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
c1​N+c2​2N+c3​3N+c4​4N+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