Author |
Topic: A Year with Gauss (Read 383 times) |
|
ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
A Year with Gauss
« on: Jan 18th, 2005, 2:49am » |
 Quote  Modify
|
If Gauss proved at least one new theorem every day, but never more than 50 new theorems in any month, prove that there was a sequence of consecutive days in a year when Gauss proved exactly 125 new theorems.
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
TenaliRaman
Uberpuzzler
I am no special. I am only passionately curious.
Gender:
Posts: 1001
|
|
Re: A Year with Gauss
« Reply #1 on: Jan 18th, 2005, 10:32am » |
Quote Modify
|
::
Let di denote the number of theorems proved by Gauss on the ith day. Let Di denote the number of theorems proved in the first i days.
First of all,
[sum] di <= 600 (50 theorem per month)
Now, gauss proves atleast one theorem per day, therefore
1<=D1<D2<...<D365<=600
Add 125 to each term of the above,
126<=D1+125<D2+125<...<D365+125<=725
Now consider the set,
G = {D1, D2,..., D365, D1+125, D2+125,..., D365+125}
#G = 730
Each of the terms takes values from 1 to 725
By Pigeonhole Principle,
There are atleast two terms which take the same value and hence proved.
::
-- AI
|
| « Last Edit: Jan 18th, 2005, 10:34am by TenaliRaman » |
IP Logged |
Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print