SEN Q ChallengeThe 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).
First Question (Spoiler in Red)
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
The number is going to have end in 5 or 0, since multiplying by 29 needs to preserve the last digit. Test odd multiples of 5 (If it ends in 0, you could just drop that and get a smaller number). Starting with 5 and multiplying by 29, you get a result of 145. Later you come across 25; 25 x 29 = 725.
Question Links
This will contain links to each question that has been answered.
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.
None.
Relatively ancient and inactive
That's.. not an AIME question.
None.
Just here for the activity... well not really
145
EDIT:
Find all odd multiples of 29, cut off the last digit, and put that over the original.
guy lifting weight (animated smiley):
O-IC
OI-C
"Oh, I see it"
Just here for the activity... well not really
lol fail, I read that as right-most. brb then.
guy lifting weight (animated smiley):
O-IC
OI-C
"Oh, I see it"
Relatively ancient and inactive
14x29 != 145 either.
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Correct!
Solution will be in the OP in just a sec.
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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;
}
None.
Relatively ancient and inactive
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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.
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Vrael: Use moar proper notation. x
1 or x
1?
Also, with all those return(0);s, how did you see the answer?
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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.
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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)?
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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.
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Or, you know, just system("PAUSE") if you want to do it in one line and stuff. That's how I do it.
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I prefer oldskool getch()
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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.
Can anyone explain that to me?
O_o
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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.
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