Difference between revisions of "1973 AHSME Problems/Problem 6"
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16 </math> | ||
| − | == | + | ==Solution== |
Write out the numbers using the definition of [[base numbers]]. | Write out the numbers using the definition of [[base numbers]]. | ||
Revision as of 13:50, 4 July 2018
Problem
If 554 is the base 
representation of the square of the number whose base representation is 24, then , when written in base 10, equals
Solution
Write out the numbers using the definition of base numbers.
![\[554_b = 5b^2 + 5b + 4\]](http://latex.artofproblemsolving.com/b/e/4/be4cb439c4b096a30dc579d5dcdf0e2e3e134f72.png)
![\[24_b = 2b+4\]](http://latex.artofproblemsolving.com/2/2/6/226ed8b9f9ee0f6b01979a8aa9b87ae64dfc5e65.png)
Since 
, we can write an equation.
![\[5b^2 + 5b + 4 = (2b+4)^2\]](http://latex.artofproblemsolving.com/5/0/a/50a74a076030848e89544a94de652ea2d04e5fcb.png)
![\[5b^2 + 5b + 4 = 4b^2 + 16b + 16\]](http://latex.artofproblemsolving.com/6/0/1/6014c8d6ddcd262b3daf255578b7de0d8618d5a9.png)
![\[b^2 - 11b - 12 = 0\]](http://latex.artofproblemsolving.com/0/2/0/020d7c07e6faefc93544f0e9d9404da39a1e8fc9.png)
![\[(b-12)(b+1) = 0\]](http://latex.artofproblemsolving.com/3/4/5/345a0162726a105dffb09838f61b2df5f4098b9b.png)
Since base numbers must be positive, in base 10 equals 
.
See Also
| 1973 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
