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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
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Posts: 36
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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 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: 12,211 Thanks: 520 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
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Posts: 3,267
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Re: Several helps needed ^^

Hello, xx3004!

These certainly are competition-level problems . . .

Quote:
 [color=beige]. . . [/color]

[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: 12,211 Thanks: 520 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
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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. January 16th, 2011, 01:44 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:

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
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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: 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.
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:
 [color=beige]. . . [/color]

[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]
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.

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:

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

Thank again! Now I learn many new things, thank you all! I'm going to search for more tricks about divisibility! January 16th, 2011, 10:21 PM #10 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 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: Tags helps, needed ### what is antilogx÷antilogy

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