Calendars: IPMAT practice questions

Solution

A year is a leap year if it is divisible by 4, but a century year must be divisible by 400.

1996 passes this test, while 1800, 1500, 1998 fail it.

Solution

Between the two dates there are 365 days, which leaves 1 odd day (each ordinary year gives 1, and each 29 February crossed adds 1 more).

Sunday + 1 = Monday.

Solution

421 ÷ 7 leaves remainder 1.

Wednesday + 1 day = Thursday.

Solution

A calendar repeats when the year starts on the same day and is of the same type (leap or ordinary).

Odd days from 1997 to 2002: 7, a multiple of 7, and 2003 is an ordinary year like 1997.

So the answer is 2003.

Solution

Use a known date: 1 January 2000 was a Saturday.

From 16 September 1952 to 1 January 2000 there are 17273 days; 17273 ÷ 7 leaves 4, so count 4 days back from Saturday.

So the day is Tuesday.

Solution

Today is 1 day after Sunday: Monday.

Three days from today is 3 days after Monday: Thursday.

Solution

Today is 2 days after Wednesday: Friday.

The day after tomorrow is 2 days after Friday: Sunday.

Solution

Days from 1 January to 8 September = 250.

250 ÷ 7 leaves 5, so the day is Friday + 5 = Wednesday.

Solution

Days from 1 January to 27 July = 207.

207 ÷ 7 leaves 4, so the day is Thursday + 4 = Monday.

Solution

20 − 5 = 15 days, which leaves 1 after removing whole weeks.

Sunday + 1 = Monday.

Solution

100 years have 76 ordinary + 24 leap years = 76 + 48 = 124 odd days → 5 odd days. So 100 → 5, 200 → 3, 300 → 1, 400 → 0 (the 400th year is a leap year).

400 years = 1 × 400 years (0 odd days each) + 0 years → 0 odd days.

Solution

Today is 3 days after Thursday: Sunday.

Tomorrow is 1 day after Sunday: Monday.

Solution

Between the two dates there are 731 days, which leaves 3 odd days (each ordinary year gives 1, and each 29 February crossed adds 1 more).

Wednesday + 3 = Saturday.

Solution

768 ÷ 7 leaves remainder 5.

Saturday + 5 days = Thursday.

Solution

A year is a leap year if it is divisible by 4, but a century year must be divisible by 400.

1600 passes this test, while 2026, 1700, 2023 fail it.

Solution

Between the two dates there are 1096 days, which leaves 4 odd days (each ordinary year gives 1, and each 29 February crossed adds 1 more).

Tuesday + 4 = Saturday.

Solution

100 years have 76 ordinary + 24 leap years = 76 + 48 = 124 odd days → 5 odd days. So 100 → 5, 200 → 3, 300 → 1, 400 → 0 (the 400th year is a leap year).

1000 years = 2 × 400 years (0 odd days each) + 200 years → 3 odd days.

Solution

Use a known date: 1 January 2000 was a Saturday.

From 8 July 1961 to 1 January 2000 there are 14056 days; 14056 ÷ 7 leaves 0, so count 0 days back from Saturday.

So the day is Saturday.

Solution

A calendar repeats when the year starts on the same day and is of the same type (leap or ordinary).

Odd days from 2007 to 2017: 14, a multiple of 7, and 2018 is an ordinary year like 2007.

So the answer is 2018.

Solution

Days from 1 January to 6 July = 187 (leap year).

187 ÷ 7 leaves 5, so the day is Sunday + 5 = Friday.

Solution

21 − 7 = 14 days, which leaves 0 after removing whole weeks.

Wednesday + 0 = Wednesday.

Solution

A calendar repeats when the year starts on the same day and is of the same type (leap or ordinary).

Odd days from 1992 to 2019: 35, a multiple of 7, and 2020 is a leap year like 1992.

So the answer is 2020.

Solution

Use a known date: 1 January 2000 was a Saturday.

From 8 September 1980 to 1 January 2000 there are 7054 days; 7054 ÷ 7 leaves 5, so count 5 days back from Saturday.

So the day is Monday.

Solution

521 ÷ 7 leaves remainder 3.

Saturday + 3 days = Tuesday.