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

$\text{Let us take each number mod 16. Then we have the following pattern:}$

$\text{ 1 3 5 7}$

$\text{15 13 11 9 }$

$\text{ 1 3 5 7}$

$\text{We can clearly see that all terms congruent to 1 mod 16 will appear in the second column. Since we can see that 1985}\equiv$ $\text{1 (mod 16) , 1985 must appear in the second column.}$

$\text{Thus, the answer is } \fbox{(B)}$

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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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These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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1985 AHSME Problems/Problem 9

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