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

m (Fixed punctuation)
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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
+
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 is the
  
 
<asy>
 
<asy>
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==Solution==
 
==Solution==
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<math>\text{Let us take each number mod 16. Then we have the following pattern:}</math>
+
Considering each integer modulo <math>16</math> gives the following pattern:
−
 
+
<asy>
−
<math>\text{     1   3  5  7}</math>
+
int i,j;
−
 
+
for(i=0; i<4; i=i+1) {
−
<math>\text{15 13 11 9  }</math>
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label(string(2*i+1), (2*i,-2*1));
−
 
+
label(string(15-2*i), (2*(i-1),-2*1.35));
−
<math>\text{    1   3  5  7}</math>
+
label(string(2*i+1), (2*i,-2*1.7));
−
 
+
label(string(15-2*i), (2*(i-1),-2*2.05));
−
<math>\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</math> <math>\text{1 (mod 16), 1985 must appear in the second column.}</math>
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}
 +
</asy>
  
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<math>\text{Thus, the answer is } \fbox{(B)}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:26, 19 March 2024

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 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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