age puzzles and tricks
Problems that ask for a person's age or, alternatively, when a person was a certain age, given several roundabout facts. They go back at least 1,500 years to the time of Metrodorus and Diophantus's riddle. A number of distinct types of age puzzle sprang up between the 16th and early 20th centuries, in most cases best solved by a little algebra. One form asks: if X is now a years old and Y is now b, when will X be c times as old as Y? The single unknown, call it x, can be found from the equation a + x = c(b + x). Another type of problem takes the form: if X is now a times as old as Y and after b years X will be c times as old as Y, how old are X and Y now? In this case the trick is to set up and solve two simultaneous equations: X = aY and X + b = c(Y + b).
Around 1900, two more variants on the age puzzle became popular. Here is an example of each for the reader to try:
Alternatively, ask the person to pick a number, multiply this by 2, add 5, and multiply by 50. If the person has already had a birthday this year and it's the year 2004, they should add 1754, otherwise they should add 1753. Each year after 2004 these numbers need to be increased by 1. Finally, the person should subtract the year they were born. The first digit(s) of the answer are the original number, while the last two digits are the person's age.
Here is one more trick. Take your age, multiply it by 7, then multiply again by 1443. The result is your age repeated 3 times. (What you have actually done is multiplied by 10101; if you multiply by 1010101 the repetition is fourfold, and so on.)
Solutions to puzzles
(2) 16½ (Mary is 27½).
Related categories• GAMES AND PUZZLES
• TIME MEASUREMENT AND PUZZLES
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