Cheapskate
When Fred graduated from high school in May 1993, after having celebrated his 18th birthday in March, he received a small check from his grandma. Being a frugal sort, he decided that the best thing to spend his money on would be something that he could use for the rest of his life, even if he lived to be 124: calendars. He figured that as long as he never wrote on them, and he didn't care about the more irregular holidays like Easter, he could use and reuse a set of calendars until he died. He had a calendar for every year since the year he was born, so he knew he wouldn't have to buy too many more. He did want to maximize his space though, so he decided to get rid of any of the ones that he already had that he didn't need. He would just buy the remaining ones that he would need as time wore on.
Can you figure out how many calendars he got rid of, how many he would need in total, and how many years it took him to finish his collection? Assume that he doesn't die before completing his collection, and that he won't buy a new one until January 1st of the year in question.
HintHe ended up with less than 30 calendars.
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Answer
Fred got rid of seven calendars: 1983, 1986, 1987, 1989, 1990, 1991, and 1993. He needed a total of fourteen calendars, but it took him seven years to buy the last two: 1996 and 2000.
Luckily for him, Fred has a penchant for esoteric knowledge. He knew that as long as every four years was a leap year until he died, he could figure out his calendar problems with a few, easy rules:
- If the year is a leap year, he could reuse its calendar in years that were multiples of 28 from that year.
- If the year is the first year after a leap year, he could reuse its calendar in years that were multiples of 6, 17, or 28 from that year.
- If the year is the second year after a leap year, he could reuse its calendar in years that were multiples of 11, 17, or 28 from that year.
- If the year is the third year after a leap year, he could reuse its calendar in years that were multiples of 11, 22, or 28 from that year.
Since 2100 will not be a leap year, Fred knows that rules will get out of whack if he lives past 124. (He wasn't too worried about having to spend extra money, though, his family typically only lives to their mid- to late-nineties.)
Fred was 18 in 1993, so he must have been born in 1975. Since he had a calendar for every year he's been alive, he started with 19 calendars. He quickly saw that the most calendars he would ever need was 28 since all calendar years repeat after 28 years. He figured that it would be a lot less, though, and, since he didn't really like math, just drew up a table to help him determine the actual number. The table consisted of the calendars he would need and what years they would represent.
1975: 1986, 1997, 2003, 2014, 2025, 2031, 2042, 2053, 2059, 2070, 2081, 2087, 2098
1976: 2004, 2032, 2060, 2088
1977: 1983, 1994, 2005, 2011, 2022, 2033, 2039, 2050, 2061, 2067, 2078, 2089, 2095
1978: 1989, 1995, 2006, 2017, 2023, 2034, 2045, 2051, 2062, 2073, 2079, 2090
1979: 1990, 2001, 2007, 2018, 2029, 2035, 2046, 2057, 2063, 2074, 2085, 2091
1980: 2008, 2036, 2064, 2092
1981: 1987, 1998, 2009, 2015, 2026, 2037, 2043, 2054, 2065, 2071, 2082, 2093, 2099
1982: 1993, 1999, 2010, 2021, 2027, 2038, 2049, 2055, 2066, 2077, 2083, 2094
1984: 2012, 2040, 2068, 2096
1985: 1991, 2002, 2013, 2019, 2030, 2041, 2047, 2058, 2069, 2075, 2086, 2097
1988: 2016, 2044, 2072
1992: 2020, 2048, 2076
1996: 2024, 2052, 2080
2000: 2028, 2056, 2084
It didn't take him too long to come up with, and at the end he had a useful resource that he could keep with his calendars. Using the table he had created, he would know which one to break out at the start of a new year.
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