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

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== Solution ==
 
== Solution ==
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You want to find the largest integer x that satisfies <math>\frac{x*(x+1)}{2}<1993</math>.
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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>.
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You then can find that value of x, which is <math>62</math>. Therefore, the next value of the sequence is <math>63</math>.
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<math>63/5</math> has a remainder of <math>3</math>.
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Therefore we want to find the smallest integer <math>n</math> that satisfies <math>\frac{n(n+1)}{2}\geq 1993</math>.
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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>.
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<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 \[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\] 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

The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of $n$'s ends at position $1+2+\dots+n$.

Therefore we want to find the smallest integer $n$ that satisfies $\frac{n(n+1)}{2}\geq 1993$.

By trial and error, the value of $n$ is $63$, and $63 \div 5$ has a remainder of $3$.

$\fbox{D}$

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

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