My Math Forum Several helps needed ^^

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 January 16th, 2011, 12:03 PM #1 Member   Joined: Dec 2010 From: Viet Nam and US Posts: 36 Thanks: 0 Several helps needed ^^ Good morning xD, I have a practice questions for my school competition, but there are some question take me 3 days and I still can't come up with any thing , I'll post them here and appreciate any help, but, please, explain how do you come up with that (not only write the answer). Thanks in advanced! 1) (THIS QUESTION I DON'T KNOW WHAT IS IT TALKING ABOUT SO I CAN'T SOLVE IT! PLEASE EXPLAIN FOR ME AND I'LL TRY ^^): The roots of the equation $x^3+x^2-6x+5=0$ are each increased by 1/3. If the new equation is written in the form $Ax^3+Bx^2+Cx+D=0$ Where A,B,C,D are relatively prime integers and A>0, then find the value of A+B+C+D. 2) Which of the following is not a factor of $3^{48}-1$? a)70 b)73 c)78 d)82 e)Other Attempt: I try to apply the theorem that if the sum of each digit of a number is divisible by 3, that number is divisible by 3, but it doesn't work! 3) Find the remainder when $7^{8^{13}}$ is divided by 11. 4) Antilogx + antilogy = ? 5) (AS QUESTION NUMBER 1, I DON'T KNOW WHAT IS IT TALKING ABOUT): Three real, positive geometric means are inserted between 8/9 and 288, what is the third one? 6) Which of the following expressions is equal to $\frac{-1+\sqrt{189}}{2}$? a) $\sqrt{17-{\sqrt{17-\sqrt{17-...}}}$ b) $\sqrt{21+{\sqrt{21+\sqrt{21+...}}}$ c)$\sqrt{47-{\sqrt{47-\sqrt{47-...}}}$ d)$\sqrt{63-{\sqrt{63-\sqrt{63-...}}}$ Attempt: I've tried to convert them into $(17-(17-....)^{\frac{1}{2})^{\frac{1}{2}}}$ but i'm stuck! 6) (THIS QUESTION I THINK RELATIVE SOMEHOW WITH QUESTION 3): What value of y make 43551234y13239136665 divisible by 11? That's all ^^!
 January 16th, 2011, 01:10 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Several helps needed ^^ A 'root' of an equation is meant as a solution, or a zero in this case. For example, the 'roots' of the equation x^2 + 5x + 6 = 0 are x = -2 and x=-3. In this one, each root is increased by 1/3. What is the new polynomial? Your approach to 2 is odd, as we know 3 does not divide 3^48 - 1. (it divides 3, but not 1).
January 16th, 2011, 01:29 PM   #3
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Joined: Dec 2010
From: Viet Nam and US

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Re: Several helps needed ^^

Quote:
 Originally Posted by DLowry A 'root' of an equation is meant as a solution, or a zero in this case. For example, the 'roots' of the equation x^2 + 5x + 6 = 0 are x = -2 and x=-3. In this one, each root is increased by 1/3. What is the new polynomial? Your approach to 2 is odd, as we know 3 does not divide 3^48 - 1. (it divides 3, but not 1).
Thanks for helping Dlowry!

1) I'm not sure what you mean, we know that 3^48 divisible by 3, so 3^48-1 is not divisible by 3, so how can I know which of the followings is not a factor of it ?

2) For the equation, for example the 3 solution is $x_1, x_2, x_3$ so will they increase by 1/3 like $x_1+\frac{1}{3}, x_2+\frac{1}{3}, x_3+\frac{1}{3}$, so how about the rest? If we rewrite the equation, are A, B, C, D still respectively the same 1, 1, -6, 5. WOW it's messed already.... I don't know what they're asking!

 January 16th, 2011, 01:52 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Several helps needed ^^ For question 6, set: $y=\sqrt{x\pm\sqrt{x\pm\sqrt{x\pm\cdots}}}$ Square both sides: $y^2=x\pm y$ Write in standard quadratic form: $y^2\mp y-x=0$ Application of the quadratic formula (ignoring the negative root as extraneous) yields: $y=\frac{\pm1+\sqrt{1+4x}}{2}$ Under the radical we require: $1+4x=189$ $x=47$ We see also that we need a negative sign in front of the 1 in the numerator for y, thus c) is the answer.
January 16th, 2011, 02:07 PM   #5
Math Team

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From: Lexington, MA

Posts: 3,267
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Re: Several helps needed ^^

Hello, xx3004!

These certainly are competition-level problems . . .

Quote:
 $\text{2) Which of the following is }not\text{ a factor of }3^{48}-1\,?$ [color=beige]. . . [/color]$(a)\;70 \;\;\;\; (b)\;73 \;\;\;\; (c)\;78 \;\;\;\; (d)\;82 \;\;\;\; (e)\text{ Other}$

$N \;=\;3^{48}\,-\,1\,\text{ is the difference of two squares: }\;3^{24}^2\,-\,1^2 \;=\;(3^{24}\,-\,1)\,(3^{24}\,+\,1)$

$3^{24}\,-\,1\text{ is a difference of squares:}$

[color=beige]. . [/color]$(3^{12}\,-\,1)(3^{12}\,+\,1) \;=\;(3^6\,-\,1)(3^6\,+\,1)(3^4\,+\,1)(3^8 \,-\,3^4\,+\,1) \;=\;(72(730)(82)(6481)" />

[color=beige]. . [/color]$=\;(2^3\cdot7\cdot13)\,(2\cdot5\cdot73)\,(2\cdot41 )\,(6481) \;=\;2^5\cdot5\cdot7\cdot13\cdot41\cdot73\cdot6481$

$3^{24}\,+\,1\text{ is a sum of cubes.}$

[color=beige]. . [/color]$(3^8\,+\,1)(3^{16}\,-\,3^8\,+\,1) \;=\;(6562)(43,040,161) \;=\;(2\cdot17\cdot193)(43,040,161)$

$\text{Hence: }\;N\;=\;3^{48}\,-\,1 \;=\;2^6\,\cdot\,5\,\cdot\,7\,\cdot\,13\,\cdot\,17 \,\cdot\,41\,\cdot\,73\,\cdot\,193\,\cdot\,6481\,\ cdot\,40,040,161$

$\text{We can see that }N\text{ is divisible by: }\;\begin{Bmatrix}70 \:=\:2\,\cdot\,5\,\cdot\,7 \\ \\ \\ \\ 82 \:=\:2\,\cdot\,41 \\ \\ \\ \\ \text{and } 73 \end{Bmatrix}$

$\text{Therefore: }\:78 \:=\:2\cdot3\cdot13\text{ is }not\text{ a factor of }N.$

That's a lot of work for one competition problem.
I hope someone can find a more efficient method.
[color=beige] .[/color]

 January 16th, 2011, 02:10 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Several helps needed ^^ For problem 2, observe that all of the factors of $3^{\small{48}}-1$ will be of the form $3n\pm1$ so a multiple of 3 cannot be a factor.
January 16th, 2011, 02:23 PM   #7
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Several helps needed ^^

Hello again, xx3004!

Here's another one . . .

Quote:
 5) Three real, positive geometric means are inserted between 8/9 and 288.[color=beige] .[/color]What is the third one?

We have 5 numbers in increasing order; they form a geometric sequence.

$\text{Let r\text{= the common ratio.}$

$\text{The the five terms are: }\;\frac{8}{9}\,,\;\;\frac{8}{9}r\,,\,\;\;\frac{8} {9}r^2\,,\;\;\frac{8}{9}r^3\,,\;\;288$

$\text{But this means: }\;\frac{8}{9}r^4 \:=\:288$

$\text{Solve for }r:\;\;r^4 \:=\:324 \;\;\;\Rightarrow\;\;\;r^2 \:=\:18 \;\;\;\Rightarrow\;\;\;r \:=\:\sqrt{18} \:=\:3\sqrt{2}$

$\text{The third term is: }\;\frac{8}{9}r^3 \;=\;\frac{8}{9}(3\sqrt{2})^3 \;=\;\frac{8}{9}(27\,\cdot\,2\sqrt{2}) \;=\;48\sqrt{2}$

January 16th, 2011, 02:44 PM   #8
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Several helps needed ^^

Hello again, xx3004!

Here's yet another one . . .

Quote:
 $\text{6. What value of }y\text{ makes }43551234y13239136665\text{ divisible by 11\,?}$

For this problem, you're expected to know a particular divisibility trick.

Given a positive integer:
[color=beige]. . [/color]add the digits in the "odd" positions (first, third, fifth, etc.),
[color=beige]. . [/color]add the digits in the "even" positions (second, fourth, sixth, etc.),
[color=beige]. . [/color]subtract these two sums.

If the result is divisible by 11, then the original number is divisible by 11.

In the odd positions, we have:
[color=beige]. . [/color]$4\,+\,5\,+\,1\,+\,3\,+\,y\,+\,3\,+\,3\,+\,1\,+\,6\ ,+\,6 \;=\;y\,+\,32$

In the even positions, we have:
[color=beige]. . [/color]$3\,+\,5\,+\,2\,+\,4\,+\,1\,+\,2\,+\,9\,+\,3\,+\,6\ ,+\,5 \;=\;40$

The difference is:[color=beige] .[/color]$(y\,+\,32)\,-\,40 \;=\;y\,-\,8$

If $y\,-\,8$ is divisible by 11, then $y \,=\,8.$

January 16th, 2011, 03:12 PM   #9
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Joined: Dec 2010
From: Viet Nam and US

Posts: 36
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Re: Several helps needed ^^

Quote:
 Originally Posted by MarkFL For question 6, set: $y=\sqrt{x\pm\sqrt{x\pm\sqrt{x\pm\cdots}}}$ Square both sides: $y^2=x\pm y$ Write in standard quadratic form: $y^2\mp y-x=0$ Application of the quadratic formula (ignoring the negative root as extraneous) yields: $y=\frac{\pm1+\sqrt{1+4x}}{2}$ Under the radical we require: $1+4x=189$ $x=47$ We see also that we need a negative sign in front of the 1 in the numerator for y, thus c) is the answer.
Wow that's a really good trick! I didn't even recognize it, thanks alot!

Quote:
Originally Posted by soroban
Hello, xx3004!

These certainly are competition-level problems . . .

Quote:
 $\text{2) Which of the following is }not\text{ a factor of }3^{48}-1\,?$ [color=beige]. . . [/color]$(a)\;70 \;\;\;\; (b)\;73 \;\;\;\; (c)\;78 \;\;\;\; (d)\;82 \;\;\;\; (e)\text{ Other}$

$N \;=\;3^{48}\,-\,1\,\text{ is the difference of two squares: }\;3^{24}^2\,-\,1^2 \;=\;(3^{24}\,-\,1)\,(3^{24}\,+\,1)$

$3^{24}\,-\,1\text{ is a difference of squares:}$

[color=beige]. . [/color]$(3^{12}\,-\,1)(3^{12}\,+\,1) \;=\;(3^6\,-\,1)(3^6\,+\,1)(3^4\,+\,1)(3^8 \,-\,3^4\,+\,1) \;=\;(72(730)(82)(6481)" />

[color=beige]. . [/color]$=\;(2^3\cdot7\cdot13)\,(2\cdot5\cdot73)\,(2\cdot41 )\,(6481) \;=\;2^5\cdot5\cdot7\cdot13\cdot41\cdot73\cdot6481$

$3^{24}\,+\,1\text{ is a sum of cubes.}$

[color=beige]. . [/color]$(3^8\,+\,1)(3^{16}\,-\,3^8\,+\,1) \;=\;(6562)(43,040,161) \;=\;(2\cdot17\cdot193)(43,040,161)$

$\text{Hence: }\;N\;=\;3^{48}\,-\,1 \;=\;2^6\,\cdot\,5\,\cdot\,7\,\cdot\,13\,\cdot\,17 \,\cdot\,41\,\cdot\,73\,\cdot\,193\,\cdot\,6481\,\ cdot\,40,040,161$

$\text{We can see that }N\text{ is divisible by: }\;\begin{Bmatrix}70 \:=\:2\,\cdot\,5\,\cdot\,7 \\ \\ \\ \\ 82 \:=\:2\,\cdot\,41 \\ \\ \\ \\ \text{and } 73 \end{Bmatrix}$

$\text{Therefore: }\:78 \:=\:2\cdot3\cdot13\text{ is }not\text{ a factor of }N.$

That's a lot of work for one competition problem.
I hope someone can find a more efficient method.
[color=beige] .[/color]
I was so closed! I factored it out for the (a^2-b^2) already, but I don't know what to do with 3^24+1 , now I obviously get it. Thank alot Soroban! But hang on, I am not allowed to use calculator, how can I come up with that big number?

Quote:
Originally Posted by soroban
Hello again, xx3004!

Here's another one . . .

Quote:
 5) Three real, positive geometric means are inserted between 8/9 and 288.[color=beige] .[/color]What is the third one?

We have 5 numbers in increasing order; they form a geometric sequence.

$\text{Let r\text{= the common ratio.}$

$\text{The the five terms are: }\;\frac{8}{9}\,,\;\;\frac{8}{9}r\,,\,\;\;\frac{8} {9}r^2\,,\;\;\frac{8}{9}r^3\,,\;\;288$

$\text{But this means: }\;\frac{8}{9}r^4 \:=\:288$

$\text{Solve for }r:\;\;r^4 \:=\:324 \;\;\;\Rightarrow\;\;\;r^2 \:=\:18 \;\;\;\Rightarrow\;\;\;r \:=\:\sqrt{18} \:=\:3\sqrt{2}$

$\text{The third term is: }\;\frac{8}{9}r^3 \;=\;\frac{8}{9}(3\sqrt{2})^3 \;=\;\frac{8}{9}(27\,\cdot\,2\sqrt{2}) \;=\;48\sqrt{2}$

Well I don't want to repeat the same word, but thank again! This problem I just don't understand what is it talking about, especially the word "means"! It turns out easier when I understand the problem, this is the basic problem ^^!

Quote:
Originally Posted by soroban
Hello again, xx3004!

Here's yet another one . . .

Quote:
 $\text{6. What value of }y\text{ makes }43551234y13239136665\text{ divisible by 11\,?}$

For this problem, you're expected to know a particular divisibility trick.

Given a positive integer:
[color=beige]. . [/color]add the digits in the "odd" positions (first, third, fifth, etc.),
[color=beige]. . [/color]add the digits in the "even" positions (second, fourth, sixth, etc.),
[color=beige]. . [/color]subtract these two sums.

If the result is divisible by 11, then the original number is divisible by 11.

In the odd positions, we have:
[color=beige]. . [/color]$4\,+\,5\,+\,1\,+\,3\,+\,y\,+\,3\,+\,3\,+\,1\,+\,6\ ,+\,6 \;=\;y\,+\,32$

In the even positions, we have:
[color=beige]. . [/color]$3\,+\,5\,+\,2\,+\,4\,+\,1\,+\,2\,+\,9\,+\,3\,+\,6\ ,+\,5 \;=\;40$

The difference is:[color=beige] .[/color]$(y\,+\,32)\,-\,40 \;=\;y\,-\,8$

If $y\,-\,8$ is divisible by 11, then $y \,=\,8.$

Thank again! Now I learn many new things, thank you all! I'm going to search for more tricks about divisibility!

 January 16th, 2011, 11:21 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Several helps needed ^^ For the first problem, if we horizontally shift the given polynomial 1/3 units to the right, the roots will all be increased by 1/3. $$$x-\frac{1}{3}$$^3+$$x-\frac{1}{3}$$^2-6$$x-\frac{1}{3}$$+5=0$ $$$x^3-x^2+\frac{1}{3}x-\frac{1}{27}$$+$$x^2-\frac{2}{3}x+\frac{1}{9}$$+$$-6x+2$$+5=0$ $x^3+0\cdot x^2-\frac{19}{3}x+\frac{191}{27}=0$ Multiply through by 27: $27x^3+0\cdot x^2-171x+191=0$ The sum of the coefficients is then: $27-171+191=47$

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