Difference between revisions of "1960 AHSME Problems/Problem 16"
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<cmath>69 = 2 \cdot 25 + 19</cmath> | <cmath>69 = 2 \cdot 25 + 19</cmath> | ||
Similarly, dividing <math>19</math> by <math>5</math> results in quotient of <math>3</math> and remainder of <math>4</math>, so rewrite the number as | Similarly, dividing <math>19</math> by <math>5</math> results in quotient of <math>3</math> and remainder of <math>4</math>, so rewrite the number as | ||
− | <cmath>69 = 2 \cdot 25 + 3 \cdot 5 + 4 \cdot 1</cmath> | + | <cmath>69 = 2 \cdot 25 + 3 \cdot 5 + 4 \cdot 1</cmath> |
Thus, the number in base <math>5</math> can be written as <math>234_5</math>, so the answer is <math>\boxed{\textbf{(C)}}</math> | Thus, the number in base <math>5</math> can be written as <math>234_5</math>, so the answer is <math>\boxed{\textbf{(C)}}</math> | ||
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==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=15|num-a=17}} | {{AHSME 40p box|year=1960|num-b=15|num-a=17}} |
Revision as of 11:03, 11 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 Similarly, dividing by results in quotient of and remainder of , so rewrite the number as 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 | |
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All AHSME Problems and Solutions |