Vol. 10 No 7 - Pi Mu Epsilon

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594 PI MU EPSILON JOURNAL PROBLEMS AND SOLUTIONS 595 c) Using the same technique as in Part (b), we see that the first k terms vanish and we have II L (i - k)(i - k + 1) ..·(i)(i + 1)(i + 2)··· i=l n-k = L (i)(i + 1)(i + 2)···
596 PI MU EPSILON JOURNAL PROBLEMS AND SOLUTIONS 597 Table 2. Solutions to the Generalized Problem. X; Since I 0 I can be written as a sum of two squares in exactly one way, I 0 I = 100 + I, and since 2jy- sj is even, then we must have 2jy- sj := IO and hence y = 0 or y = I 0. Therefore, there are no integer solutions with I:$;; y ~ 9. 2 3 4 8c + I 8c + 3 8c + 2 64c 2 + 24c + I 64c 2 + 40c + 5 64c 2 + 32c + 3 5I2c 3 + 320c 2 5I2c 3 + 448c 2 5I2c 3 + 384c 2 + 48c + I + II2c + 7 + 80c + 4 Also solved by Paul S. Bruckman, Highwood, /L, Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Mark Evans, Louisville, KY, Richard I. Hess, Rancho Palos Verdes, CA, Murray S. Klamkin, University of Alberta, Canada, William H. Peirce, Delray Beach, FL. H.-J. Seiffert, Berlin, Germany, and the Proposer. Klamkin referred to his article "Perfect Squares of the form (m 2 - I)~ + 1," Math. Mag., 40(I969)III-113. 898. [Fall 1996] Proposed by PaulS. Bruckman, Edmonds, Washington. An n-digit number N is defined to be a base 10 Armstrong number of order n if n-1 n-1 N = E dklok = E dk", k=O k=O where the dk are decimal digits, with dn-J > 0. (See Miller and Whalen, "Armstrong Numbers: I 53= I 3 +5 3 + 3 3 ," Fibonacci Quarterly30.3, (I992), pp. 22I-224.) Prove that there are no base ten Armstrong numbers of order 2; that is, prove the impossibility of the equation I Oy + x = x 2 + y, where x and y are integers with 0 :$;; x :$;; 9 and I :$;; y :$;; 9. I. Solution by Charles R. Diminnie, San Angelo, Texas. By completing the squares, we reduce the equation to (2x- I) 2 + 4(y- 5) 2 = IOI. II. Solution by Bob Prielipp, University of Wisconsin-Oshkosh, Oshkosh, Wisconsin. Since I Oy - x(x - 1) = y, then y is even, so y is even. Thus y ~ 2, 4, 6, or 8. If y = 2 or 8, then the given equation becomes r =X+ I6, that is, (2x- I) 2 = 65. When x = 4 or 6, the equation becomes x 2 = x + 24, that is, (2x- I) 2 = 97. Since neither 65 nor 97 is a square number, there is no solution with the given constraints. III. Solution by James Campbell, University of Missouri, Columbia, Missouri. We must have (10- y)y = x(x- I). Now 9 S (10- y)y ~ 25 whenever 1 :$;; y :$;; 9 and x(x - 1) is an increasing function, so if there is a solution, x = 4 or x = 5. Substituting these values in for x, we get the equations y - 1 Oy + 12 = 0 and y - 1 Oy + 20 = 0. In neither case is the discriminant a perfect square, so no integer solution exists. IV. Solution by Robert C. Gebhardt, Hopatcong, New Jersey. By the quadratic formula, y = 5 ± V25 - x 2 + X. The radicand, 25 - x 2 + x, is a perfect square only if x = 0 or x = I. Therr y = 0 or y = 10, neither of which is permitted. Thus there is no solution.
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