Difference between revisions of "1985 AHSME Problems/Problem 9"

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==Problem==
 
==Problem==
The odd positive integers <math> 1, 3, 5, 6, \cdots </math>, are arranged into five columns continuing with the pattern shown on the right. Counting from the left, the column in which <math> 1985 </math> appears in is the
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The odd positive integers <math> 1, 3, 5, 7, \cdots </math>, are arranged into five columns continuing with the pattern shown on the right. Counting from the left, the column in which <math> 1985 </math> appears in is the
  
 
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<asy>

Revision as of 17:07, 15 December 2011

Problem

The odd positive integers $1, 3, 5, 7, \cdots$, are arranged into five columns continuing with the pattern shown on the right. Counting from the left, the column in which $1985$ appears in is the

[asy] int i,j; for(i=0; i<4; i=i+1) { label(string(16*i+1), (2*1,-2*i)); label(string(16*i+3), (2*2,-2*i)); label(string(16*i+5), (2*3,-2*i)); label(string(16*i+7), (2*4,-2*i)); } for(i=0; i<3; i=i+1) { for(j=0; j<4; j=j+1) { label(string(16*i+15-2*j), (2*j,-2*i-1)); }} dot((0,-7)^^(0,-9)^^(2*4,-8)^^(2*4,-10)); for(i=-10; i<-6; i=i+1) { for(j=1; j<4; j=j+1) { dot((2*j,i)); }}[/asy]

$\mathrm{(A)\ } \text{first} \qquad \mathrm{(B) \ }\text{second} \qquad \mathrm{(C) \  } \text{third} \qquad \mathrm{(D) \  } \text{fourth} \qquad \mathrm{(E) \  }\text{fifth}$

Solution

Notice that the pattern repeats every $8$ terms. Therefore, we only have to consider the first $8$ terms. Also, since each term adds $2$ to the previous term, the pattern repeats every $16$ natural numbers. Thus, notice that all numbers $\equiv 15(\text{mod }16)$ is in the first row, all numbers $\equiv 1, 13(\text{mod }16)$ are in the second row, and so on.

Notice that since $1985\equiv 1(\text{mod }16)$, it's in the same row as $1$, which is the fourth row, $\boxed{\text{D}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AHSME Problems and Solutions