Author |
Topic: Many sevens (Read 466 times) |
|
BNC
Uberpuzzler
Gender: 
Posts: 1732
|
Look at this addition:
7 + 72 + 73 + .... + 749
Without calculating, what are the two LSB digits?
|
|
 IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
 |
Re: Many sevens
« Reply #1 on: Jul 9th, 2003, 4:44pm » |
 Quote  Modify
|
I assume LSB means Least Significant Binary? If so, then...
::
Since 7 = -1 (mod 8), 7odd power = 1112 (mod 8) and 7even power = 12 (mod 8)
Therefore 7 + ... + 748 = 02 (mod 8)
Hence 7 + ... + 749 = 1112 (mod 8)
So the last three LSB digits are 111.
::
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
BNC
Uberpuzzler
Gender:
Posts: 1732
|
|
Re: Many sevens
« Reply #2 on: Jul 9th, 2003, 9:20pm » |
Quote Modify
|
I actually meant the 2 least significant decimal figures  but NuckH's method works there as well
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
wowbagger
Uberpuzzler
 
Gender:
Posts: 727
|
|
Re: Many sevens
« Reply #3 on: Jul 10th, 2003, 2:32am » |
Quote Modify
|
on Jul 9th, 2003, 9:20pm, BNC wrote:| I actually meant the 2 least significant decimal figures |
|
As far as I know, LSB stands for "least significant bit", so strictly speaking there's no such thing as a decimal LSB digit. Perhaps "least significant decimal digit" would be more appropriate. I thought that your intention was to aim at decimal digits, but your original statement can be misleading - as was proven by "NuckH"'s answer.
|
|
IP Logged |
"You're a jerk, <your surname>!"
|
|
|
Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261
|
|
Re: Many sevens
« Reply #4 on: Aug 20th, 2003, 10:30am » |
Quote Modify
|
"LSB" has been loosely used to signify the "last" digits in whatever "base" system you are talking.
Anyways in decimal system we will notice that 7 + 7^2 + 7^3 + 7^4 mod 100 comes to zero.
This is true for subsequent quadruples too (following NickH's solution)
Subsequently we are left with 7^49 which belongs to 7^(4n-3) group or whose first element is 7 hence giving the answer to be 7^49 mod 100 = 7
so last two decimals ditis are "07"
This process made me think:
Can you generalise on last two digits or at least last digit for the sum (where n is any number):
n^1 + n^2 + ... + n^k for a base "b" ?
|
|
IP Logged |
"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print