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January 16th, 2011, 11:03 AM   #1
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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 ^^!
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January 16th, 2011, 12:10 PM   #2
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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).
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January 16th, 2011, 12:29 PM   #3
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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!
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January 16th, 2011, 12:52 PM   #4
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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.
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January 16th, 2011, 01:07 PM   #5
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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]
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January 16th, 2011, 01:10 PM   #6
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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.
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January 16th, 2011, 01:23 PM   #7
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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.













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January 16th, 2011, 01:44 PM   #8
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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

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January 16th, 2011, 02:12 PM   #9
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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!
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January 16th, 2011, 10:21 PM   #10
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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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