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January 16th, 2011, 11:03 AM  #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 are each increased by 1/3. If the new equation is written in the form 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 ? 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 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 ? a) b) c) d) Attempt: I've tried to convert them into 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, 12:10 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, 12:29 PM  #3  
Member Joined: Dec 2010 From: Viet Nam and US Posts: 36 Thanks: 0  Re: Several helps needed ^^ Quote:
1) I'm not sure what you mean, we know that 3^48 divisible by 3, so 3^481 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 so will they increase by 1/3 like , 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, 12:52 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs  Re: Several helps needed ^^
For question 6, set: Square both sides: Write in standard quadratic form: Application of the quadratic formula (ignoring the negative root as extraneous) yields: Under the radical we require: 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, 01:07 PM  #5  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Several helps needed ^^ Hello, xx3004! These certainly are competitionlevel problems . . . Quote:
[color=beige]. . [/color](730)(82)(6481)" /> [color=beige]. . [/color] [color=beige]. . [/color] 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, 01:10 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs  Re: Several helps needed ^^
For problem 2, observe that all of the factors of will be of the form so a multiple of 3 cannot be a factor.

January 16th, 2011, 01:23 PM  #7  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Several helps needed ^^ Hello again, xx3004! Here's another one . . . Quote:
We have 5 numbers in increasing order; they form a geometric sequence.  
January 16th, 2011, 01:44 PM  #8  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Several helps needed ^^ Hello again, xx3004! Here's yet another one . . . Quote: 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] In the even positions, we have: [color=beige]. . [/color] The difference is:[color=beige] .[/color] If is divisible by 11, then  
January 16th, 2011, 02:12 PM  #9  
Member Joined: Dec 2010 From: Viet Nam and US Posts: 36 Thanks: 0  Re: Several helps needed ^^ Quote:
Quote:
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January 16th, 2011, 10:21 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 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. Multiply through by 27: The sum of the coefficients is then: 

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