Difference between revisions of "1993 AHSME Problems/Problem 16"

(Solution)
(Solution)
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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>63/5</math> has a remainder of <math>3</math>.
+
<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 \[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\] in which the $n^{th}$ positive integer appears $n$ times. The remainder when the $1993^{rd}$ term is divided by $5$ is

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$

Solution

You want to find the largest integer x that satisfies $\frac{x*(x+1)}{2}<1993$. You then can find that value of x, which is $62$. Therefore, the next value of the sequence is $63$. $\frac{63}{5}$ has a remainder of $3$. $\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
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
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