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

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, $\boxed{\text{D}}$ .

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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1985 AHSME Problems/Problem 9

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