Problem 5

List

A

contains the decimal numbers

10^k

for integers

k\ge1

. Lists

B

and

C

contain these same numbers written in bases

2

and

5

, respectively. Prove that for every integer

n>1

, exactly one number in exactly one of

B

C

has exactly

n

digits.

Step 1 of 4: Count binary digits

2^{b_k-1}\le10^k<2^{b_k}\Longrightarrow b_k=\lfloor k\log_2 10\rfloor+1

Detailed analysis

If $b_k$ is the number of binary digits of $10^k$ , then $2^{b_k-1}\le10^k<2^{b_k}$ . Taking logarithms gives $b_k=\lfloor k\log_2 10\rfloor+1$ .