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Problem 1. We are given $2\le a\le b\le c\le d\le16,$$(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2,\text{ and}$$(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2$and 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 $y=\pm1$ cannot have a finite solution. For $y=0$, $x=-6$ is a solution. For $x\ge5$, $1<|y|<2$, so it suffices just to check $1\le x\le4$. This gives only the solutions $(1,7)$ and $(4,-2)$, for a total of 3 solutions as desired.

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.

@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?

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.

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 $\begin{aligned} xy^2-y^2-x+y &= 10 \\ (xy^2-x) - (y^2-y) &= 10 \\ x(y-1)(y+1) - y(y-1) &= 10 \\ (y-1)(xy+x-y) &= 10 \end{aligned}$ Since $y \ge 1,$ we must have $y - 1 \in \{1, 2, 5, 10\}.$

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 $x = \dfrac87,$ 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. $\square$

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?

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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestProblem 1. We are given $2\le a\le b\le c\le d\le16,$$(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2,\text{ and}$$(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2$and 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 $y=\pm1$ cannot have a finite solution. For $y=0$, $x=-6$ is a solution. For $x\ge5$, $1<|y|<2$, so it suffices just to check $1\le x\le4$. 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

asolution. Also, 10 is incorrect, as it passes through (-10,0). Also, I'm posting solutions at my leisure.Log in to reply

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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.....

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

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Never mind...

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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.

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What is BMO1?

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British Mathematical Olympiad Round 1

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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.

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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 $\begin{aligned} xy^2-y^2-x+y &= 10 \\ (xy^2-x) - (y^2-y) &= 10 \\ x(y-1)(y+1) - y(y-1) &= 10 \\ (y-1)(xy+x-y) &= 10 \end{aligned}$ Since $y \ge 1,$ we must have $y - 1 \in \{1, 2, 5, 10\}.$

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 $x = \dfrac87,$ 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. $\square$

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This is exactly how I solved.

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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?

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It can't be determined exactly...

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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.

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

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