Algebra (1st math Thailand POSN 2014)

1.) Find the value of xx such that

x+11100+x+12100+x+13100++x+99100=761\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 (x2y2)2+(2xy)2=(x2+y2)2(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 x2+y2=34(xy)x^{2} + y^{2} = 34(x-y)

3.) Let α,β\alpha, \beta be the roots of 3x2(α2+1)x+3=6(α+β)3x^{2} - (\alpha^{2} + 1)x + 3 = 6(\alpha + \beta), find the value of α+β\alpha + \beta

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

  • 2(1+P(x))=P(x1)+P(x+1)2(1+P(x)) = P(x-1) + P(x+1)
  • P(0)=91,P(5)=166P(0) = 91, P(5) = 166

Find P(45)P(45).

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

Note by Samuraiwarm Tsunayoshi
4 years, 10 months ago

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1) 16920=8.45\frac{169}{20} = 8.45

2.1) Just expand

2.2) 12 integral solutions

3) 0\displaystyle 0

4) 25662566

Are these right?

Krishna Sharma - 4 years, 10 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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I don't know the answer too!

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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@Samuraiwarm Tsunayoshi Limited time cant post full solution and I suck in latex

Krishna Sharma - 4 years, 10 months ago

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@Krishna Sharma Just a guideline is okay.

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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@Samuraiwarm Tsunayoshi 1st question there is a theorem

[x]+[x+1n]+[x+2n][x+n1n]=[nx]\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 α=1,β=13\alpha = 1, \beta = \frac{-1}{3}

4) Is a quadratic equation consider

ax2+bx+c\displaystyle ax^2 + bx + c

Given

c = 91 And 2 other equations you can solve

Krishna Sharma - 4 years, 10 months ago

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@Krishna Sharma 4) Why is P(x)P(x) quadratic?

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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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.

John Ashley Capellan - 4 years, 10 months ago

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Somehow the calculator said that the answer for 1) is

8.49x<8.5\displaystyle 8.49 \leq x < 8.5

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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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.

John Ashley Capellan - 4 years, 10 months ago

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Trying 2.2)

(x2+y2)2=1156(xy)2(x^{2}+y^{2})^{2} = 1156(x-y)^{2}

(x2y2)2+4x2y2=1156(xy)2(x^{2}-y^{2})^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}

(xy)2(x+y)2+4x2y2=1156(xy)2(x-y)^{2}(x+y)^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}

4x2y2=(1156(x+y)2)(xy)24x^{2}y^{2} = (1156-(x+y)^{2})(x-y)^{2}

2xy=(342(x+y)2)(xy)2xy = \sqrt{(34^{2}-(x+y)^{2})}(x-y)

Checking 2<x+y<342 < x+y < 34 is going to be hardcore for me lol.

Calculator gives me x+y=16,30x+y = 16, 30

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

(x+15)(y15)=225(x+15)(y-15) = -225 which gives (x,y)=(10,6)(x,y) = (10,6).

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

(x+8)(y8)=64(x+8)(y-8) = -64 which gives (x,y)=(24,6)(x,y) = (24,6).

Samuraiwarm Tsunayoshi - 4 years, 10 months ago

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

avn bha - 4 years, 10 months ago

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For 2.1 a neater way would be to use complex numbers. x2+y2=(x+iy)(xiy)x^2+y^2 = (x +iy)(x-iy) . Hence squaring it would give us RHS=(x2y2+i2xy)(x2y2i2xy) RHS = (x^2 - y^2 + i2xy)(x^2 - y^2 - i2xy) . I believe what follows is obvious.

Sudeep Salgia - 4 years, 10 months ago

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  1. We have α2+1=3(α+β) \alpha^2 +1 = 3( \alpha + \beta) and 12(α+β)=αβ 1 - 2(\alpha + \beta) = \alpha \beta . Rewriting the second equation we have, 23(α+β)=(1α)(1β)2(α2+1)=(1α)(1β)1+α=1β 2 - 3( \alpha + \beta) = (1 - \alpha)(1 - \beta) \Rightarrow 2 - (\alpha^2 +1) = (1 - \alpha)(1 - \beta) \Rightarrow 1 + \alpha = 1 - \beta . Thus the required value is zero.

Sudeep Salgia - 4 years, 10 months ago

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

Joel Tan - 4 years, 10 months ago

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

Sudeep Salgia - 4 years, 10 months ago

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