# 1985 AHSME Problems/Problem 9

## 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 $\text{Let us take each number mod 16. Then we have the following pattern:}$ $\text{ 1 3 5 7}$ $\text{15 13 11 9 }$ $\text{ 1 3 5 7}$ $\text{We can clearly see that all terms congruent to 1 mod 16 will appear in the second column. Since we can see that 1985}\equiv$ $\text{1 (mod 16), 1985 must appear in the second column.}$ $\text{Thus, the answer is } \fbox{(B)}$

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