Number of Tuesdays in five consecutive calendar years taken together is exactly 260. Which day of the week was 1st Jan. of the first of five years?

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Number of Tuesdays in five consecutive calendar years taken together is exactly 260. Which day of the week was 1st January of the first of these five years?

I don't know at all how to do this question. I know that in leap year there are 2 odd days and in ordinary year there is 1 odd day.

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Since $5*52=260$, the question is equivalent to asking what day of the week the first year must begin on if none of five consecutive years has 53 Tuesdays.

A year will have 53 Tuesdays if the dominical letter is F, FE, or GF.

Examining the possibilities for the dominical letters (including the cases involving non-leap century years) gives the following:

  • AG, F, E, D, CB (second year has 53 Tuesdays)
  • GF, E, D, C, BA (first year has 53 Tuesdays)
  • FE, D, C, B, AG (first year has 53 Tuesdays)
  • ED, C, B, A, GF (fifth year has 53 Tuesdays)
  • DC, B, A, G, FE (fifth year has 53 Tuesdays)
  • CB, A, G, F, ED (fourth year has 53 Tuesdays)
  • BA, G, F, E, DC (third year has 53 Tuesdays)
  • A, GF, E, D, C (second year has 53 Tuesdays)
  • G, FE, D, C, B (second year has 53 Tuesdays)
  • F, ED, C, B, A (first year has 53 Tuesdays)
  • E, DC, B, A, G
  • D, CB, A, G, F (fifth year has 53 Tuesdays)
  • C, BA, G, F, E (fourth year has 53 Tuesdays)
  • B, AG, F, E, D (third year has 53 Tuesdays)
  • A, G, FE, D, C (third year has 53 Tuesdays)
  • G, F, ED, C, B (second year has 53 Tuesdays)
  • F, E, DC, B, A (first year has 53 Tuesdays)
  • E, D, CB, A, G
  • D, C, BA, G, F (fifth year has 53 Tuesdays)
  • C, B, AG, F, E (fourth year has 53 Tuesdays)
  • B, A, GF, E, D (third year has 53 Tuesdays)
  • A, G, F, ED, C (third year has 53 Tuesdays)
  • G, F, E, DC, B (second year has 53 Tuesdays)
  • F, E, D, CB, A (first year has 53 Tuesdays)
  • E, D, C, BA, G
  • D, C, B, AG, F (fifth year has 53 Tuesdays)
  • C, B, A, GF, E (fourth year has 53 Tuesdays)
  • B, A, G, FE, D (fourth year has 53 Tuesdays)
  • $2096, 2097, 2098, 2099, 2100$ ($2097$ has 53 Tuesdays)
  • $2097, 2098, 2099, 2100, 2101$ ($2097$ has 53 Tuesdays)
  • $2098, 2099, 2100, 2101, 2102$
  • $2099, 2100, 2101, 2102, 2103$
  • $2100, 2101, 2102, 2103, 2104$ ($2104$ has 53 Tuesdays)
  • $2196, 2197, 2198, 2199, 2200$ ($2199$ has 53 Tuesdays)
  • $2197, 2198, 2199, 2200, 2201$ ($2199$ has 53 Tuesdays)
  • $2198, 2199, 2200, 2201, 2202$ ($2199$ has 53 Tuesdays)
  • $2199, 2200, 2201, 2202, 2203$ ($2199$ has 53 Tuesdays)
  • $2200, 2201, 2202, 2203, 2204$
  • $2296, 2297, 2298, 2299, 2300$
  • $2297, 2298, 2299, 2300, 2301$ ($2301$ has 53 Tuesdays)
  • $2298, 2299, 2300, 2301, 2302$ ($2301$ has 53 Tuesdays)
  • $2299, 2300, 2301, 2302, 2303$ ($2301$ has 53 Tuesdays)
  • $2300, 2301, 2302, 2303, 2304$ ($2301$ has 53 Tuesdays)

For the cases that do not involve non-leap century years, one can see that the first year must be a non-leap year beginning on a Wednesday.

For the cases that involve non-leap century years, the above $15$ cases are the only ones that one need to check for because the Gregorian calendar repeats every $400$ years. Of those $15$ cases, $4$ of them do not contain a year with $53$ Tuesdays. The years $2098, 2200,$ and $2296$ begin on Wednesdays, while the year $2099$ begins on a Thursday.

So, the answer is that the first year must usually begin on a Wednesday. Once every $400$ years, there is a solution beginning with a year that begins on a Thursday (namely, the years $2099$ to $2103$).