Vol. 10 No 3 - Pi Mu Epsilon

PROBLEMS AND SOLUTIONS836. [Fall 19941 Proposed by the editor.Solve this base ten holiday additionalphametic. Since the coming year 1995 is an oddyear, you are asked to find that solution such thatA is an odd digit.MANYNEWNEWYEARSIim 21 nxdx = lim 2[xInx -XI = -2 - lim 2elne = -2,€ â €so the integral I converges. Now we set x = a cos 6 and z = a cos 4, thenadd an equal term to 1 by replacing 6 by TT - 6, gettingIâ€Solution by Alma College Problem Solving Group, Alma College, AlmaMichigan.Since we can carry at most 1 from the hundreds column, we see thatYE = 10 and M = 9. Since E = 0, we must carry 1 from the units columnand R = N + 1. Thus we carry 0 to the hundreds column. From the twoA's in that column, we see that N = 5, so R = 6. Since W > 4, we haveW = 7 or 8. Because W = 7 leads to the contradiction S = N = 5, thenW = 8 and S = 7. We know A is an odd digit and therefore A = 3, theonly remaining odd digit. Thus our solution is 9351 + 508 + 508 =10367.Also solved by Charles Ashbacher, Scott H. Brown, Paul S. Bruckman,Sandra Reni Chandler, William Chau, Mark Evans, Victor G. Feser,Stephen I. Gendler, Sergey Gershtein, Richard I. Hess, Bill Hooper, CarlLibis, Henry S. Lieberman, Raymond Medley, Yoshinobu Murayoshi,Michael R. Pinter, Mike Saparov, Leslie J. Upton, Rex H. Wu, and theProposer.Since both integrand summands have period TT, we can replace 6 by 6 - 4,or by 6 + 4, and getwhere we used sin 2 y = (1 - cos 2y)/2 and the formula for cos (u + r).Since837. [Fall 19941 Proposed by J. Sutherland Frame, Michigan StateUniversity, East Lansing, Michigan.Evaluate in closed form the integralI = x- Injz - x\ dx, jzj < a.-aSolution by the Proposer.Note that I is an improper integral because x = z at one point inside theinterval of integration. In a small neighborhood of that point the quantityI72 is essentially a nonzero constant and the integral is equivalentto- 2)