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Consider the following sequence:

11, 6, 11, 11, 6, 11, 11, 6, 11, 6, 6, 11, 6, 11, 11, ...

Although it looks like a sequence of 6s and 11s at first, it will in fact eventually contain a 12. Also, it is periodic.

But where could the 12s possibly be within the repeating block?

Here are some hints:

The two consecutive 6s have to do with "00" years not divisible by 400 (2100, 2200, 2300, ...) not being leap years.

It will take longer if the two consecutive 6s involve 2300 than if they involve 2100 or 2200, as 2400 is a leap year.

You only need to search within the first 43 terms, as all days of the week except Tuesday and Thursday start common years 43 times in 400 years.

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    $\begingroup$ It is often hard to choose the right hints to make a problem interesting. Without the hints, this sequence is impossible to understand. With the hints given, it becomes too straightforward. Yet it feels like there was the potential for a great problem from this idea... $\endgroup$
    – Evargalo
    Commented Jul 20, 2023 at 13:04

3 Answers 3

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Not sure if I completely understood the puzzle. You have a sequence and are basically asking to continue the sequence, i.e. finding the rule. You already gave some clear hints...

So, I guess the rule is...

The number of years between years which are starting and ending on a Monday. Starting from the year 2210.

Years starting and ending on a Monday are: 2210,2221,2227,2238,2249,2255,2266,2277,2283,2294, 2300,2306,2317,2323,2334,2345,2351,2362,2373,2379, 2390,2401,2407,2418,2429,2435,2446,2457,2463,2474, 2485,2491,2503,2514,2525,2531,2542,2553,2559,2570, 2581,2587,2598,2610

...and the periodic sequence is...

defined by the differences between these numbers: 11,6,11,11,6,11,11,6,11,6,6,11,6,11,11,6,11,11,6,11,11,6, 11,11,6,11,11,6,11,11,6,12,11,11,6,11,11,6,11,11,6,11,12

So, there are 12s at...

the 32nd and 43rd position.

However, I haven't checked yet whether there are other possible definitions for the rule (disregarding periodicity).

Addition:

It looks like there are multiple solutions. There are other possible definitions of the rule:

The difference in years between years starting and ending on:

- Monday, starting from 2210. With 12s on positions 32 and 43

- Wednesday, starting from 2110. With 12s on positions 21 and 43

- Friday, starting from 2010. With 12s on positions 21 and 32

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Using the tip my hypothesis is that the double 6s in the sequence is related to the year 2100, which will start on a Friday and end on one, because it is not a leap year. This will break the usual spacing between the years that start and end on Fridays.

To test this hypothesis, I will examine the sequence of years that start and end on Fridays and their intervals. The sequence is:

1. 2010 - 2021 : 11 
2. 2021 - 2027 : 6  
3. 2027 - 2038 : 11 
4. 2038 - 2049 : 11 
5. 2049 - 2055 : 6  
6. 2055 - 2066 : 11 
7. 2066 - 2077 : 11 
8. 2077 - 2083 : 6
9. 2083 - 2094 : 11 
10. 2094 - 2100 : 6 
11. 2100 - 2106 : 6 
12. 2106 - 2117 : 11
13. 2117 - 2123 : 6 
14. 2123 - 2134 : 11    
15. 2134 - 2145 : 11

If this hypothesis is correct, then the sequence should continue as follows::

16. 2145 – 2151 : 6 
17. 2151 – 2162 : 11    
18. 2162 – 2173 : 11    
19. 2173 – 2179 : 6
20. 2179 – 2190 : 11    
21. 2190 – 2202 : 12

Answer: Finishing on item 21 of the sequence, in 2190 the year begins and ends on a Friday, and this only happens again in 2202, with a spacing of 12 years.

Explanation: A year has 365 days, except for leap years which have 366 days. A week has 7 days. If we divide the number of days in a year by the number of days in a week, we get a quotient and a remainder. The remainder tells us how many days are left after completing the last full week of the year. For example, if the remainder is 1, it means that the last day of the year is one day after the last Sunday of the year.

The remainder depends on whether the year is a leap year or not. For normal years, the remainder is 1, because 365 divided by 7 gives a remainder of 1. For leap years, the remainder is 2, because 366 divided by 7 gives a remainder of 2.

We can use this fact to figure out what day of the week a year starts on, based on the previous year. If the previous year was a normal year, then the next year will start on the day after the previous year ended. For example, if 2020 ended on a Thursday, then 2021 started on a Friday. If the previous year was a leap year, then the next year will start on the second day after the previous year ended. For example, if 2024 ended on a Tuesday, then 2025 started on a Thursday.

This means that only normal years can start and end on the same day of the week, because they have a remainder of 1. Leap years always shift the day of the week by one more than normal years. We can use this logic to create a sequence of years and their starting days of the week, if we know the starting day of the first year in the sequence.

The rule for leap years states that a year is a leap year if it is divisible by 4, except if it is also divisible by 100, unless it is also divisible by 400. For example, 2000 was a leap year, but 2100 will not be a leap year, even though both are divisible by 4 and 100. This exception affects the pattern of the years that start and end on the same day of the week.

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The first year is 2102 which starts on day number 11 (Thursday) of Zeller's congruence. Then you add years every 6 years until 2170 (number of the week, day 12, Sunday) which will be in position 29. After the forty-third number, the sequence is repeated, and the number 12 (Sunday) occurs again.

Therefore, the number 12 (Sunday) occurs in the 29th number of the sequence, which corresponds to the year 2170. This sequence occurs every 400 years, for it I used only number theory and Zeller's congruence itself. I hope I helped, bole it ia right haha.

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