Tuymaada 2010, Junior League, Problem 5

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Tuymaada 2010, Junior League, Problem 5

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Octav

#1 h Jul 18, 2010, 9:21 AM • 2 Y

Y by Adventure10, Mango247

We have a set $M$ of real numbers with $|M|>1$ such that for any $x\in M$ we have either $3x-2\in M$ or $-4x+5\in M$ .
Show that $M$ is infinite.

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#2 h Jul 18, 2010, 11:21 PM • 2 Y

Y by Adventure10, Mango247

This is a solution, I don't think it is the best.
$M$ admits at least one element, let it be $x_0$ , then we have two cases:
1) If $x_0 \geq 1$ , then we found easily that : $3x_0 - 2 \geq x_0$ , and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$ , with $n$ to $\infty$ , we see that $M$ doesn't accept a bigger element, so M is infinite.
2) If $x_0 \leq 1$ , we proceed the same way : $-4x_0+5\geq x_0$ , and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite.
Done.

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#3 h Jul 19, 2010, 5:26 AM • 2 Y

Y by Adventure10, Mango247

oussama1305 wrote:

This is a solution, I don't think it is the best.
$M$ admits at least one element, let it be $x_0$ , then we have two cases:
1) If $x_0 \geq 1$ , then we found easily that : $3x_0 - 2 \geq x_0$ , and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$ , with $n$ to $\infty$ , we see that $M$ doesn't accept a bigger element, so M is infinite.
2) If $x_0 \leq 1$ , we proceed the same way : $-4x_0+5\geq x_0$ , and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite.
Done.

Your idea is good, but I think your solution should be fixed, because the problem says "either $3x-2\in M$ or $-4x+5\in M$ ."

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#4 h Jul 19, 2010, 6:17 AM • 2 Y

Y by Adventure10, Mango247

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Octav

#5 h Jul 19, 2010, 12:11 PM • 6 Y

Y by djb86, jmclaus, Adventure10, bin_sherlo, and 2 other users

The solution which I found during the contest and it seems the easiest

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#6 h Jul 19, 2010, 1:13 PM • 2 Y

Y by Adventure10, Mango247

vinhhop wrote:

oussama1305 wrote:

This is a solution, I don't think it is the best.
$M$ admits at least one element, let it be $x_0$ , then we have two cases:
1) If $x_0 \geq 1$ , then we found easily that : $3x_0 - 2 \geq x_0$ , and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$ , with $n$ to $\infty$ , we see that $M$ doesn't accept a bigger element, so M is infinite.
2) If $x_0 \leq 1$ , we proceed the same way : $-4x_0+5\geq x_0$ , and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite.
Done.

Your idea is good, but I think your solution should be fixed, because the problem says "either $3x-2\in M$ or $-4x+5\in M$ ."

Ah! So I need just to add one more case in each case, and discussing the smaller element, like discussing the fact that $-4x_0+5\leq x_0$ or $3x_0 - 2 \geq x_0$ in the first case for example, and it's done.

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#7 h Jul 23, 2010, 4:13 PM • 2 Y

Y by Adventure10, Mango247

hi octav
can you explain me with details your solution please?

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#8 h Jul 2, 2013, 3:07 PM • 1 Y

Y by Adventure10

To Octav you can't say that |3x-3|=3|x-1| counter example x=2.5

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#9 h Jul 2, 2013, 3:21 PM • 2 Y

Y by Adventure10, Mango247

proof: f(x)=3x-2 g(x)=-4x+5 then we can see that if (not necessarily order of f,g...) A= f(g(...g(x)...))=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 where * product of number and k number of g, t number of f. and we can se that if k+t don't equal to p+q where p,q number of f,g A not equal B where B=(-1)^q*4^q*3^p*x - (-1)^p*4^p*3^q + 1 so M is infinite.

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#10 h Jul 2, 2013, 3:43 PM • 1 Y

Y by Adventure10

akhan98 wrote:

To Octav you can't say that |3x-3|=3|x-1| counter example x=2.5

In fact, you can say that and the counterexample is invalid: $|3x-3|=|3\cdot 2.5-3|=|4.5|=4.5$ and $3|x-1|=3|2.5-1|=3\cdot 1.5=4.5$ . Actually, for any reals (or even complex) $a,b$ the following holds: $|a\cdot b|=|a| \cdot |b|$ .

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#11 h Jul 2, 2013, 7:24 PM • 2 Y

Y by Adventure10, Mango247

i thought that |a| means that the greatest integer less than or equal to x

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#12 h Jul 2, 2013, 7:30 PM • 2 Y

Y by Adventure10, Mango247

Sorry)

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#13 h Jul 2, 2013, 7:31 PM • 2 Y

Y by Adventure10, Mango247

to phantom and what do you think about my solution???

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#14 h Jul 2, 2013, 10:12 PM • 2 Y

Y by Adventure10, Mango247

@akhan98: No problem

. The integral part symbol is kind of confusing around here as far as I've noticed, so it is understandable.

EDIT: Regarding the solution, I don't really get what you meant with "number of f".

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#15 h Jul 3, 2013, 6:19 AM • 2 Y

Y by Adventure10, Mango247

@PhantomR: if A= f(g(...g(x)...)) in the expression there are k times g and t times f so A=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 because we can see that coefficient near the x is (-1)^k*4^k*3^t*x and if x=1 A=1 what means that A=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 and do you understand what I mean?

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