# Difference between revisions of "1993 AHSME Problems/Problem 7"

## Problem

The symbol $R_k$ stands for an integer whose base-ten representation is a sequence of $k$ ones. For example, $R_3=111,R_5=11111$, etc. When $R_{24}$ is divided by $R_4$, the quotient $Q=R_{24}/R_4$ is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in $Q$ is:

$\text{(A) } 10\quad \text{(B) } 11\quad \text{(C) } 12\quad \text{(D) } 13\quad \text{(E) } 15$

## Solution

Note $R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}$.

Therefore $\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}$.

We can recognize this is also the formula for the sum of a geometric series $1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}$

Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the $10^5$, $10^6$ and $10^7$ places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives $5\times 3=15$ zeros altogether.

The answer is $\fbox{E}$