Difference between revisions of "1993 AHSME Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | + | The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of <math>n</math>'s ends at position <math>1+2+\dots+n</math>. | |
| − | + | Therefore we want to find the smallest integer <math>n</math> that satisfies <math>\frac{n(n+1)}{2}\geq 1993</math>. | |
| + | |||
| + | By trial and error, the value of <math>n</math> is <math>63</math>, and <math>63 \div 5</math> has a remainder of <math>3</math>. | ||
<math>\fbox{D}</math> | <math>\fbox{D}</math> | ||
Latest revision as of 21:18, 27 May 2021
Problem
Consider the non-decreasing sequence of positive integers
![\[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\]](http://latex.artofproblemsolving.com/9/b/7/9b743ef0ec1ebde7d9edc21ad8a51d362c5dc481.png)
in which the 
positive integer appears 
times. The remainder when the 
term is divided by 
is
Solution
The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of 's ends at position 
.
Therefore we want to find the smallest integer that satisfies 
.
By trial and error, the value of is 
, and 
has a remainder of 
.
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 16 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
