N is a base ten positive integer formed by writing the digit 7 precisely 2010 times, that is N = 77....77 (2010 times).
Determine the
digital root of [N/199].
Note: [x] denotes the greatest integer ≤ x.
*** For an extra challenge, solve this problem without using a computer program.
N mod 9 is easy to compute: N mod 9 = 2010*7 mod 9 = 21*7 mod 9 = 3.
N mod 199 is trickier without a computer. First express N as a sum:
N = Sigma{i=0 to 699} 777*1000^i
777 mod 199 = 180, 1000 mod 199 = 5. Then:
N = Sigma{i=0 to 669} 180*5^i mod 199
The closed form of this sum is:
N = 180*(5^670 - 1)/4 mod 199
Since 199 is prime Fermat's Little Theorem implies x^198 = 1 mod 199. With 670 = 3*198+76, this implies:
N = 45*(5^76-1) mod 199
This is small enough to evaluate by hand. Then N = 186 mod 199.
From the Chinese Remainder Theorem: N mod (9*199) = 199*3 + 9*87 = 1380.
Then floor[N/199] mod 9 = floor[1380/199] mod 9
1380 = 6*199 + 186 and 199 mod 9 = 1. Therefore floor[N/199] mod 9 = 6*1 = 6.
Analytic Answer
