USAMO qualifying 2 questions

  • #1
:surprise: Suppose we are playing a board game with the following rules:

1. The board has 5 spaces on it.

2. Your playing piece starts out on space number 1.

3. Each move consists of moving the piece to a randomly chosen space other than the one which you currently occupy (for example, if you are on space number 2, you randomly move to either space 1, 3, 4, or 5, with each space having equal chance of being selected, but you cannot stay on space number 2).

4. You win if, after seven moves, you are back on space 1.

The probability of winning this game is p. Find 4096p.




:eek: Suppose log.4a (40*sqrt(3)) = log.3a (45)
(Log.m (n) denotes log base m of n. sqrt(x) denotes the square root of x.)
Find a^3 (a^3 denotes "a cubed.")
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
a^3 = 75.

Will work on Q1 now...
 
  • #3
129
0
hmmmm... 4096 = 2^12 = (2^2)^6 = 4^6 = (5-1)^(7-1)

interesting question.
 
Last edited:
  • #4
695
0
Ehm, 4096 = 2^12.

For the second question, consider rewriting both logs with the same base.
 
  • #5
uart
Science Advisor
2,776
9
Answer to Q1 is p=819/4096
 
  • #6
uart
Science Advisor
2,776
9
You know I had no problem with Q1. But in Q2 I couldn't do it without pluging in numbers and doing what parts I could do on a calculator (windows calc). So instead of getting a=75^(1/3) as Gokul did I just got whole bunch of messy numerical intermediates before getting a final answer of a=4.21716332650874621422854875703.

Actually what I had was :
a = 10^(log(k)/(m-1))

where m= log.45(40r3) and k=4/(3^m) were the numerical intermediates.


I still cant figure why it comes out to 75^(1/3), does this mean I'd fail the test :cry:
 
Last edited:
  • #7
977
1
hello3719 said:
hmmmm... 4096 = 2^12 = (2^2)^6 = 4^6 = (5-1)^(7-1)

interesting question.
Muzza said:
Ehm, 4096 = 2^12.

For the second question, consider rewriting both logs with the same base.
You do realize you were asked to find 4096p, in other words find p (the probability of winning the game)?
 
  • #8
695
0
Yes, but thanks for clarifying (we're not morons)... hello3719 originally wrote 4096 = 2^14 (but has since edited his post), and I thought I'd correct it.
 
Last edited:
  • #9
129
0
In some contests you don't always have time to solve the problem completely so you must use some shortcuts. I didn't claim i've solved the problem only getting the feel of it by wondering why they want us to find 4096p and not EXPLICTLY p. Everyone knows how to multiply by 4096 when they get p, but sometimes that isn't the goal of the problem.
 
  • #10
10
0
why it is 75

Hmm... #2 is fairly simple. I'll write my solution as a good LaTeX practice...

[tex] \log_{4a}40\sqrt3 = \log_{3a}45 [/tex]
[tex] \frac{\log40\sqrt3}{\log4a} = \frac{\log45}{\log3a} [/tex]
[tex] \frac{\log40\sqrt3}{\log45} = \frac{\log4a}{\log3a} [/tex]
[tex] \log_{45}{40\sqrt3} = \log_{3a}{4a} [/tex]
[tex] (3a)^{\log_{45}{40\sqrt3}} = 4a [/tex]
[tex] (3a)^{\log_{45}{40\sqrt3}-1} = \frac{4}{3} [/tex]
[tex] a^3 = (\frac{4}{3^{\log_{45}{40\sqrt3}}})^{ \frac{3}{\log_{45}{40\sqrt3} - 1} [/tex]
after obvious modification i am too lazy to type up
[tex] y = \log_{45}{40\sqrt3} - 1 [/tex]
[tex] a^3 = (\frac{4}{3})^{\frac{3}{y}}(1/27) [/tex]
here is the annoying part
[tex] y = \log_{45}{40\sqrt3} - 1 = \log_{45}{(40/45)\sqrt3} [/tex]
[tex] = \log_{45}{(8/9)\sqrt3} = \log_{45}{4^{3/2}/3^{3/2}} [/tex]
[tex] = 3/2 \log_{45}{4/3} [/tex]
then magically,
[tex] a^3 = (\frac{4}{3})^{\frac{3}{3/2 \log_{45}{4/3}}}(1/27) [/tex]
[tex] = (\frac{4}{3})^{2\log_{4/3}{45}}(1/27) [/tex]
[tex] = 45^2(1/27) = 81\times25\times\frac{1}{27} = 75 [/tex]
Whoopdeedoo.

I hate this question.
Contests are a waste of time.
 
  • #11
uart
Science Advisor
2,776
9
Thanks for the details AmirSafavi. I went through the first three steps much the same as yourself, but once I had one side of the equation purely numerical it was just too irresistable for me to finish it using floating point approx.
 
  • #12
I'd say that about takes care of these 2 questions. lol!
 
  • #13
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
Here's another way to solve Q2.

Let log.4a (40*sqrt(3)) = x = log.3a (45)
Then, (4a)^x = 40*sqrt(3) and (3a)^x = 45. Now take logs on both sides...
So, x*log(4a) = log(40*sqrt(3)) and x*log(3a) = log(45). Subtract one of these equations from the other...
You get, x*log(4/3) = log(40*sqrt(3)/45). But 40*sqrt(3)/45 = 8*sqrt(3)/9 = (4*sqrt(4)/3*sqrt(3) = (4/3)^1.5
So x = 1.5 , plug this into (4a)^x = 40*sqrt(3) to get 8*a^1.5 = 40*sqrt(3) or a^1.5 = 5*sqrt(3)

Hence a^3 = 25*3 = 75

Simple, wot ?
 
  • #14
That explanation hurts my eyes, but after reading it, that is a more simpler way of solving this problem.
 

Related Threads on USAMO qualifying 2 questions

  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
7
Views
3K
Replies
13
Views
7K
Replies
7
Views
880
Replies
6
Views
938
Top