Difference between revisions of "1960 AHSME Problems/Problem 16"
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| − | == Problem | + | == Problem == |
In the numeration system with base <math>5</math>, counting is as follows: <math>1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots</math>. | In the numeration system with base <math>5</math>, counting is as follows: <math>1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots</math>. | ||
Revision as of 23:04, 10 May 2018
Problem
In the numeration system with base 
, counting is as follows: 
.
The number whose description in the decimal system is 
, when described in the base system, is a number with:
Solution
Since 
, divide by 
. The quotient is 
and the remainder is 
, so rewrite the number as
![\[69 = 2 \cdot 25 + 19\]](http://latex.artofproblemsolving.com/b/1/a/b1a4295d340d73388a356126c67a112fc8ffab99.png)
Similarly, dividing by results in quotient of 
and remainder of 
, so rewrite the number as
![\[69 = 2 \cdot 25 + 3 \cdot 5 + 4 \cdot 1\]](http://latex.artofproblemsolving.com/7/5/2/752158eab205ffaef039b882053c8911843a4b7c.png)
.
Thus, the number in base can be written as 
, so the answer is 
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
