Difference between revisions of "1975 AHSME Problems/Problem 18"
Jiang147369 (talk | contribs) (Created page with "== Problem 18 == A positive integer <math>N</math> with three digits in its base ten representation is chosen at random, with each three digit number having an equal chance...") |
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The probability of choosing <math>128</math>, <math>256</math>, or <math>512</math> from <math>900</math> numbers is <math>\frac{3}{900} = \frac{1}{300}</math>, which gives us the answer <math>\boxed{\textbf{(D)}\ 1/300}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369] | The probability of choosing <math>128</math>, <math>256</math>, or <math>512</math> from <math>900</math> numbers is <math>\frac{3}{900} = \frac{1}{300}</math>, which gives us the answer <math>\boxed{\textbf{(D)}\ 1/300}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369] | ||
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+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:20, 19 January 2021
Problem 18
A positive integer with three digits in its base ten representation is chosen at random, with each three digit number having an equal chance of being chosen. The probability that is an integer is
Solution
In order for to be an integer, has to be an integer power of . Since is a three-digit number in decimal form, can only be . From to , there are three-digit numbers in total.
The probability of choosing , , or from numbers is , which gives us the answer . ~jiang147369
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 | ||
All AHSME Problems and Solutions |
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