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It's a Puzzlement: Four Degrees of Matthew Rodermund
by John P. Robertson
In 1999 the Actuarial Review and this column begin their 26th year. Matthew Rodermund was the first editor of the Actuarial Review (actually called CAS Newsletter for the first few issues). There are 13 letters in his name. Putting all this together, we offer the following puzzle. Find a fourth degree polynomial, f(x)=ax4+bx3+cx2+dx+e, with integer coefficients, so that f(25)=13 and f(1974)=1999, or show that no such polynomial exists.
Multiply While You Shop
The problem was to find four prices of items that add to $7.11 and multiply to $7.11. The unique solution has prices 1.20, 1.25, 1.50, and 3.16. These add to 7.11 and multiply to 7.11 without any rounding.
Gary Venter's solution went more or less as follows. Working in cents, the sum is 711 and the product is 711,000,000 = 26×32×56 ×79. As 56 > 711, at least two prices are divisible by 5. If only two prices are divisible by 5, then either one is divisible by 625 and the other is divisible by 25, or both are divisible by 125. If one is divisible by 625, it must be 625. The three remaining prices add to 86. No matter how the 86 is split up into three parts, the product of all four prices will fall short of 711,000,000. If two prices are divisible by 125, then each is 125, 250, 375, or 500. One of the remaining two prices must be a multiple of 79. A little experimentation with the various possibilities will show that none of these give a solution. Therefore at least three items have prices divisible by 5. So the sum of these three must be a multiple of 5. If one of the three is also a multiple of 79, then it must be 395 = 79×5. But no price can be 395, as can be seen by noting that one of the remaining prices would have to be a multiple of 125, another a multiple of 25, and trying the cases. So three prices are multiples of 5 and the fourth price is a multiple of 79. The three prices that are multiples of 5 must add to a multiple of 79, so they must add to 395. This makes the fourth price 316. The other three are multiples of 51, 52, and 53. Testing cases shows that the solution given above is the only one.
Many solvers did computer searches, usually after narrowing down the number of cases to consider. A few used the "Solver" feature of Excel. In some cases this gave the exact result above, and in other cases it gave an approximate solution. Charlie Orlowicz used Solver in a particularly clever way, having Solver find the exponents of 2, 3, and 5 in the prices (exactly one price will have a factor of 791).
Solutions were also sent in by Nilgun A. Akgul, Amy Angell, Ed Bouchie, Oscar Chow, Alfred Commodore, Rosemary Gabriel, Bob Hallstrom, John Herder, Ira Kaplan, Frank Karlinski, Joe Kilroy, Greg Kissel, Blair Laddusaw, Turhan Murguz, Melissa Neidlinger, Rich Newell, Khanh Nguyen, Randy Nordquist, Richard Pitbladdo, Jean-Denis Roy, Nathan Schwartz, Chris Throckmorton, and Walter Wright.