Difference between revisions of "2020 AIME II Problems/Problem 5"
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==Problem== | ==Problem== | ||
| + | For each positive integer <math>n</math>, left <math>f(n)</math> be the sum of the digits in the base-four representation of <math>n</math> and let <math>g(n)</math> be the sum of the digits in the base-eight representation of <math>f(n)</math>. For example, <math>f(2020) = f(133210_{\text{four}}) = 10 = 12_{\text{eight}}</math>, and <math>g(2020) = \text{the digit sum of }12_{\text{eight}} = 3</math>. Let <math>N</math> be the least value of <math>n</math> such that the base-sixteen representation of <math>g(n)</math> cannot be expressed using only the digits <math>0</math> through <math>9</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. | ||
| + | ==Solution== | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=4|num-a=6}} | {{AIME box|year=2020|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:09, 7 June 2020
Problem
For each positive integer 
, left 
be the sum of the digits in the base-four representation of and let 
be the sum of the digits in the base-eight representation of . For example, 
, and 
. Let 
be the least value of such that the base-sixteen representation of cannot be expressed using only the digits 
through 
. Find the remainder when is divided by 
.
Solution
See Also
| 2020 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
