# 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


Because the only digits we are using are 1s and zeroes (they say that in question) we essentially perform the operation in binary and convert back to base 10 to get the answer.

Notice then that $R(n)_2=(2^n-1)_{10}$. It follows that $R(24)=2^{24}-1$ and $R(4)=2^4-1$
Notice to compute $\frac{2^{24}-1}{2^4-1}$ we take advantage of the fact that $x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)$ Our quotient then is just $2^{20}+2^{16}+2^{12}+2^{8}+2^4+1$

Notice then this just a $21$ digit in binary with $5$ $1$s which occupy the $2^a$ slots for the $6$ $a$ we have.
Our answer then is just $21-6=\boxed{15}$ $\fbox{E}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 