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 $\boxed{\textbf{(D)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1961 AHSME Problems/Problem 17

Problem

Solution

See Also