Difference between revisions of "2020 CIME II Problems/Problem 4"
(Created page with "==Problem== The probability a randomly chosen positive integer <math>N<1000</math> has more digits when written in base <math>7</math> than when written in base <math>8</math>...") |
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==Similar Problems== | ==Similar Problems== | ||
[[2009 AIME I Problems/Problem 6|2009 AIME I Problem 6]] | [[2009 AIME I Problems/Problem 6|2009 AIME I Problem 6]] | ||
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+ | The intuition behind the above solution: | ||
+ | consider 7^3 = 343 <= N < 8^3 = 512 ; notice that any N in this range will have 4 digits in base 7 ( since 1*7^3 <= N) and since 1*8^3 > N , we have almost 3 digits in base 8 representation of N | ||
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+ | its easy to see 3 can be replaced by any positive integer k , where k = 1,2,3 | ||
==See also== | ==See also== |
Latest revision as of 23:49, 18 December 2020
Contents
[hide]Problem
The probability a randomly chosen positive integer has more digits when written in base than when written in base can be expressed in the form , where and are relatively prime positive integers. Find .
Solution
If a positive integer has more digits in base than base , then for some positive integer . There are positive integers that satisfy this condition for every positive integer . If , will be greater than , so we only need to consider . The number of possible values of is The requested probability is and so the answer is .
Similar Problems
The intuition behind the above solution:
consider 7^3 = 343 <= N < 8^3 = 512 ; notice that any N in this range will have 4 digits in base 7 ( since 1*7^3 <= N) and since 1*8^3 > N , we have almost 3 digits in base 8 representation of N
its easy to see 3 can be replaced by any positive integer k , where k = 1,2,3
See also
2020 CIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |
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