ⧇ āϝāϤ āĻŦ⧇āĻļāĻŋ āĻ­āĻžāĻŦ⧇ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āϏāĻ‚āĻ–ā§āϝāĻž āĻŦāĻžāύāĻžāϤ⧇ āĻĒāĻžāϰāĻŦ⧇ āϏ⧇ āϜāĻŋāϤāĻŦ⧇āĨ¤ āϕ⧇ āϜāĻŋāϤāĻŦ⧇ āĻāĻŦāĻ‚ āϕ⧇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…āĻ‚āϕ⧇āϰ āĻāĻ•āϟāĻŋ āϏāĻ‚āĻ–ā§āϝāĻž, 1*3 ; āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋ 11 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āĻšāϞ⧇ * āϚāĻŋāĻšā§āύāĻŋāϤ āĻ¸ā§āĻĨāĻžāύ⧇āϰ āĻ…āĻ‚āĻ•āϟāĻŋ āĻ•āϤ⧋ āϏ⧇āϟāĻž āĻĒā§āϰāĻŽāĻžāĻŖ āϏāĻš āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ The three digit number 1 * 3 is divisible by 11. Find, with proof, the missing digit (represented by the asterisk).

⧍. āĻāĻ•āχ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻŦāĻŋāĻļāĻŋāĻˇā§āϟ āĻāĻ•āϟāĻŋ āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§āϰ āĻāĻŦāĻ‚ āĻāĻ•āϟāĻŋ āĻŦāĻ°ā§āĻ—āĻ•ā§āώ⧇āĻ¤ā§āϰ⧇āϰ āĻŽāĻ§ā§āϝ⧇ āϕ⧋āύāϟāĻŋāϰ āĻĒāϰāĻŋāϏ⧀āĻŽāĻž āĻŦāĻĄāĻŧ āϏ⧇āϟāĻŋ āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖ āϏāĻš āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ A rectangle and a square have the same area, find, with proof, which one has a greater perimeter.

ā§Š. āϟāĻŽ āĻŦāϏ⧇ āĻāĻ•āĻĻāĻŋāύ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāϝāĻŧ⧇ āϖ⧇āϞāĻ›āĻŋāϞāĨ¤ āϟāĻŽ 1 āĻĨ⧇āϕ⧇ 22 āĻĒāĻ°ā§āϝāĻ¨ā§āϤ āϏāĻŦ āϏāĻ‚āĻ–ā§āϝāĻžāϕ⧇ āĻāĻ•āĻŦāĻžāϰ āĻ•āϰ⧇ āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻ•āϰ⧇ āĻŽā§‹āϟ 11 āϟāĻŋ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļ āϞāĻŋāϖ⧇āϛ⧇āĨ¤ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāϗ⧁āϞ⧋āϰ āϞāĻŦ āĻ“ āĻšāϰ⧇ āϝ⧇ āϕ⧋āύ āϏāĻ‚āĻ–ā§āϝāĻž āϏ⧇ āĻŦāϏāĻžāϤ⧇ āĻĒāĻžāϰ⧇āĨ¤ āĻāϗ⧁āϞ⧋āϰ āĻŽāĻžāĻā§‡ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āĻ•āϤāϗ⧁āϞ⧋ āĻĒā§‚āĻ°ā§āĻŖāϏāĻ‚āĻ–ā§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. āĻāĻŽāύ āĻ•ā§āώ⧁āĻĻā§āϰāϤāĻŽ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ āϝ⧇āύ āϏ⧇āϟāĻŋāϕ⧇ 4, 6 āĻ…āĻĨāĻŦāĻž 9 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ 1 āĻ…āĻŦāĻļāĻŋāĻˇā§āϟ āĻĨāĻžāϕ⧇ āĻāĻŦāĻ‚ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋ 13 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§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) āĻāϰ āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻĒā§‚āĻ°ā§āĻŖ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āϕ⧋āύ āϕ⧋āύ āĻŽāĻžāύ⧇āϰ āϜāĻ¨ā§āϝ \[M^3 + 1331 = n^3 \] āϏāĻŽā§€āĻ•āϰāĻŖāϟāĻŋ āĻļ⧁āĻĻā§āϧāĨ¤ Find all pairs of positive integers (m, n) which satisfy \[M^3 + 1331 = n^3 \]

ā§Ŧ. āύāĻ¨ā§āĻŸā§‡ āφāϰ āĻĢāĻ¨ā§āĻŸā§‡āϰ āĻŽāĻžāĻā§‡ āĻĒā§āϰāϤāĻŋāϝ⧋āĻ—āĻŋāϤāĻž āĻšāĻšā§āϛ⧇, āĻ“āĻĻ⧇āϰāϕ⧇ āĻ•āϤāϗ⧁āϞ⧋ āϏāĻ‚āĻ–ā§āϝāĻž āĻĻāĻŋāϝāĻŧ⧇ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇, āĻ“āϰāĻž āϏ⧇āĻ–āĻžāύ āĻĨ⧇āϕ⧇ āĻĻ⧁āϟāĻŋ āĻ•āϰ⧇ āϏāĻ‚āĻ–ā§āϝāĻž āύ⧇āĻŦ⧇ āϝ⧇āύ āϐ āϏāĻ‚āĻ–ā§āϝāĻž āĻĻ⧁āϟāĻŋāϰ āϝ⧋āĻ—āĻĢāϞ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āĻšāϝāĻŧāĨ¤ āύāĻ¨ā§āĻŸā§‡āϕ⧇ 1 āĻĨ⧇āϕ⧇ 40 āĻāϰ āĻŽāĻ§ā§āϝ āĻĨ⧇āϕ⧇ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇, āĻĢāĻ¨ā§āĻŸā§‡āϕ⧇ 1 āĻĨ⧇āϕ⧇ 100 āĻāϰ āĻŽāĻ§ā§āϝ⧇ āĻŦāĻŋāĻœā§‹āĻĄāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāϗ⧁āϞ⧋ āĻĻ⧇āĻ“āϝāĻŧāĻž āĻšāϝāĻŧ⧇āϛ⧇āĨ¤ āϝ⧇ āϝāϤ āĻŦ⧇āĻļāĻŋ āĻ­āĻžāĻŦ⧇ 3 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ āϏāĻ‚āĻ–ā§āϝāĻž āĻŦāĻžāύāĻžāϤ⧇ āĻĒāĻžāϰāĻŦ⧇ āϏ⧇ āϜāĻŋāϤāĻŦ⧇āĨ¤ āϕ⧇ āϜāĻŋāϤāĻŦ⧇ āĻāĻŦāĻ‚ āϕ⧇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. āĻāĻ•āϟāĻŋ āϤāϞ⧇ 25 āϟāĻŋ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ āφāϛ⧇, āĻāϰ āĻŽāĻ§ā§āϝ⧇ āĻāĻ•āχ āϰ⧇āĻ–āĻžāϝāĻŧ āϤāĻŋāύāϟāĻŋ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āĻ“ āĻāĻ•āχ āϏāϰāϞāϰ⧇āĻ–āĻžāϝāĻŧ āύ⧇āχāĨ¤ āϏāĻŦ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϕ⧇ āφāϞāĻžāĻĻāĻž āĻ•āϰ⧇ āϰāĻžāĻ–āϤ⧇ āĻšāϞ⧇ āĻ•āĻŽāĻĒāĻ•ā§āώ⧇ āĻ•āϝāĻŧāϟāĻŋ āϰ⧇āĻ–āĻž āϞāĻžāĻ—āĻŦ⧇?

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. āĻŸā§āϰāĻžāĻĒāĻŋāϜāĻŋāϝāĻŧāĻžāĻŽ āĻšāϞ⧋ āϏ⧇āχ āϚāϤ⧁āĻ°ā§āϭ⧁āϜ āϝāĻžāϰ āĻĻ⧁āχāϟāĻŋ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻŦāĻžāĻšā§ āĻĒāϰāĻ¸ā§āĻĒāϰ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āĻ•āĻŋāĻ¨ā§āϤ⧁ āĻ…āĻ¨ā§āϝ āĻĻ⧁āχāϟāĻŋ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āύāϝāĻŧāĨ¤ āĻāĻ•āϟāĻŋ āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āĻŸā§āϰāĻžāĻĒāĻŋāϜāĻŋāϝāĻŧāĻŽā§‡āϰ āĻ•āĻ°ā§āĻŖ āĻāϟāĻŋāϕ⧇ āĻĻ⧁āχāϟāĻŋ āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡ āĻ­āĻžāĻ— āĻ•āϰ⧇āϛ⧇āĨ¤ āĻŸā§āϰāĻžāĻĒāĻŋāϜāĻŋāϝāĻŧāĻžāĻŽā§‡āϰ āϕ⧋āĻŖāϗ⧁āϞ⧋āϰ āĻŽāĻžāύ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ⧋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. āϟāĻŽ āĻ“ āĻœā§‡āϰ⧀āϰ āĻ•āĻžāϛ⧇ āĻŽā§‹āϟ 14 āϟāĻŋ āϟāĻžāχāϞāϏ āφāϛ⧇āĨ¤ āĻāϰ āĻŽāĻ§ā§āϝ⧇ ā§Ē āϟāĻŋ āύ⧀āϞ āĻ“ 6 āϟāĻŋ āϞāĻžāϞāĨ¤ āϤāĻžāϰāĻž āĻāϗ⧁āϞ⧋ āĻāĻ• āϞāĻžāχāύ⧇ āĻāĻŽāύāĻ­āĻžāĻŦ⧇ āϏāĻžāϜāĻžāϤ⧇ āϚāĻžāϝāĻŧ āϝ⧇, āĻĒā§āϰāϤāĻŋ āĻĻ⧁āχāϟāĻŋ āϞāĻžāϞ āϟāĻžāχāϞāϏ⧇āϰ āĻŽāĻžāĻāĻ–āĻžāύ⧇ āĻ•āĻŽāĻĒāĻ•ā§āώ⧇ āĻāĻ•āϟāĻŋ āύ⧀āϞ āϟāĻžāχāϞāϏ āĻĨāĻžāĻ•āĻŦ⧇āĨ¤ āϏāĻŽā§āĻ­āĻžāĻŦā§āϝ āĻ•āϤāĻ­āĻžāĻŦ⧇ āϟāĻŽ āĻ“ āĻœā§‡āϰ⧀ āĻāχ āĻ•āĻžāϜ āĻ•āϰāϤ⧇ āĻĒāĻžāϰāĻŦ⧇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 āĻāĻ•āϟāĻŋ āϚāϤ⧁āĻ°ā§āϭ⧁āϜāĨ¤ āĻāϰ AB, BC āĻ“ CD āĻŦāĻžāĻšā§āϰ āĻŽāĻ§ā§āϝāĻŦāĻŋāĻ¨ā§āĻĻ⧁ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ P, Q āĻ“ R | āϝāĻĻāĻŋ PQ = 3, QR= 4 āĻāĻŦāĻ‚ PR = 5 āĻšāϝāĻŧ āϤāĻŦ⧇ ABCD āϚāϤ⧁āĻ°ā§āϭ⧁āĻœā§‡āϰ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻŦ⧇āϰ āĻ•āϰ⧋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§§. āĻāĻ•āϟāĻŋ āĻĒāĻžāĻ°ā§āϟāĻŋāϤ⧇ āϤ⧁āĻŽāĻŋ āĻ›āĻžāĻĄāĻŧāĻžāĻ“ āφāϰ⧋ 20 āϜāύ āϞ⧋āĻ• āϰāϝāĻŧ⧇āϛ⧇āĨ¤ āϐ āĻĒāĻžāĻ°ā§āϟāĻŋāϤ⧇ āϤ⧁āĻŽāĻŋ āϝāϤāϜāύāϕ⧇ āĻšā§‡āύ, āφāĻŦāĻžāϰ āĻ āĻŋāĻ• āϤāϤāϜāύāϕ⧇āχ āĻšā§‡āύāύāĻžāĨ¤ āϤ⧁āĻŽāĻŋ āĻ•āϤāϜāύāϕ⧇ āĻšā§‡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§ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ B āϕ⧋āĻŖāϟāĻŋ āϏāĻŽāϕ⧋āĻŖ āĻāĻŦāĻ‚ AB = 3āĨ¤ āĻāϰ āϝ⧇āϕ⧋āύ āĻāĻ•āϟāĻŋ āĻļā§€āĻ°ā§āώāϕ⧇ āϕ⧇āĻ¨ā§āĻĻā§āϰ āĻ•āϰ⧇ āĻāĻ•āϟāĻŋ āĻāĻ•āĻ• āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ⧇āϰ āĻŦ⧃āĻ¤ā§āϤ āφāρāĻ•āĻž āĻšāϞāĨ¤ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ āϝ⧇ āĻ…āĻ‚āĻļ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻ…āĻ¨ā§āϤāĻ°ā§āĻ­ā§‚āĻ•ā§āϤ āύāϝāĻŧ āϤāĻžāϰ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻŦ⧇āϰ āĻ•āϰāĨ¤

 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. āĻŦāĻžāĻ¸ā§āϤāĻŦ āϏāĻ‚āĻ–ā§āϝāĻžāϝāĻŧ āϏāĻŽāĻžāϧāĻžāύ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāσ \[\frac{|x^2 – 1|}{x – 2}= x\]

 [āϏāĻžāĻšāĻžāĻ¯ā§āϝāσ ax2 + bx + c = 0 āϏāĻŽā§€āĻ•āϰāϪ⧇āϰ āϏāĻŽāĻžāϧāĻžāύ āĻšāϞ x = \[\frac{ – b Âą \sqrt{b^2 – – 4ac} }{2a}\]

Solve for real x: \[\frac{|x^2 – 1|}{x – 2}= x\]

 [Hint: The solutions to the equation ax2 + bx + c = 0 are x = \[\frac{ – b Âą \sqrt{b^2 – – 4ac} }{2a}\]

ā§Ē. āĻāĻ•āϟāĻŋ āϧāĻžāϰāĻžāϕ⧇ āύāĻŋāĻšā§‡āϰ āϏāĻ‚āĻœā§āĻžāĻž āĻ…āύ⧁āϝāĻžāϝāĻŧā§€ āĻ—āĻ āύ āĻ•āϰāĻž āĻšāϞāσ

A(1) = 1

A(n) = f(m) āϏāĻ‚āĻ–ā§āϝāĻ• f(m) āĻāĻŦāĻ‚ āϤāĻžāϰ āĻĒāϰ f(m) āϏāĻ‚āĻ–ā§āϝāĻ• 0; āĻāĻ–āĻžāύ⧇ 111 āĻšāĻšā§āϛ⧇ A(n-1) āĻāϰ āĻ…āĻ‚āĻ• āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤

A(30) āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ āωāĻ˛ā§āϞ⧇āĻ–ā§āϝ, 11 āϕ⧇ 9 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ āϝ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ āĻĨāĻžāϕ⧇ āϏ⧇āϟāĻŋāχ 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 āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ B āϕ⧋āĻŖāϟāĻŋ āϏāĻŽāϕ⧋āĻŖāĨ¤ ∠BAC āĻāϰ āϏāĻŽāĻĻā§āĻŦāĻŋāĻ–āĻŖā§āĻĄāĻ• BC āϕ⧇ D āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āϛ⧇āĻĻ āĻ•āϰ⧇āĨ¤ G āĻāχ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ āĻŽāĻ§ā§āϝāĻŽāĻžāĻ¤ā§āϰāϝāĻŧ⧇āϰ āϛ⧇āĻĻāĻŦāĻŋāĻ¨ā§āĻĻ⧁⧎ GD|| AB āĻšāϞ⧇ ∠C āĻāϰ āĻŽāĻžāύ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤

Triangle ABC is right angled at B. The bisector of ∠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 ∠C.

ā§Ŧ. āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖāϏāĻš āĻāĻŽāύ āϏāĻ•āϞ āĻĒā§‚āĻ°ā§āĻŖāĻŦāĻ°ā§āĻ— āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ āϝāĻžāϰāĻž āϚāĻžāϰāϟāĻŋ āĻ•ā§āϰāĻŽāĻŋāĻ• āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻŦ⧇āĻœā§‹āĻĄāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āϗ⧁āĻŖāĻĢāϞāĨ¤

Find, with proof, all the perfect squares each of which is the product of four consecutive odd natural numbers.

ā§­. āϝāĻĨāĻžāϝāĻĨ āĻĒā§āϰāĻŽāĻžāĻŖāϏāĻš āĻ›āϝāĻŧ āĻ…āĻ‚āϕ⧇āϰ 13**45* āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋāϤ⧇ āĻĒā§āϰāĻ¤ā§āϝ⧇āĻ•āϟāĻŋ āϤāĻžāϰāĻ•āĻžāϚāĻŋāĻšā§āύāϕ⧇ (*) āĻ­āĻŋāĻ¨ā§āύ āĻ­āĻŋāĻ¨ā§āύ āĻ…āĻ‚āĻ• āĻĻā§āĻŦāĻžāϰāĻž āĻĒā§āϰāϤāĻŋāĻ¸ā§āĻĨāĻžāĻĒāĻŋāϤ āĻ•āϰ āϝ⧇āύ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋ 792 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§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 āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ A āϕ⧋āĻŖāϟāĻŋ āϏāĻŽāϕ⧋āĻŖāĨ¤ BC āĻāϰ āωāĻĒāϰ D āĻāĻ•āϟāĻŋ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āĨ¤ AC āĻāĻŦāĻ‚ AB āĻāϰ āϏāĻžāĻĒ⧇āĻ•ā§āώ⧇ D āĻāϰ āĻĒā§āϰāϤāĻŋāĻĢāϞāύ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ E āĻāĻŦāĻ‚ F āĨ¤ āĻĻ⧇āĻ–āĻžāĻ“ āϝ⧇, [ABC] â‰Ĩ [DEF] āĻāĻŦāĻ‚ āϏāĻŽāϤāĻž āĻšāĻŦāĻžāϰ āϜāĻ¨ā§āϝ D āĻāϰ āϏāĻ•āϞ āϏāĻŽā§āĻ­āĻžāĻŦā§āϝ āĻ…āĻŦāĻ¸ā§āĻĨāĻžāύ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ ([x] āĻĻā§āĻŦāĻžāϰāĻž x āĻāϰ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āύāĻŋāĻ°ā§āĻĻ⧇āĻļ āĻ•āϰāĻž āĻšāĻšā§āϛ⧇)

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] â‰Ĩ [DEF]. Find all possible positions of D for equality. (Here [x] denotes the area of x)

 

math Olympiad questions

⧝. āĻāĻŽāύ āϏāĻ•āϞ āĻŽā§ŒāϞāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž p āĻāĻŦāĻ‚ āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻĒā§‚āĻ°ā§āĻŖ āϏāĻ‚āĻ–ā§āϝāĻž a, b āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ āϝ⧇āύ \[ p^a + p^b\] āĻāĻ•āϟāĻŋ āĻĒā§‚āĻ°ā§āĻŖ āĻŦāĻ°ā§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 āϟāĻŋ āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻĒā§‚āĻ°ā§āĻŖāϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻāĻ•āϟāĻŋ āϏ⧇āĻŸā§‡ āϝ⧇āϕ⧋āύ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻŽā§ŒāϞāĻŋāĻ• āĻ‰ā§ŽāĻĒāĻžāĻĻāĻ•āϗ⧁āϞ⧋ 42 āĻāϰ āĻšā§‡āϝāĻŧ⧇ āϛ⧋āϟāĨ¤ āĻĻ⧇āĻ–āĻžāĻ“ āϝ⧇, āĻāχ āϏ⧇āϟ āĻĨ⧇āϕ⧇ āĻāĻŽāύ āϚāĻžāϰāϟāĻŋ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāĻ°ā§āĻŦāĻžāϚāύ āĻ•āϰāĻž āϝāĻžāĻŦ⧇ āϝ⧇āύ āϤāĻžāĻĻ⧇āϰ āϗ⧁āĻŖāĻĢāϞ āĻāĻ•āϟāĻŋ āĻĒā§‚āĻ°ā§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 āϕ⧇ 2 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ āĻ•āϤ āĻšāĻŦ⧇ āϤāĻž āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ Let S = \[ 1^1 + 2^2 + 3^3 + ….. + 2010^{2010} \]. What is the remainder when S is divided by 2?

⧍. ABC āϏāĻŽāĻĻā§āĻŦāĻŋāĻŦāĻžāĻšā§ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ B āϕ⧋āĻŖāϟāĻŋ āϏāĻŽāϕ⧋āĻŖ āĻāĻŦāĻ‚ AB = 3āĨ¤ āĻāϰ āϝ⧇āϕ⧋āύ āĻāĻ•āϟāĻŋ āĻļā§€āĻ°ā§āώāϕ⧇ āϕ⧇āĻ¨ā§āĻĻā§āϰ āĻ•āϰ⧇ āĻāĻ•āϟāĻŋ āĻāĻ•āĻ• āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ⧇āϰ āĻŦ⧃āĻ¤ā§āϤ āφāρāĻ•āĻž āĻšāϞāĨ¤ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ āϝ⧇ āĻ…āĻ‚āĻļ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻ…āĻ¨ā§āϤāĻ°ā§āϭ⧁āĻ•ā§āϤ āύāϝāĻŧ āϤāĻžāϰ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻŦ⧇āϰ āĻ•āϰāĨ¤

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. āĻāĻ•āϟāĻŋ āϧāĻžāϰāĻžāϕ⧇ āύāĻŋāĻšā§‡āϰ āϏāĻ‚āĻœā§āĻžāĻž āĻ…āύ⧁āϝāĻžāϝāĻŧā§€ āĻ—āĻ āύ āĻ•āϰāĻž āĻšāϞāσ

A(1) = 1

A(n) = f(m) āϏāĻ‚āĻ–ā§āϝāĻ• f(m) āĻāĻŦāĻ‚ āϤāĻžāϰ āĻĒāϰ f(m) āϏāĻ‚āĻ–ā§āϝāĻ• ā§Ļ;

āĻāĻ–āĻžāύ⧇ 111 āĻšāĻšā§āϛ⧇ A(n-1) āĻāϰ āĻ…āĻ‚āĻ• āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤

A(30) āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤ āωāĻ˛ā§āϞ⧇āĻ–ā§āϝ 1 āϕ⧇ 9 āĻĻā§āĻŦāĻžāϰāĻž āĻ­āĻžāĻ— āĻ•āϰāϞ⧇ āϝ⧇ āĻ­āĻžāĻ—āĻļ⧇āώ āĻĨāĻžāϕ⧇ āϏ⧇āϟāĻŋāχ 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. āĻāĻ•āϟāĻŋ āĻŦ⧃āĻ¤ā§āϤ  āĻāϰ āϭ⧇āϤāϰ⧇ P āĻāĻ•āϟāĻŋ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĨ¤ āĻāχ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĻĻāĻŋāϝāĻŧ⧇ āĻĻ⧁āϟāĻŋ āĻœā§āϝāĻž āφāρāĻ•āĻž āĻšāϞ āϝāĻžāϰāĻž āĻĒāϰāĻ¸ā§āĻĒāϰ⧇āϰ āωāĻĒāϰ āϞāĻŽā§āĻŦāĨ¤ āĻāχ āĻœā§āϝāĻž āĻĻ⧁āϟāĻŋ āĻŦ⧃āĻ¤ā§āϤāϕ⧇ āϘāĻĄāĻŧāĻŋāϰ āĻ•āĻžāϟāĻžāϰ āĻĻāĻŋāϕ⧇ a; b; c; d āĻāχ āϚāĻžāϰāϟāĻŋ āĻ­āĻžāϗ⧇ āĻ­āĻžāĻ— āĻ•āϰ⧇ āϝāĻžāϰ āĻŽāĻ§ā§āϝ⧇ a āĻ…āĻ‚āĻļāϟāĻŋ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āϕ⧇āĻ¨ā§āĻĻā§āϰāϕ⧇ āϧāĻžāϰāĻŖ āĻ•āϰ⧇āĨ¤ āĻĻ⧇āĻ–āĻžāĻ“ āϝ⧇, [a] + [c] â‰Ĩ [b] + [d], āĻāĻ–āĻžāύ⧇ [x] āĻĻā§āĻŦāĻžāϰāĻž x āĻāϰ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āύāĻŋāĻ°ā§āĻĻ⧇āĻļ āĻ•āϰ⧇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. āĻāĻ•āϟāĻŋ 2010 āĻŦāĻžāĻšā§ āĻŦāĻŋāĻļāĻŋāĻˇā§āϟ āϏ⧁āώāĻŽ āĻŦāĻšā§āϭ⧁āĻœā§‡āϰ āĻļā§€āĻ°ā§āώāϗ⧁āϞ⧋ āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻ•āϰ⧇ āĻ•āϤāϗ⧁āϞ⧋ āϏ⧁āώāĻŽ āĻŦāĻšā§āϭ⧁āϜ āφāρāĻ•āĻž āϏāĻŽā§āĻ­āĻŦ? (2010 āĻŦāĻžāĻšā§ āĻŦāĻŋāĻļāĻŋāĻˇā§āϟ āϏ⧁āώāĻŽ āĻŦāĻšā§āϭ⧁āĻœā§‡āϰ āĻļā§€āĻ°ā§āώāĻŦāĻŋāĻ¨ā§āĻĻ⧁āϗ⧁āϞ⧋āϰ āĻ•ā§āϰāĻŽ āĻŽā§āĻ–ā§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 āĻāĻŦāĻ‚ b āĻĻ⧁āϟāĻŋ āĻĒā§‚āĻ°ā§āĻŖ āϏāĻ‚āĻ–ā§āϝāĻž āϝ⧇āύ 1 ≤ a, b < 2010 āĻāĻŦāĻ‚ a + b; āĻāĻŽāύ āĻ•āϤāϗ⧁āϞ⧋ āĻ•ā§āϰāĻŽāĻœā§‹āĻĄāĻŧ (a, b) āφāϛ⧇ āϝ⧇āĻ–āĻžāύ⧇ \[ a^2 + b^2 \], 5 āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ­āĻžāĻœā§āϝ? a + b āĻāϰ āĻŽāĻžāύ āϕ⧋āύ āĻŽāĻžāύ⧇āϰ āϜāĻ¨ā§āϝ \[ a^2 + b^2 \] āϏāĻŦāĻšā§‡āϝāĻŧ⧇ āĻŦāĻĄāĻŧ āĻšāĻŦ⧇?

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 āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ AC > AB āĨ¤ āĻāϰ BC āĻŦāĻžāĻšā§āϰ āϞāĻŽā§āĻŦāϏāĻŽāĻĻā§āĻŦāĻŋāĻ–āĻŖā§āĻĄāĻ• āĻāĻŦāĻ‚ CAB āϕ⧋āĻŖ āĻāϰ āĻ…āĻ¨ā§āϤāϏāĻŽāĻĻā§āĻŦāĻŋāĻ–āĻŖā§āĻĄāĻ• P āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āϛ⧇āĻĻ āĻ•āϰ⧇āĨ¤ P āĻĨ⧇āϕ⧇ AB āĻāĻŦāĻ‚ AC āĻŦāĻžāĻšā§āϰ āωāĻĒāϰ⧇ āĻ…āĻ™ā§āĻ•āĻŋāϤ āϞāĻŽā§āĻŦāĻĻā§āĻŦāϝāĻŧ āĻŦāĻžāĻšā§āĻĻ⧁āϟāĻŋāϕ⧇ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ X āĻāĻŦāĻ‚ Y āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āϛ⧇āĻĻ āĻ•āϰ⧇āĨ¤ XY āĻāĻŦāĻ‚ BC āĻāϰ āϛ⧇āĻĻāĻŦāĻŋāĻ¨ā§āĻĻ⧁ Z, \[ \frac{BZ}{ZC} \] āĻāϰ āĻŽāĻžāύ āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤

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. āĻāĻŽāύ āϏāĻ•āϞ āĻŽā§ŒāϞāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž p āĻāĻŦāĻ‚ āĻĒā§‚āĻ°ā§āĻŖ āϏāĻ‚āĻ–ā§āϝāĻž a, b (āĻ‹āĻŖāĻžāĻ¤ā§āĻŽāĻ•āĻ“ āĻšāϤ⧇ āĻĒāĻžāϰ⧇) āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰ āϝ⧇āύ \[ p^a + p^b \] āĻāĻ•āϟāĻŋ āĻŽā§‚āϞāĻĻ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻŦāĻ°ā§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

⧝. \[ (x + y)^{2010} \]  āĻāϰ āĻŦāĻŋāĻ¸ā§āϤ⧃āϤāĻŋāϤ⧇ āĻŦāĻŋāĻœā§‹āĻĄāĻŧ āϏāĻšāϗ⧇āϰ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻŋāĻ°ā§āĻŖāϝāĻŧ āĻ•āϰāĨ¤

Find the number of odd coefficients in expansion of \[ (x + y)^{2010} \].

10. \[a_1,a_2,…,a_k,…,a_n \] āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻŦāĻžāĻ¸ā§āϤāĻŦ āϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻāĻŽāύ āĻāĻ•āϟāĻŋ āϧāĻžāϰāĻž āϝāĻžāϰ āϕ⧋āύ āĻĻ⧁āϟāĻŋ āĻĒāĻĻāχ āϏāĻŽāĻžāύ āύāϝāĻŧ āĻāĻŦāĻ‚ \[a_1<a_2<…<a_k \] āĻāĻŦāĻ‚ \[ a_k>a_{k+1}>…>a_n \] āĨ¤ āĻāĻ•āϟāĻŋ āϘāĻžāϏāĻĢāĻĄāĻŧāĻŋāĻ‚ āĻŽā§‚āϞāĻŦāĻŋāĻ¨ā§āĻĻ⧁ O āĻĨ⧇āϕ⧇ āĻŦāĻžāĻ¸ā§āϤāĻŦ āĻ…āĻ•ā§āώ āĻŦāϰāĻžāĻŦāϰ āĻĄāĻžāύāĻĻāĻŋāĻ• āĻŦāϰāĻžāĻŦāϰ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ \[a_1,a_2,…,a_n \] āĻĻā§‚āϰāĻ¤ā§āĻŦāϗ⧁āϞ⧋ āĻ…āϤāĻŋāĻ•ā§āϰāĻŽ āĻ•āϰ⧇ āϞāĻžāĻĢāĻŋāϝāĻŧ⧇ āϞāĻžāĻĢāĻŋāϝāĻŧ⧇ āϚāϞ⧇āĨ¤ āĻĒā§āϰāĻŽāĻžāĻŖ āĻ•āϰ āϝ⧇, āϏāĻŦāĻžāϰ āĻĄāĻžāύ⧇ āĻĒ⧌āρāϛ⧇ āĻ—āĻŋāϝāĻŧ⧇ āϏ⧇ a1, 2, â€ĸâ€ĸâ€ĸ, āĻŦā§€ āϧāĻžāϰāĻžāϰ āĻĒāĻĻāϗ⧁āϞ⧋āϕ⧇ āϕ⧋āύ āĻāĻ• āĻ•ā§āϰāĻŽā§‡ āĻ…āύ⧁āϏāϰāĻŖ āĻ•āϰ⧇ āĻŦāĻžāĻŽāĻĻāĻŋāĻ• āĻŦāϰāĻžāĻŦāϰ āϞāĻžāĻĢāĻŋāϝāĻŧ⧇ āϞāĻžāĻĢāĻŋāϝāĻŧ⧇ āϏ⧇āĻ–āĻžāύ āĻĨ⧇āϕ⧇ āĻŽā§‚āϞāĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āĻĢāĻŋāϰ⧇ āφāϏāϤ⧇ āĻĒāĻžāϰ⧇ āϝ⧇āύ āϝāĻžāĻŦāĻžāϰ āĻĒāĻĨ⧇ āϏ⧇ āϝ⧇ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϗ⧁āϞ⧋āϤ⧇ āύ⧇āĻŽā§‡āĻ›āĻŋāϞ āφāϏāĻžāϰ āϏāĻŽāϝāĻŧ āϏ⧇āϗ⧁āϞ⧋āϰ āϕ⧋āύāϟāĻŋāϤ⧇āχ āύāĻž āύ⧇āĻŽā§‡ āĻĨāĻžāϕ⧇ (āĻŽā§‚āϞāĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĻāĻŦāĻ‚ āϏāĻ°ā§āĻŦāĻĄāĻžāύ⧇āϰ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĻšāϞ āĻŦā§āϝāϤāĻŋāĻ•ā§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)

 

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