The string of the base ten numbers 4k3, 4k2, 4k1 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)
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)=6f(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)=5f(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)=1f(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
r(2008)=6r(2009)=3r(2010)=0r(2011)=4Adding 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 20110517 17:25:57 