Difference between revisions of "1956 AHSME Problems/Problem 31"
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| + | == Problem 31== | ||
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| + | In our number system the base is ten. If the base were changed to four you would count as follows: | ||
| + | <math>1,2,3,10,11,12,13,20,21,22,23,30,\ldots</math> The twentieth number would be: | ||
| + | |||
| + | <math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110 </math> | ||
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==Solution== | ==Solution== | ||
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~JustinLee2017 | ~JustinLee2017 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1956|num-b=30|num-a=32}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:32, 12 February 2021
Problem 31
In our number system the base is ten. If the base were changed to four you would count as follows:
The twentieth number would be:
Solution
The 
number will be the value of 
in base 
. Thus, we see
![\[20_{10} = (1) \cdot 4^2 + (1) \cdot 4^1 + 0 \cdot 4^0\]](http://latex.artofproblemsolving.com/9/2/8/9286e58edcb29d1e12e9817b6705f799cbf5f9a2.png)
![\[= 110_{4}\]](http://latex.artofproblemsolving.com/3/e/5/3e5f2635f76d4ac2dcbf8f76190f0d4a701ca60d.png)
~JustinLee2017
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
