Problem:
Let N be the positive integer 7777…777, a 313-digit number where each digit is a 7. Let f(r) be the leading digit of the r th root of N. What is
f(2)+f(3)+f(4)+f(5)+f(6)?
Answer Choices:
A. 8
B. 9
C. 11
D. 22
E. 29
Solution:
We can rewrite N as 97​⋅9999…999=97​⋅(10313−1). When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it. When we take the power of ten out of the square root, we'll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of f(r) will be equal to the leading digit of r97​⋅10313(modr)​.
Then f(2) is the first digit of 97​⋅(10)​=970​​=7……​≈2.