1992 Paper 1 Q1

Year: 1992
Paper: 1
Question Number: 1

Course: LFM Pure
Section: Proof

Difficulty: 1484.0 Banger: 1500.0

proof modular arithmetic calendar calculation divisibility leap year systematic reasoning

Problem

Today's date is June 26th 1992 and the day of the week is Friday. Find which day of the week was April 3rd 1905, explaining your method carefully. {[}30 days hath September, April, June and November. All the rest have 31, excepting February alone which has 28 days clear and 29 in each leap year.{]}

Solution

There are \(87\) years between 1905 and 1992. Of those years, every 4th is a leap years, starting with 1908, and ending with 1992, so there are 22 leap years. There are 30 days between Apr 3 and May 3, 31 days between May 3 and Jun 3 and a further 23 days to the 26th. Therefore \(87 \times 365 + 22 \cdot 1 + 61 + 23\) total days. \(\pmod{7}\) this is \(3 \times 1 + 1 + 5 + 2 = 3\), therefore it is \(4\) week days before Friday, ie Monday.

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Difficulty Rating: 1484.0

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Banger Rating: 1500.0

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Problem source

Today's date is June 26th 1992 and the day of the week is Friday.
Find which day of the week was April 3rd 1905, explaining your method \textbf{carefully. }
{[}30 days hath September, April, June and November. All the rest have 31, excepting February alone which has 28 days clear and 29 in each leap year.{]}

Solution source

There are $87$ years between 1905 and 1992.
Of those years, every 4th is a leap years, starting with 1908, and ending with 1992, so there are 22 leap years. 


There are 30 days between Apr 3 and May 3, 31 days between May 3 and Jun 3 and a further 23 days to the 26th.


Therefore $87 \times 365 + 22 \cdot 1 + 61 + 23$ total days. $\pmod{7}$ this is $3 \times 1 + 1 + 5 + 2 = 3$, therefore it is $4$ week days before Friday, ie Monday.

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