Home > Problem of the Week > Archive List > Detail

<< Prev 5/7/2006 Next >>

Age-Old Old Age Problem

A man (living now) states that he was x years old in the year x2. He adds, if to the number of my years you add the number of my month, it equals the square of the date (i.e. the day of the month) of my birthday. When was he born?

Note: Use date problem was posed and that the man was "living now." Is this information useful? Necessary? That is, are other answers possible without the phrase "living now"?


Source: Mathematics Magazine, Problem 130, March 1937

Hint: Focus on the year first....what years are perfect squares?


Solution Commentary: Answer given by Lucille Meyer in 1937:

A man living now could not have been 43 years old in 1949, that is 432. Therefore, the man must have been 44 years old in 1936. From the conditions given, 44 + m = d2 and 0 < m < 13, whence m = 5 is the only integral solution, and d = 7. The man was born May 7, 1892.

Now, using this same reasoning process, "find" other solutions after removing the restriction "living now."