This brainteaser was written by Derrick Niederman. A calendar year is typically referred to as a four‐digit number, as in 2008, or as a two‐digit number, as in ’08. Sometimes, the two‐digit number divides evenly into the four‐digit number, with no remainder. How many times did this happen during the twentieth century? Resources for Teaching Math
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Solution: 12 times. We’re looking for the number of times that a two‐digit number of the form xy divides evenly into the four‐digit number 19xy. (Just for clarity, note that xy is meant to represent the two‐digit number equal to 10x + y, not the product of x and y; and, 19xy is meant to represent the four‐digit number equal to 1900 + 10x + y, not the product of 19, x and y.) But if the number 19xy is a multiple of xy, then so is 19xy – xy = 1900, and vice versa. The prime factorization of 1900 is 22 × 52 × 19. Hence, the years in question are formed by finding all one‐ or two‐digit numbers that can be obtained using these factors and then putting a 19 in front. Starting with the trivial solution of 1901, these are: 1901 1902 1904 1905 1910 1919 1920 1925 1938 1950 1976 1995 That’s a total of 12 years during the twentieth century. Of course, you may be wondering whether we should count 1900 or 2000 as part of the twentieth century. Different sources include one or the other of these as part of the twentieth century. But for this puzzle, it doesn’t matter either way. The last two digits of both numbers are zero, so neither 1900 nor 2000 could be a solution since division by 0 is undefined. Resources for Teaching Math
© 2009 National Council of Teachers of Mathematics, Inc.
http://illuminations.nctm.org