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
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Nupur Prasad
7 years, 6 months ago
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Problem 1. We are given 2≤a≤b≤c≤d≤16,(d−1)2=(a−1)2+(b−1)2+(c−1)2, and(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) are both solutions.
Problem 3. c=6. Clearly y=±1 cannot have a finite solution. For y=0, x=−6 is a solution. For x≥5, 1<∣y∣<2, so it suffices just to check 1≤x≤4. This gives only the solutions (1,7) and (4,−2), for a total of 3 solutions as desired.
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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?
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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.
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For question 3, it says "in positive integers". So 6 doesn't work, and 10 does.
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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?
nupur, you are 21 years old and you are allowed to sit for olympiads?? which olympiad by the way??
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I am training my Brother, 15, so I am posting on behalf of him. Good observation.....
wew.... I think i cant reach this level.. I wish I can...
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Never mind...
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.
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What is BMO1?
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British Mathematical Olympiad Round 1
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.
Problem 3 (Yes I know Cody already did a solution to this problem. Bad Cody.)
We claim that c=10 works.
Manipulating the given equation, we get xy2−y2−x+y(xy2−x)−(y2−y)x(y−1)(y+1)−y(y−1)(y−1)(xy+x−y)=10=10=10=10 Since y≥1, we must have y−1∈{1,2,5,10}.
Case 1: y−1=1 and xy+x−y=10
From the first EQ, we get y=2. substituting into the second EQ and solving gives x=4, giving one solution in this case.
Case 2: y−1=2 and xy+x−y=5
From the first EQ, we get y=3. substituting into the second EQ and solving gives x=2, giving one solution in this case.
Case 3: y−1=5 and xy+x−y=2
From the first EQ, we get y=6. substituting into the second EQ and solving gives x=78, which is not an integer; thus there are no solutions in this case.
Case 4: y−1=10 and xy+x−y=1
From the first EQ, we get y=11. substituting into the second EQ and solving gives x=1, giving one solution in this case.
In conclusion, setting c=10 gives three solutions, as requested. □
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This is exactly how I solved.
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?
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It can't be determined exactly...
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.
Which book are you referring for RMO? I need help for my olympiad.