SEN Q Challenge

The SEN Q Challenge is a series of academic questions that are asked to be answered competitively. In order to post a question to be answered by us awesome SEN'ers, you have to successfully answer the current question being asked!


Rules for Posting a Question:
The first person to successfully complete the question currently being asked is required to post the next question ASAP. When posting a question, you must include a link to the previous question asked in your post (to make things more organized). Diagrams and pictures are encouraged. When someone answers your question successfully, modify your question post to show the solution as well as an explanation to the solution in a drop down box. Also, state who answered your question successfully in your post. The questions can be from any of the following: Math, Physics, Biology, Chemistry (These are open for suggestions).




Link to Current Question: http://www.staredit.net/topic/8721/1/ Post #12

Post has been edited 7 time(s), last time on Oct 5 2009, 12:06 am by CecilSunkure.

That's.. not an AIME question.

Yes it is silly

145

EDIT:
Find all odd multiples of 29, cut off the last digit, and put that over the original.

145/45 = 3.22

45:145 != 1:29

lol fail, I read that as right-most. brb then.

14x29 != 145 either.

725

Correct!

Solution will be in the OP in just a sec.

Here was my solution:

#include <iostream>

int main()
{
using namespace std;

double uno = 1;
double dos = 29;
double moose = uno/dos;

cout << "1/29: " << moose << endl;

double temp;
for(int i = 1; i <= 1000; i++)
{
if(i < 100)
{
temp = i%10/double(i);
cout << i << "\t" << temp << endl;
if(temp == moose)
{
return 0;
}
}
else if(i >= 100 && i <= 1000)
{
temp = i%100/double(i);
cout << i << "\t" << temp << endl;
if(temp == moose)
{
return 0;
}
}
}

return 0;

}

Category: Math

The approximate value of log_2 (9), which you can obtain with a calculator, is 3.16993. . . Explain how you can figure out that 3 < log_2 (9) < 3.5 by integer multiplication or division (i.e. with pencil and paper and no calculator, reducing it to a computation with integers.)
Write x1 = log_2 (9) = n1 + y1 where n1 is the integer part of x1 and y1 is in (0,1) is the fractional part. Explain how you get the integer n1 by integer multiplication and division.
Define x2 = 1/y1. Interpret x2 as a logarithm.
Note that x2 > 1 so you can decompose x2 again into its integer and fractional part x2 = n2 + y2. How do you get n2 by integer multiplication and division?
You can now continue with x3 = 1/y2 and so on. After k steps you get bored and stop. How do you get an approximate value of x1 = log_2(9) from the integers n1, n2,..,nk, starting with the case k = 2?
You have found this way an algorithm using only multiplication and division for computing log_a (b) with arbitrary precision.

Edit: Note that log_2 (9) is simply the example given. The correct answer is a method that will work for any log_a (b).
Hint: the answer shares some similarity with infinite series, almost like an Euler product.

Post has been edited 1 time(s), last time on Oct 7 2009, 7:21 am by Vrael.

Vrael: Use moar proper notation. x¹ or x₁?

Also, with all those return(0);s, how did you see the answer?

x1 = x subscript 1.
n2 = n subscript 2, ect.
subscript or superscript notation is irrelevant, use whatever you like.

When a program in C++ hits a return statement during int main(), it ends the program. Basically,
if(temp ==moose)
{
return 0;
}

Says "if we found the right answer, end the program" and it stopped listing terms when i = 725, the correct answer.

Yeah, I know, I know some C++, but what I was asking is how did you know that the program ended at 725? Does whatever fancy environment you have not close the window at a return(0)?

Err yeah. I compile with Microsoft Visual C++ 2008. You could always do something like

cin.clear();
cin.ignore(255, '\n');
cin.get();
return 0;

instead of just return 0; if your compiler doesn't keep the screen open automatically.

Or, you know, just system("PAUSE") if you want to do it in one line and stuff. That's how I do it.

I prefer oldskool getch()

Can anyone explain that to me?
O_o

Not yet, because no one's answered it yet

Well, I can, but I'm going to wait until someone posts the answer first so as not to defeat the point of the challenge.