Help me in my Olympiad preparation

Hi everybody, I solved this paper just now. I want to check my answer. Please try these questions and tell me the answer with complete logic. Thanks

Note by Nupur Prasad
5 years, 11 months ago

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12 votes

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Problem 1. We are given 2abcd16,2\le a\le b\le c\le d\le16,(d1)2=(a1)2+(b1)2+(c1)2, and(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2,\text{ and}(d+1)2+(a+1)2=(b+1)2+(c+1)2(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2and we want to check if there is one unique integer solution. Except there is not, as (a,b,c,d)=(3,6,15,16),(3,7,10,12)(a,b,c,d)=(3,6,15,16),(3,7,10,12) are both solutions.

Problem 3. c=6c=6. Clearly y=±1y=\pm1 cannot have a finite solution. For y=0y=0, x=6x=-6 is a solution. For x5x\ge5, 1<y<21<|y|<2, so it suffices just to check 1x41\le x\le4. This gives only the solutions (1,7)(1,7) and (4,2)(4,-2), for a total of 3 solutions as desired.

Cody Johnson - 5 years, 11 months ago

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I have found it to be 10 @ (4,2) (2,3) and (1,11), is it correct?? And what have you found in Questions 1, 2, 4 and 5?

Kumar Ashutosh - 5 years, 11 months ago

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Multiple solutions exist. It just asked you to find a solution. Also, 10 is incorrect, as it passes through (-10,0). Also, I'm posting solutions at my leisure.

Cody Johnson - 5 years, 11 months ago

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@Cody Johnson For question 3, it says "in positive integers". So 6 doesn't work, and 10 does.

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin Yes!! This is exactly what I was thinking. Thanks I found out 18 to be okay. Is there any sequence of numbers which satisfy the given conditions?

Kumar Ashutosh - 5 years, 11 months ago

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nupur, you are 21 years old and you are allowed to sit for olympiads?? which olympiad by the way??

Divyaanand Sinha - 5 years, 11 months ago

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I am training my Brother, 15, so I am posting on behalf of him. Good observation.....

Nupur Prasad - 5 years, 11 months ago

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wew.... I think i cant reach this level.. I wish I can...

Unfaithful Angel - 5 years, 11 months ago

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Never mind...

Nupur Prasad - 5 years, 11 months ago

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I think you should buy "mathematical olympiad primer" if you want to prepare for BMO1. I have it and this BMO1 paper is included there with complete solutions.

Arkan Megraoui - 5 years, 11 months ago

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What is BMO1?

A Former Brilliant Member - 5 years, 11 months ago

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British Mathematical Olympiad Round 1

Arkan Megraoui - 5 years, 11 months ago

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No, I am Indian and I am going to give RMO(Indian exam similar to BMO) but just for practice I solved a BMO paper.

Nupur Prasad - 5 years, 11 months ago

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Problem 3 (Yes I know Cody already did a solution to this problem. Bad Cody.)

We claim that c=10c = 10 works.

Manipulating the given equation, we get xy2y2x+y=10(xy2x)(y2y)=10x(y1)(y+1)y(y1)=10(y1)(xy+xy)=10\begin{aligned} xy^2-y^2-x+y &= 10 \\ (xy^2-x) - (y^2-y) &= 10 \\ x(y-1)(y+1) - y(y-1) &= 10 \\ (y-1)(xy+x-y) &= 10 \end{aligned} Since y1,y \ge 1, we must have y1{1,2,5,10}.y - 1 \in \{1, 2, 5, 10\}.

Case 1: y1=1y - 1 = 1 and xy+xy=10xy+x-y = 10

From the first EQ, we get y=2.y = 2. substituting into the second EQ and solving gives x=4,x = 4, giving one solution in this case.

Case 2: y1=2y - 1 = 2 and xy+xy=5xy+x-y = 5

From the first EQ, we get y=3.y = 3. substituting into the second EQ and solving gives x=2,x = 2, giving one solution in this case.

Case 3: y1=5y - 1 = 5 and xy+xy=2xy+x-y = 2

From the first EQ, we get y=6.y = 6. substituting into the second EQ and solving gives x=87,x = \dfrac87, which is not an integer; thus there are no solutions in this case.

Case 4: y1=10y - 1 = 10 and xy+xy=1xy+x-y = 1

From the first EQ, we get y=11.y = 11. substituting into the second EQ and solving gives x=1,x = 1, giving one solution in this case.

In conclusion, setting c=10c = 10 gives three solutions, as requested. \square

Michael Tang - 5 years, 11 months ago

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This is exactly how I solved.

Kumar Ashutosh - 5 years, 11 months ago

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Can anybody give me the hint how I should proceed to solve this question? Q. In a very hotly fought battle, at least 70% of the soldiers lost an eye, at least 75% lost an ear ,at least 80% lost an arm ,at least 85% lost a leg.How many lost all four limbs?

kanak rani - 5 years, 11 months ago

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It can't be determined exactly...

Nupur Prasad - 5 years, 11 months ago

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Hints: 1) Bounding. 2) Similar triangles. In fact, first find an expression for tan(<APX)/tan(<PAX) to see what you need to prove is constant. 3) Experiment: First see if $c=p$ for a prime $p$ works. Then try $c=pq$ (p,q primes), etc. You will find $(p,q)=(2,5)$ works. 4) Bound! It boils down to showing that such $r$ exists for all $n\le 27$ (if i remember correctly) 5) What can $f(1)$ be? Once you've got that sorted, you'll need a clever insight, but once you find it the problem is just computation.

Arkan Megraoui - 5 years, 11 months ago

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Which book are you referring for RMO? I need help for my olympiad.

Himanshu Dhawale - 5 years, 11 months ago

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