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Solution to November 2005 Problem
The Problem:
Professor Adams wrote on the blackboard a polynomial, f(x), with integral coefficients and said, "Today is my son's birthday, and when we substitute his age A for x, then f(A) = A. You will also note that f(0) = p, and that p is a prime number greater than A." How old is Professor Adams' son?
This month's problem was found in an Ontario Secondary School Mathematics Bulletin from the late 60's.
We received correct solutions from
Anurag Agarwal (USA) |
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Xavier Hecquet (France) |
Saïd Amghibech (Québec) |
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Wolfgang Kais (Germany) |
Daniel Bitin (internet) |
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Normand Laliberté (Ontario) |
Pierre Bornsztein (France) |
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Matthew Lim (internet) |
| Bernard Carpentier (France) |
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Arne Loosveldt (Belgium) |
| Stéphane Charpentier (France) |
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Juan Mir Pieras (Spain) |
| K.A. Chandrashekara (internet) |
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Anonymous |
| Zdravko Cvetkovski (Macedonia) |
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Mark Pilloff (USA) |
| Sasir Kumar Das (India) |
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Jose Luis Requena (Spain) |
| Sébastien Dumortier (France) |
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A. Teitelman (Israel) |
| Philippe Fondanaiche (France) |
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William D. Volterman (internet) |
| Giovanni Gherdovich (Italy) |
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Vincent Bardoux (France) |
| Aleksandar Ilic (Serbia and Montenegro) |
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Solution:
We can write the polynomial as
f(x) = anxn + ··· + a1x + a0
with the coefficients ai all integers. Since f(0) = p, we must have a0 = p; and since f(A) = A our polynomial must satisfy
f(A) = anAn + ··· + a1A + p = A.
Because A represents the son's age, it is a positive integer. From the last equation we deduce that A divides p (because it divides all the other terms). But p is a prime that is greater than A. The only positive integer less than p that divides it must be 1. We conclude that A = 1 — the child is one year old. Happy Birthday.
Note that f(x) can be any polynomial having constant term a prime p, and integer coefficients that sum to 1. A simple example from Mark Pilloff:
f(x) = 3 – 2x.
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