1961 AHSME Problems/Problem 17

Problem

In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$. In the Land of Mathesis, however, numbers are written in the base $r$. Jones purchases an automobile there for $440$ monetary units (abbreviated m.u). He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The base $r$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 12$

Solution

If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then \[440_r + 340_r = 1000_r\] This equation can be rewritten as \[4r^2 + 4r + 3r^2 + 4r = r^3\] Bring all of the terms to one side to get \[r^3 - 7r^2 - 8r = 0\] Factor to get \[r(r-8)(r+1)=0\] Since base numbers are always positive, base $r$ is $8$, which is answer choice $\boxed{\textbf{(D)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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