Difference between revisions of "1985 AHSME Problems/Problem 9"
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==Problem== | ==Problem== | ||
− | 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 | + | 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> | ||
Line 23: | Line 23: | ||
==Solution== | ==Solution== | ||
− | <math> | + | Considering each integer modulo <math>16</math> 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> | ||
− | <math>\ | + | 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 19:26, 19 March 2024
Problem
The odd positive integers , are arranged into five columns continuing with the pattern shown on the right. Counting from the left, the column in which
appears is the
Solution
Considering each integer modulo gives the following pattern:
We therefore observe that all numbers congruent to will appear in the second column, and since
, the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.