Difference between revisions of "1993 AHSME Problems/Problem 16"
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You want to find the largest integer x that satisfies <math>\frac{x*(x+1)}{2}<1993</math>. | You want to find the largest integer x that satisfies <math>\frac{x*(x+1)}{2}<1993</math>. | ||
You then can find that value of x, which is <math>62</math>. Therefore, the next value of the sequence is <math>63</math>. | You then can find that value of x, which is <math>62</math>. Therefore, the next value of the sequence is <math>63</math>. | ||
| − | <math>63 | + | <math>\frac{63}{5}</math> has a remainder of <math>3</math>. |
<math>\fbox{D}</math> | <math>\fbox{D}</math> | ||
Revision as of 15:42, 3 August 2019
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
You want to find the largest integer x that satisfies 
.
You then can find that value of x, which is 
. Therefore, the next value of the sequence is 
.
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. 
