AIME Marathon

[AwesomeToad]

NP

The number can be written as where and are positive integers. Find

This is classified as (Easy AIME)

[jxl28]

Solution

We compare parts and concluded that

Thus

Multiplying them together yields

Since , , and are positive, our answer is simply

NP

An infinite geometric series has sum . A new series, obtained by squaring each term of the original series, has times the sum of the original series. The common ratio of the original series is where and are relatively prime integers. Find . (AIME 2005 II)

[remy1140]

solution

We are given that


Using the formula for infinite geometric series, we obtain the followings:


We rearrange these equations and find:

Since , we cancel .

Therefore,

NP

In parallelogram , point is on so that and point is on so that Let be the point of intersection of and . Find .

[AwesomeToad]

We need to get back to this again, this lasted so shortly.

Solution

We can make the whole parallelogram a straight line, for this problem. We let and have length so

New Problem

Let and such that Find

[crazypianist1116]

[mathwizarddude]

Seriously I don't see the whole point of reposting AIME problems when their solutions are already accessible (in the resource section)... are you expecting a better solution then?
Here's the next problem () which I believe is not in the resource section:

Let be the sum of the digits of when written in binary. Let be the sum of the digits of in its usual base 10 representation.

For which values of does ?

[Caelestor]

I made that point earlier. Instead, why don't people collaborate and make a mock AIME? (Unfortunately, I can't write problems past #10...)

[andersonw]

A few of us that went to AMP (Zhero, aneo., fwolth, and abacadaea who was not in AMP) are already in the process of writing a mock AIME/USAJMO/USAMO. We're about 2/3 done, although that was mainly during the summer, and we really haven't discussed it since then. We'll probably be finished by the time AMC season comes around, though.

[mfn]

For which values of n < 100 does d(n) = b(n)?
Click to reveal hidden content

The maximum of the function b(n) in the interval 0 < n < 100 is 6, which first happens when n = 63. This means that d(n) has to be equal to or less than 6, which in turn implies that n is equal to or less than 60.

I just did a bunch of trial and error from here..

Case: b(n) = 6. Impossible since b(n) never equals 6 for n < 63.

Case: b(n) = 5. This requires that n > 31. Listing all integers n with n > 31 and d(n) = 5 we get n = 50, 41, 32. b(n) does not equal 5 for any of these.

Case: b(n) = 4. This requires that n > 15. Listing all integers n with n > 15 and d(n) = 4 we get n = 40, 31, 22. b(n) does not equal 4 for any of these.

Case: b(n) = 3. n > 7, from 30, 21, and 12, only 21 works.
Case: b(n) = 2. n > 3, from 20 and 11, only 20 works.

n = 1 also works, so the solutions for b(n) = d(n) are n = 1, 20, and 21.

[mathwizarddude]

Here's a new problem for those interested in keeping the marathon going:

At the basement of a building with 5 floors, Adam, Bob, Cindy, Diana and Ernest entered the elevator. The elevator goes only up and doesn 't come back, and each person gets out of the elevator at one of the five floors. In how many ways can the five people leave the elevator in such a way that at no time are there a male and a female alone in the elevator?

[Ihatepie]

Am I right in assuming only one person gets off at a floor?
Solution

Case 1: The three boys get out first and the two girls get out in the following floors.
There are ways for the boys to get out and then for the girls, giving total ways.

Case 2: The two girls and one guy get out on the first three floors and the two other guys get out on the last 2.
There are ways to choose the boy, and then ways to arrange the guy/2 girls through the first three floors. Then, there are ways for the last two guys to get off. This gives a total of ways.
So the answer is .

NP:
This is a problem (Easy AIME)

A man is walking on the coordinate plane. He starts at and moves in steps of lengths one along the lines of the grid. Each direction of left, up, right, and down are equally likely. What is the probability that the man makes it to in 6 or fewer steps?

[mathwizarddude]

solution?

The equivalant probability is the probability of reaching the desired point at least once within 6 steps. The only possibilities are first reaching the point in 4 steps and first reaching the point in 6 steps. There are ways for reaching it in step (with whatever directions for the rest two steps).

To reach in 4 steps, we must have exactly 2 UPs and 2 RIGHTs. To reach in 6 steps, we must additional two directions added, with the directions being opposites (UP & DOWN and LEFT & RIGHT are opposites). In reaching in 4 steps, each of the paths pass through 5 points, and we can have ways to choose 2 of these with the first being inserted the additional direction and the second being inserted , or the opposite, to offset it. However, in the case with both of them being the desired point , these combinatios are included in the case of exactly 4 steps, and so we have to subtract them.

Thus the desired probability is

Is this correct? What is the desired solution (from whatever the source is)? Frankly it took me a while to figure out the case of first reaching the point in 6 steps.

If this is correct, here's the next problem (taken from the MIMC, but I believe the offered solution isn't rigorous enough):
On a road, we decide to give a name to each car that goes by, where a name consists of any set of letters. Each letter may only be used at most once. No two different cars may be given the same name. Two names which are merely permutations of each other are not considered different. Finally, each new name must have at least one leter in common with each of the names already given to cars before it. Let be the number of cars it takes for us to run out of names. Find the number of digits of .

[sdkudrgn88]

Inquiry

Just as a matter of interest, is the answer to the license-plate problem greater than ?

[mathwizarddude]

response

Definitely, since we can pick a letter (say A) so that it appears in every name, and there are subsets of the rest 25 letters.

By the way is my solution to Ihatepie's problem correct?

[Brut3Forc3]

[Ihatepie]

[mathwizarddude]

Whoops, I just realized my should be changed to , since first of all, even with my original logic, is incorrect when it should have been ; secondly, this doesn't work when we have something like RIGHT(RIGHT)RIGHT & RIGHT RIGHT(RIGHT), which are the same but are counted as different.

By the way for the bus problem, what if we are looking for the exact number instead of just the number of digit?

[sup3rcrash3r]

I think all those problems have been solved...

New Problem

A boy builds a 12-block-high tower, and wants to paint it with red and blue paint. However, he wants the tower to meet his aesthetic requirements, so he decides that he can have no more than three red blocks in a row. He also does not want any blue blocks to touch. Assuming that all the blocks are either red or blue, how many ways can the boy paint the tower?

[mathwizarddude]

messive ugly casework bash

Case 1 (6 blue): trivially there are ways
Case 2 (5 blue): , since after fixing 4 red inside the 4 gaps recreated by the 5 blue, there are 3 left, and there are only 4 cases with 4 or more red together (i.e, inside one of the 4 gaps between the blue)
Case 3 (4 blue): ; almost the same as the previous case using complementary counting; after fixing the first 7, there are 5 red left, with first case of having 4 on either end with another 5 possible positions for the last one, and with the second case of having a 3 in one of he 3 gaps, and ways for the other 2
Case 4 (3 blue):
_*_*_*_ where * = blue. again use complementay counting - there are 12-5 = 7 red left, and in order to have 4 or more red together, we can either have 4 or more red on one of the two ends, or a 3 or more in one of the 2 gaps:

1) 4 or more on either end:
2) 3 or more in the gap :

we have to subtract the cse with both 1) and 2) satisfied and the case where there are exactly 3 each for the two gaps: 4+2=6.

So the total number of ways is if I did not make any dumb mistake... is there a better way?

On a road, we decide to give a name to each car that goes by, where a name consists of any set of letters. Each letter may only be used at most once. No two different cars may be given the same name. Two names which are merely permutations of each other are not considered different. Finally, each new name must have at least one leter in common with each of the names already given to cars before it. Let N be the number of cars it takes for us to run out of names. Find a way to calculate the exact value of N.

For those that want another one, try problem #8 on this page
http://web.mit.edu/hmmt/www/datafiles/problems/1998/ADVTOP98.pdf

[sup3rcrash3r]

I used recursion to solve my problem. I find it to be easier and less time consuming. Also, you don't risk missing or overcounting a case.

Solution

Count the number of nth-block-high towers where the nth block is blue.

: Obviously, it can be blue or not, so there is only tower where it can be blue.

: Now, because no two blue blocks can touch, if the second block is blue, then the first must be red, so there is tower where the second block is blue.

: If the third block is blue, then the first block can be red or blue. possibilities

: If the fourth block is blue, then the third block is red. Then, the second block is red or blue. If it is red, the first block can be blue or red. If it is blue, the first block must be red. possibilities

Notice that for a given blue block block , the previous blue block can be , , or (it can't be , that block is red!). So the number of distinct towers that blue block can be in is the sum of the towers that all the potential previous blue blocks can be in.

From this, we get the recursion = , where is the number of number of distinct towers that meet our requirements.
We have the first four terms, so we can easily compute it up to .





Note that the solution is not . We must add up the those four terms of our recursion, because any of them could be the last blue block in our stack. We get .

EDIT: I just noticed a similar problem. AIME 2001 I number 14 is based on the same concept.
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001