We publish here the solution to the divertimento A problem of ages.
The divertimento:
After some time without seeing each other, Alberto and Manolo meet in the street and talk about their ages. “When you reach my age,” says Manolo, “I’ll be twice as old as you were when I was your age.
Years later they meet again and Alberto comments now: Manolo, if we could add a year to your age, what you told me at our last meeting would still be true. It’s true,” replies Manolo, “and besides, my age would be a palindromic number.
How old were you the year of your first meeting and how many years did it take you to see each other again?
The solution:
Let us denote by \(x\) and \(y\) the ages of Alberto and Manolo, respectively, the first time they meet. From the conversation between the two it is clear that \(y > x\). On one hand, Alberto will be Manolo’s age after \(y – x\) years have passed since their meeting, when Manolo’s age will be \(y + (y – x) = 2y – x\). On the other hand, Manolo was Alberto’s age \(y – x\) years before meeting, when Alberto’s age was \(x – (y – x) = 2x – y\). According to Manolo, \(2y – x = 2(2x – y)\), or in other words, $$5x – 4y = 0 \qquad (1)$$
Let us denote by \(d\) the number of years until the two friends meet again, at which time Alberto has real age \(x+d\) and Manolo fictitious age (real age increased by one unit) \(y + d + 1\). From Alberto’s statement it follows that these two ages are also solutions of (1). Substituting, we have $$ 5x – 4y + (d – 4) = 0 \qquad (2)$$
The compatibility of (1) and (2) is only possible if \(d = 4\). On the other hand, the natural solutions of (1) are \(x = 4n\) e \(y = 5n\), with \(n\) being natural. So, the substitutions \(y = 5n\) and \(d = 4\) cause Manolo’s fictitious age to be written as \(5(n + 1)\), which is a palindromic number (and in the human age range) only if \(n = 10\). And from this, \(x = 40, y = 50\).
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