1.) Find the value of \(x\) such that

\[\displaystyle \left\lfloor{x + \frac{11}{100}}\right\rfloor + \left\lfloor{x + \frac{12}{100}}\right\rfloor + \left\lfloor{x + \frac{13}{100}}\right\rfloor + \cdots + \left\lfloor{x + \frac{99}{100}}\right\rfloor = 761\]

2.) Do the following.

2.1) Prove the identity \((x^{2} - y^{2})^{2} + (2xy)^{2} = (x^{2}+y^{2})^{2}\)

2.2) Use the identity (2.1) to find all the positive integer solutions of \[x^{2} + y^{2} = 34(x-y)\]

3.) Let \(\alpha, \beta\) be the roots of \(3x^{2} - (\alpha^{2} + 1)x + 3 = 6(\alpha + \beta)\), find the value of \(\alpha + \beta\)

4.) Let \(P(x)\) be the polynomial with real coefficients such that

- \(2(1+P(x)) = P(x-1) + P(x+1)\)
- \(P(0) = 91, P(5) = 166\)

Find \(P(45)\).

This is the part of Thailand 1st round math POSN problems.

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## Comments

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TopNewestSomehow the calculator said that the answer for 1) is

\(\displaystyle 8.49 \leq x < 8.5\)

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For number 2, 2.1 can simply be proven by expanding but for 2.2 (somehow not using 2.1), x^2 - 34x + y^2 + 34y = 0, we can use completing the square to get (x - 17)^2 + (y + 17)^2 = 578 and express (x - 17)^2 = 578 - (y + 17)^2. Since x and y are positive, y > 0 and 289 < (y + 17)^2 <= 578. And this is true when 17 < y + 17 < 25. By checking one-by-one, the equation has integral solutions if y = 6 and gives x = 10 or 24. (10, 6) and (24, 6) are the only solutions.

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1) \(\frac{169}{20} = 8.45\)

2.1) Just expand

2.2) 12 integral solutions

3) \(\displaystyle 0\)

4) \(2566\)

Are these right?

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Forgot to say that I need full solutions. Sorry about that. =..="

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I don't know the answer too!

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\(\displaystyle [x] + [x + \frac{1}{n}] + [x + \frac{2}{n}] \ldots [x + \frac{n -1}{n}] = [nx]\)

2.2. Draw the circle(passes through origin)

3) 2 eq. 2 variable I got \(\alpha = 1, \beta = \frac{-1}{3}\)

4) Is a quadratic equation consider

\(\displaystyle ax^2 + bx + c\)

Given

c = 91 And 2 other equations you can solve

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There's an error. 2-3(a+b)=(1-a)(1-b)

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Yeah sorry about that. I have edited the same however the answer remains the same.

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For 2.1 a neater way would be to use complex numbers. \(x^2+y^2 = (x +iy)(x-iy) \). Hence squaring it would give us \( RHS = (x^2 - y^2 + i2xy)(x^2 - y^2 - i2xy) \). I believe what follows is obvious.

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In the 3 rd i got alpha + beta as 0 ? am i correct would anyone tell ?

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Trying 2.2)

\((x^{2}+y^{2})^{2} = 1156(x-y)^{2}\)

\((x^{2}-y^{2})^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}\)

\((x-y)^{2}(x+y)^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}\)

\(4x^{2}y^{2} = (1156-(x+y)^{2})(x-y)^{2}\)

\(2xy = \sqrt{(34^{2}-(x+y)^{2})}(x-y)\)

Checking \(2 < x+y < 34 \) is going to be hardcore for me lol.

Calculator gives me \(x+y = 16, 30\)

Case1: \(2xy = 30(x-y)\)

\((x+15)(y-15) = -225\) which gives \((x,y) = (10,6)\).

Case2: \(2xy = 16(x-y)\)

\((x+8)(y-8) = -64\) which gives \((x,y) = (24,6)\).

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For number 1, the answer is any real value of x: 7.58 <= x < 7.59

From this equation, there n of the terms in which the value is equal to x and 99 - n of the terms equivalent to x + 1. we bring up the equation: nx + (99 - n)(x + 1) = nx + 99x + 99 - nx - n = 761. And rewriting it into, 99x + 99 - n = 761 and 99x = 662 + n. Since x is integral, 99|(662 + n) and this is only possible when n = 31 and so.. x = 7. From here, we find out that 31 of terms in the sequence above has the floor of 7 and 68 of them has floor of 8. We compare x + 41/100 and x + 42/100 since x + 41/100 is the 31st term and x + 42/100 is the 32nd term where x + 41/100 < 8 <= x + 42/100. Rewriting the inequality gives 379/100 <= x < 759/100.

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