All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Sum Remainder Deduction (Posted on 2011-05-14) Difficulty: 3 of 5
The string of the base ten numbers 4k-3, 4k-2, 4k-1 and 4k written side by side without commas or spaces is denoted by F(k). For example:

F(1) = 1234, F(2) = 5678, F(3) = 9101112, F(4) = 13141516 - and, so on.

Determine the remainder when this expression is divided by 7. What is the remainder when this expression is divided by 13?

F(1) + F(2) + F(3) + ..... + F(2011)

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Long Solution Comment 1 of 1
Upon reflection the following analysis could have been shortened a bit.  I will leave it to show how it was reasoned out.

define r(k) as the remainder upon dividing f(k) by 7.

f(1) and f(2) are the only values with four digits.  r(1)=2, r(2)=1

f(3) is the only value with 7 digits.  r(3)=6

f(4) through f(24) each have 8 digits.  More importantly each differs by 4040404.  r(4)=3 and 4040404 has a remainder of 4, so the remainders of successive terms increase by 4 (mod7):
3,0,4,1,5,2,6 this cycle of seven numbers repeats also the sum of these numbers is 0 (mod 7).  Conveniently there are 21 terms so this cycle repeats 3 times so the total of r(4) to r(24) is 0.

f(25) is the only value with nine digits.  r(25)=5

f(26) through f(249) each have 12 digits.  More importantly each differs by 4004004004 which is actually divisible by 7. r(26)=2 which is also true for all the terms to 249.  Conveniently there are 224 terms which is divisible by 7 so the total of r(26) to r(249) is 0.

f(250) is the only value with 13 digits.  r(250)=1

f(251) through f(2011) each have 16 digits.  r(251)=6 More importantly each differs by 4000400040004 which is 4(mod 7).   So we again get a cycle of 7 terms and the sum of these 7 terms is 0 (mod 7).  There are an inconvenient 1761 terms which leaves a remainder of 4 when divided by 7.  So we have 4 leftover terms of the cycle 6,3,0,4,1,5,2.  i.e. we just need

Adding these boldface remainders gives 2+1+6+0+5+0+1+6+3+0+4 = 28

So the remainder is 0

  Posted by Jer on 2011-05-17 17:25:57
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information