Difference between revisions of "1985 AHSME Problems/Problem 9"
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<math>\text{ 1 3 5 7}</math> | <math>\text{ 1 3 5 7}</math> | ||
| − | <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> | + | <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> |
<math>\text{Thus, the answer is } \fbox{(B)}</math> | <math>\text{Thus, the answer is } \fbox{(B)}</math> | ||
Revision as of 23:47, 2 April 2018
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 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]](http://latex.artofproblemsolving.com/d/f/5/df560371df560804d3112eae004031cd301a3f76.png)
Solution
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. 
