1.) Find the value of \(x\) such that
⌊x+10011⌋+⌊x+10012⌋+⌊x+10013⌋+⋯+⌊x+10099⌋=761
2.) Do the following.
2.1) Prove the identity (x2−y2)2+(2xy)2=(x2+y2)2
2.2) Use the identity (2.1) to find all the positive integer solutions of x2+y2=34(x−y)
3.) Let α,β be the roots of 3x2−(α2+1)x+3=6(α+β), find the value of α+β
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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2^{34}
a_{i-1}
\frac{2}{3}
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\sum_{i=1}^3
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\boxed{123}
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Top Newest1) 20169=8.45
2.1) Just expand
2.2) 12 integral solutions
3) 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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[x]+[x+n1]+[x+n2]…[x+nn−1]=[nx]
2.2. Draw the circle(passes through origin)
3) 2 eq. 2 variable I got α=1,β=3−1
4) Is a quadratic equation consider
ax2+bx+c
Given
c = 91 And 2 other equations you can solve
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P(x) quadratic?
4) Why isLog in to reply
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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Somehow the calculator said that the answer for 1) is
8.49≤x<8.5
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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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Trying 2.2)
(x2+y2)2=1156(x−y)2
(x2−y2)2+4x2y2=1156(x−y)2
(x−y)2(x+y)2+4x2y2=1156(x−y)2
4x2y2=(1156−(x+y)2)(x−y)2
2xy=(342−(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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In the 3 rd i got alpha + beta as 0 ? am i correct would anyone tell ?
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For 2.1 a neater way would be to use complex numbers. x2+y2=(x+iy)(x−iy). Hence squaring it would give us RHS=(x2−y2+i2xy)(x2−y2−i2xy). I believe what follows is obvious.
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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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