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

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==Problem==
 
==Problem==
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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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The odd positive integers <math>1, 3, 5, 7, \ldots</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 is the
  
 
<asy>
 
<asy>
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==Solution==
 
==Solution==
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Notice that the pattern repeats every <math> 8 </math> terms. Therefore, we only have to consider the first <math> 8 </math> terms. Also, since each term adds <math> 2 </math> to the previous term, the pattern repeats every <math> 16 </math> natural numbers. Thus, notice that all numbers <math> \equiv 15(\text{mod }16) </math> is in the first row, all numbers <math> \equiv 1, 13(\text{mod }16) </math> are in the second row, and so on.
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Considering each integer modulo <math>16</math> gives the following pattern:
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<asy>
 +
int i,j;
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for(i=0; i<4; i=i+1) {
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label(string(2*i+1), (2*i,-2*1));
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label(string(15-2*i), (2*(i-1),-2*1.35));
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label(string(2*i+1), (2*i,-2*1.7));
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label(string(15-2*i), (2*(i-1),-2*2.05));
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}
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</asy>
  
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Notice that since <math> 1985\equiv 1(\text{mod }16) </math>, it's in the same row as <math> 1 </math>, which is the fourth row, <math> \boxed{\text{D}} </math>.
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We therefore observe that all numbers congruent to <math>1 \pmod{16}</math> will appear in the second column, and since <math>1985 \equiv 1 \pmod{16}</math>, the answer is <math>\boxed{\text{(B)} \ \text{second}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=8|num-a=10}}
 
{{AHSME box|year=1985|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 21:08, 19 March 2024

Problem

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

Considering each integer modulo $16$ gives the following pattern: [asy] int i,j; for(i=0; i<4; i=i+1) { label(string(2*i+1), (2*i,-2*1)); label(string(15-2*i), (2*(i-1),-2*1.35)); label(string(2*i+1), (2*i,-2*1.7)); label(string(15-2*i), (2*(i-1),-2*2.05)); } [/asy]

We therefore observe that all numbers congruent to $1 \pmod{16}$ will appear in the second column, and since $1985 \equiv 1 \pmod{16}$, the answer is $\boxed{\text{(B)} \ \text{second}}$.

See Also

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
All AHSME Problems and Solutions

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