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AIME Marathon
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AwesomeToad
Yang-Mills Theory
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#1
AIME Marathon
Everyone take a look at this
To get to Olympiad, you get through AIME, a key step for many people.

I have decided to start a marathon covering problems at these levels:

1) Hard AMC 12 level (such as 20-25)
2) Easy AIME (1-5)
3) More difficult AIME Problems (6-9 or 10)
4) Very hard AIME Problems (basically the last 5)
5) Beginning Olympiad

Please do not post problems that frequently appear on national or international olympiads, and do not post too easy problems(such as AMC 10 1-5)

If possible please classify which category a problem is when you post it (listed above)

First Problem: Easy AIME
Determine 3x_4+2x_5 if x_1, x_2, x_3, x_4, and x_5 satisfy the system of equations below.
2x_1+x_2+x_3+x_4+x_5=6
x_1+2x_2+x_3+x_4+x_5=12
x_1+x_2+2x_3+x_4+x_5=24
x_1+x_2+x_3+2x_4+x_5=48
x_1+x_2+x_3+x_4+2x_5=96

PostPosted: Thu Sep 03, 2009 8:26 am  Back to top 
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RoFlLoLcOpT
Yang-Mills Theory
Yang-Mills Theory

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#2
How about just use stars to represent the difficulty of the problem? For example, yours would be (\star\star).

Solution

Setting S=x_1+x_2+x_3+x_4+x_5, we have \begin{align*}x_1+S&=6\\x_2+S&=12\\x_3+S&=24\\x_4+S&=48\\s_5+S&=96.\end{align*}. Summing these 5 equations gives us \begin{align*}(x_1+x_2+x_3+x_4+x_5)+5S&=186\\6S&=186\\S&=31.\end{align*}. Finally, \begin{align*}3x_4+2x_5&=3(48-31)+2(96-31)\\&=3\cdot17+2\cdot65\\&=51+130\\&=\boxed{181}.


Problem 2

(\star\star)

Find x^2+y^2 if x and y are positive integers such that \begin{align*}xy+x+y&=71\\x^2y+xy^2&=880.\end{align*}.

PostPosted: Thu Sep 03, 2009 8:59 am  Back to top 
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Caelestor
Riemann Hypothesis
Riemann Hypothesis

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#3
I swear I've seen these problems before.

Solution #2

Note (x+1)(y+1)=72=2^3\cdot3^2 and xy(x+y)=880=11\cdot2^4\cdot5. It's easy to see WLOG x=11, y=5. Thus x^2+y^2=\boxed{146}.


Someone put up another problem; I'd like my original problems to go into mock AMC 12s.
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A collection of problems @ CC&II!

Try a Mock AMC: 9/09, 11/09 under construction.

PostPosted: Thu Sep 03, 2009 9:22 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#4
RoFlLoLcOpT wrote:
How about just use stars to represent the difficulty of the problem? For example, yours would be (\star\star).


Good idea.Smile

Another thing: Please do not look at past tests to see the answers.

[hide=\star\star]Find the number of pairs (a,b) so no digits of a or b are zero and a+b=100[/hide]

PostPosted: Thu Sep 03, 2009 12:58 pm  Back to top 
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RoFlLoLcOpT
Yang-Mills Theory
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#5
Oh, I don't think you can put the stars in a hide tag... I just put it above the problem. Smile

Also, I think you should include the problem number in the hide tag.

PostPosted: Thu Sep 03, 2009 1:58 pm  Back to top 
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Ihatepie
Navier-Stokes Equations
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#6
Alternate Solution for two
Let xy=a and x+y=b. So we have a+b=71 and ab=880. Solving, we get either (a,b) as either (55,16) or (16,55). Obviously, the latter is impossible with integers, so it is the former. We can then see that (x,y) is either (5,11) or (11,5) which both yield the same answer of 146.


@toad, I assume you mean positive integers?
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2010 Goals: ARML-7 AMC10- 144 AMC12- 126 AIME- 8 USAJMO-14?

PostPosted: Thu Sep 03, 2009 3:35 pm  Back to top 
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AIME15USAMO
Poincare Conjecture
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#7
Solution
So we have to divide 100 into 2. There are \binom{101}{1} = 101 ways to do this. Out of these 101 solutions, we can't use the ones where a = 0,10,20,30,40,50,60,70,80,90,100 That is 11 solutions. 101 - 11 = \boxed{90}

(\star\star\star\star) why do my stars look weird?
Problem 4
The decimal representation of \dfrac{m}{n} where m and n are relatively prime positive integers and m < n, contains the digits, 2, 5, and 1 consecutively, and in that order. Find the smallest possible value of n for which this is possible.
_________________
Goals: | AMC10 : 150 | AMC12 : 132+ | AIME : 9+ | USAMO : score positive
\sim \text{AIME15USAMO}\sim

PostPosted: Thu Sep 03, 2009 7:22 pm  Back to top 
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basketball9
Hodge Conjecture
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#8
Solution to Problem 4
Plase correct me if i am wrong

Since the decimal representation is 251 in that order m/n is equal to some non negative integer.251. Let call this integer x. Now if this x is positive then m>n whch it cannot be and it cannot be negative anyways. Therefore x equals 0 and m/n is equal to 0.251 which equals 251/1000. 251 and 1000 are relatively prime so n must equal 1000.


If I am right I dont understand why bvecause shouldnt AIME problems be from 0 to 999

Anyways heres a problem
ILL rate it three stars just because its like a number 10 but
...This should be easy(I can solve it)

In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum,N . If told the value of , the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if N=3194

PostPosted: Fri Sep 04, 2009 1:46 am  Back to top 
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themorninglighttt
Riemann Hypothesis
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#9
AIME15USAMO wrote:

(\star\star\star\star) why do my stars look weird?
Problem 4
The decimal representation of \dfrac{m}{n} where m and n are relatively prime positive integers and m < n, contains the digits, 2, 5, and 1 consecutively, and in that order. Find the smallest possible value of n for which this is possible.


i too cannot find a value of n less than 1000 that fits it. maybe it's m > n. or i just am not thinking about it hard enough.

PostPosted: Fri Sep 04, 2009 5:48 am  Back to top 
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basketball9
Hodge Conjecture
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#10
..
but then again if m>n then cant there be a million different decimal representations

PostPosted: Fri Sep 04, 2009 6:05 am  Back to top 
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RoFlLoLcOpT
Yang-Mills Theory
Yang-Mills Theory

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#11
Does it say it has to be exactly 0.251? Unless I am mistaken, it's possible for the fraction to be \frac{m}{n}=\frac{251}{999}, although that is very unlikely to be the smallest value of n.

PostPosted: Fri Sep 04, 2009 7:19 am  Back to top 
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Caelestor
Riemann Hypothesis
Riemann Hypothesis

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#12
I've seen something like this before, so I'm at an unfair advantage Ninja Unfortunately, I shouldn't solve it, since I couldn't solve it by myself (this is an AIME #14 and I looked at the solution).

To answer a few questions, no it just says it has to contains 251 somewhere. (Heck, it doesn't need to have a repeating sequence of 251). Becomes much harder now, no?
_________________
A collection of problems @ CC&II!

Try a Mock AMC: 9/09, 11/09 under construction.

PostPosted: Fri Sep 04, 2009 8:10 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#13
Ihatepie wrote:
@toad, I assume you mean positive integers?

Yes, sorry for not clarifying Blush

New Problem
An integer is called snakelike if its decimal representation a_1a_2a_3\cdots a_k satisfies a_i<a_{i+1} if i is odd and a_i>a_{i+1} if i is even. How many snakelike integers between 1000 and 9999 have four distinct digits?


This is a \star\star\star problem (Medium AIME)

Please do not try to look up the solution!

PostPosted: Fri Sep 04, 2009 8:13 am  Back to top 
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Ihatepie
Navier-Stokes Equations
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#14
Caelestor wrote:
I've seen something like this before, so I'm at an unfair advantage Ninja Unfortunately, I shouldn't solve it, since I couldn't solve it by myself (this is an AIME #14 and I looked at the solution).

To answer a few questions, no it just says it has to contains 251 somewhere. (Heck, it doesn't need to have a repeating sequence of 251). Becomes much harder now, no?
Doesn't it? (because all fractions are repeating decimals.)

@toad we still have two problems going

Edit: I got 909 as a denominator, but I really doubt that is the lowest.
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2010 Goals: ARML-7 AMC10- 144 AMC12- 126 AIME- 8 USAJMO-14?

PostPosted: Fri Sep 04, 2009 1:38 pm  Back to top 
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sdkudrgn88
Riemann Hypothesis
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#15
Ihatepie wrote:
Edit: I got 909 as a denominator, but I really doubt that is the lowest.


I agree. I got 800 easily (201/800 = 0.25125).

PostPosted: Fri Sep 04, 2009 5:44 pm  Back to top 
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Ihatepie
Navier-Stokes Equations
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#16
How did you find that out?
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2010 Goals: ARML-7 AMC10- 144 AMC12- 126 AIME- 8 USAJMO-14?

PostPosted: Fri Sep 04, 2009 5:53 pm  Back to top 
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sdkudrgn88
Riemann Hypothesis
Riemann Hypothesis

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#17
Ihatepie wrote:
How did you find that out?


Actually, that's not even the answer. But, it was easy, the two zeros at the end of 800 let the first two digits be anything I want it to be, and the 8 allows me to do 0.125.

In fact, I can make an even better number, 151/600 (0.251666666...).

PostPosted: Fri Sep 04, 2009 5:55 pm  Back to top 
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Ihatepie
Navier-Stokes Equations
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#18
Re: Solution to Problem 4
basketball9 wrote:
In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum,N . If told the value of , the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if N=3194

Solution
So all of it added up is equal to 3194. When you add abc to it, you get a multiple of 222. So we just try out subsequent abcs so 3194+abc=222x. Starting too low obviously won't work, so let's try abc=358 and x=3552. We find out this works so the answer is 358.
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2010 Goals: ARML-7 AMC10- 144 AMC12- 126 AIME- 8 USAJMO-14?

PostPosted: Fri Sep 04, 2009 6:11 pm  Back to top 
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sdkudrgn88
Riemann Hypothesis
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#19
NP, then?

\star \star \star \star \star

Find the largest integer, if it exists, that cannot be written as the sum of some number of distinct powers of 3 and 4. (3^0 and 4^0 are considered distinct.)

(For example, 29 = 3^2 + 3^1 + 3^0 + 4^2.)

I can't even solve this... Sad

If that's too hard: (\star \star \star)

Find all triples of positive integers (a, b, c) that satisfy a+b+c = abc.

PostPosted: Sat Sep 05, 2009 7:17 am  Back to top 
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Ihatepie
Navier-Stokes Equations
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#20
AwesomeToad wrote:
New Problem
An integer is called snakelike if its decimal representation a_1a_2a_3\cdots a_k satisfies a_i<a_{i+1} if i is odd and a_i>a_{i+1} if i is even. How many snakelike integers between 1000 and 9999 have four distinct digits?


This is a \star\star\star problem (Medium AIME)

Please do not try to look up the solution!
I remember doing this in a few minutes the first time I saw it in the Intro to C&P book. I hope that doesn't mean I shouldn't do it. maybe
Solution

Case 1: Let's first assume all of the digits are non-0. For simplicity sake, let's first assume that the digits are 1,2,3, and 4. There are 5 ways to arrange them so it works. 1324, 2314, 1423, 2413, and 3412. There are \binom{9}{4} ways to choose the digits.
Case 2: There are \binom{9}{3} ways to choose and 3 ways to arrange.
The answer is 5\binom{9}{4}+3\binom{9}{3}=630+252=882.

_________________
2010 Goals: ARML-7 AMC10- 144 AMC12- 126 AIME- 8 USAJMO-14?

PostPosted: Sat Sep 05, 2009 8:44 pm  Back to top 
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