Hello, jackGee!
We find that:
. .
Therefore: .
n . . . . . . . . . . . . .
n . . . . . . . . . . . . .
. . . . . . . . . . . . . .
You can use Soroban's method without needing a calculator. The trick is that if you know the mod(21) value for a certain power of ten, you can easily find mod(21) for the next higher power of ten - just multiply 10 by the remainder and find teh mod(21) value of that. Here you know 1 = 1 (mod 21). Multiply by the remainder of 1 by 10 and you find 10^1 = 10(mod 21). Now multiply th eremainder of 10 by 10 again and you get 10^2 = 16 mod 21. Continue for 10^3: 10 x 16 = 13 (mod 21). For 10^4: 10 x 13 = 4 mod(21). For 10^5: 10x4 = 19 mod(21). Then for 10^6: 190 = 1 mod(21). And then the pattern repeats.