Problem:
In a rectangular array of points, with 5 rows and N columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through N, the second row is numbered N+1 through 2N, and so forth. Five points, P1,P2,P3,P4, and P5, are selected so that each Pi is in row i. Let xi be the number associated with Pi. Now renumber the array consecutively from top to bottom, beginning with the first column. Let yi be the number associated with Pi after renumbering. It is found that x1=y2,x2=y1, x3=y4,x4=y5, and x5=y3. Find the smallest possible value of N.
Solution:
Suppose that Pi is in row i and column ci. It follows that
x1=c1,x2=N+c2,x3=2N+c3,x4=3N+c4,x5=4N+c5
and
y1=5c1−4,y2=5c2−3,y3=5c3−2,y4=5c4−1,y5=5c5
The Pi have been chosen so that
c1N+c22N+c33N+c44N+c5=5c2−3=5c1−4=5c4−1=5c5=5c3−2
Use the first two equations to eliminate c1, obtaining 24c2=N+19. Thus N=24k+5, where k=c2−1. Next use the remaining equations to eliminate c3 and c4, obtaining 124c5=89N+7. Substitute for N to find that 124c5=2136k+452, and hence 31c5= 534k+113=31(17k+3)+7k+20. In other words, 7k+20=31m for some positive integer m. Now 7k=31m−20=7(4m−2)+3m−6. Since 7 must divide 3m−6, the minimum value for m is 2, and the smallest possible value of k is therefore 6, which leads to N=24⋅6+5=149. It is not difficult to check that c2=7,c1=32,c5=107, c4=5c5−3N=88, and c3=5c4−1−2N=141 define an acceptable placement of points Pi. The numbers associated with the points are x1=32,x2=156,x3=439,x4=535, and x5=703.
Note: Modular arithmetic can be used to simplify this solution.
The problems on this page are the property of the MAA's American Mathematics Competitions