Question

Let $T$ be the set of powers $9^k,$ where $k$ is an integer with $0\leq k\leq4000.$ Given that $9^{4000}$ has $3817$ digits and that its first (leftmost) digit is $9,$ how many elements of $T$ have $9$ as their leftmost digit?

Answer

Answer: 184

Concepts:digits exponent invariant

Difficulty rating: 2230

Small Hint:

Compare the number of digits of $9^k$ with that of $9^{k-1}$

Big Hint:

A multiplication by $9$ produces a leading $9$ exactly when the digit count does not increase

Solution:

For $k\geq1,$ the number $9^k$ begins with $9$ exactly when it has the same number of digits as $9^{k-1}.$ Indeed, if $9^{k-1}$ has $d$ digits and multiplication by $9$ creates no new digit, then $9^k\geq9\cdot10^{d-1},$ so its leading digit is $9.$ If multiplication does create a new digit, then $9^k\lt9\cdot10^d,$ so its leading digit is at most $8.$

Starting from the one-digit number $9^0,$ the digit count reaches $3817$ after $4000$ multiplications. Thus it increases on $3816$ steps and stays unchanged on $4000-3816=184$ steps. Since $9^0=1$ does not begin with $9,$ exactly $184$ elements of $T$ do.

1990 AIME Problem 13

Answer: 184

Difficulty rating: 2230

Small Hint:

Big Hint:

Solution:

Problem 13 in Other Years