Calendar Problems for Competitive Exams: Speed Tricks
Four question types, one day-of-week formula, and three traps that cost time in TCS NQT, AMCAT, and Mu Sigma MuApt calendar sections.
Calendar questions in campus placement tests follow four distinct patterns, and naming the pattern within 20 seconds of reading the question is more than half the work.
This article covers fast pattern recognition and the shortcuts that apply at exam speed. The foundations guide covers odd-day derivations and the formula proof from scratch. If your exam is next week, start here. If you want to understand why the month codes are what they are, start there.
Spot the Question Type First
Four question types appear in Indian placement aptitude tests. Matching the question to its type before picking up a pencil saves 30 to 60 seconds per question.
| Question type | What is given | What is asked |
|---|---|---|
| Day-of-week | A specific date (day, month, year) | Which day of the week |
| Same-calendar year | A given year | Next year with an identical calendar |
| Day-count | Two specific dates | Number of days between them |
| Leap-year check | A year | Whether that year is a leap year |
Day-of-week and same-calendar-year are the two types that appear most often in TCS NQT, AMCAT Quantitative, and Mu Sigma MuApt. Prioritise those two in your first pass through the question paper.
The Four Tables to Memorise
Four lookup tables are the only things you need in memory on exam day. Everything else is arithmetic.
Odd-days table
| Period | Odd days |
|---|---|
| Ordinary year (365 days) | 1 |
| Leap year (366 days) | 2 |
| 100 years | 5 |
| 200 years | 3 |
| 300 years | 1 |
| 400 years | 0 |
Month codes (non-leap year)
| Month | Code | Month | Code |
|---|---|---|---|
| January | 0 | July | 6 |
| February | 3 | August | 2 |
| March | 3 | September | 5 |
| April | 6 | October | 0 |
| May | 1 | November | 3 |
| June | 4 | December | 5 |
For leap years only: January becomes 1 (not 0) and February becomes 4 (not 3). All other months are unchanged.
Century codes
| Century range | Code |
|---|---|
| 1600s / 2000s | 6 |
| 1700s / 2100s | 4 |
| 1800s / 2200s | 2 |
| 1900s / 2300s | 0 |
The sequence descends: 6, 4, 2, 0 as you step forward by 100 years.
Day-of-week output map
| Remainder (mod 7) | Day |
|---|---|
| 0 | Sunday |
| 1 | Monday |
| 2 | Tuesday |
| 3 | Wednesday |
| 4 | Thursday |
| 5 | Friday |
| 6 | Saturday |
Day-of-Week Formula in Three Steps
The formula takes four memorised values plus the date and produces a remainder that maps directly to a weekday.
Formula: (Y + floor(Y ÷ 4) + M + C + D) mod 7
Where Y = last two digits of the year, M = month code, C = century code, D = date (day of month).
Steps:
- Step 1: Extract Y from the year. Compute floor(Y ÷ 4). Add the two values together.
- Step 2: Look up M (use the leap-year column for January or February if the year is a leap year). Look up C.
- Step 3: Add M, C, and D to the step-1 total. Take mod 7. Match the remainder to the output map.
Worked example: 15 August 1947
- Y = 47. floor(47 ÷ 4) = floor(11.75) = 11. Step-1 total = 47 + 11 = 58.
- M = 2 (August, no leap-year adjustment). C = 0 (1900s). D = 15.
- Sum = 58 + 2 + 0 + 15 = 75. 75 mod 7 = 5.
- Remainder 5 = Friday. India’s first Independence Day was a Friday.
The only derivations during the exam are a single floor-division and a final modulo. Both take under 10 seconds on scratch paper.
Three traps that flip the answer
- Trap 1: Wrong January or February code in a leap year. If the question involves a date in January or February of a leap year, use codes 1 and 4 respectively. Using 0 and 3 gives a result that is off by one day. Check whether the year is a leap year before looking up the month code.
- Trap 2: Wrong century code. 1900s = 0, 2000s = 6. Students often swap these. A mnemonic: the current century (2000s) has the highest code (6). Codes decrease by 2 as you go back in time.
- Trap 3: Taking floor incorrectly. floor(47 ÷ 4) = 11, not 12. The floor function truncates toward zero. On a rough scratch calculation, write the division result and drop everything after the decimal point.
Same-Calendar Year Problems
Two years share the same calendar when they start on the same day of the week AND have the same leap or non-leap status. A leap-year calendar and a non-leap-year calendar can never match, even if both start on Monday.
The method:
- Step 1: Start from the given year and add its odd days to a running total.
- Step 2: Continue year by year (1 per ordinary year, 2 per leap year).
- Step 3: When the running total is divisible by 7, check the next year’s type.
- Step 4: If the next year matches the starting year’s type, that is the answer. If not, continue.
Worked example: year after 2010 with the same calendar
2010 is an ordinary (non-leap) year. The table tracks cumulative odd days:
| Year | Type | Odd days | Running total | Check |
|---|---|---|---|---|
| 2010 | Ordinary | 1 | 1 | — |
| 2011 | Ordinary | 1 | 2 | — |
| 2012 | Leap | 2 | 4 | — |
| 2013 | Ordinary | 1 | 5 | — |
| 2014 | Ordinary | 1 | 6 | — |
| 2015 | Ordinary | 1 | 7 (0 mod 7) | Next year 2016 is leap — no match |
| 2016 | Leap | 2 | 2 | — |
| 2017 | Ordinary | 1 | 3 | — |
| 2018 | Ordinary | 1 | 4 | — |
| 2019 | Ordinary | 1 | 5 | — |
| 2020 | Leap | 2 | 0 mod 7 | Next year 2021 is ordinary — match |
Answer: 2021.
When the running total hits 0 mod 7 after 2015, the next candidate year is 2016. Since 2016 is a leap year and 2010 is not, they cannot share a calendar. The accumulation continues until after 2020, where 2021 qualifies as an ordinary year starting on the same weekday as 2010.
Speed shortcut for non-leap years
Memorising the position of the year relative to the most recent leap year gives a faster path:
- Year immediately after a leap year (e.g., 2009, 2013): same calendar recurs after 6 years.
- Year 2 or 3 years after a leap year (e.g., 2010, 2011): same calendar recurs after 11 years.
2010 is 2 years after 2008 (a leap year), so same calendar = 2010 + 11 = 2021. The table confirms this.
For leap years, the same calendar repeats after 28 years.
Exam Attack Order and Time Budget
Not all calendar question types take the same time. Setting an attack order before the exam prevents last-minute scrambling.
Attempt in first pass (quick wins)
- Leap-year check: single divisibility test, under 15 seconds.
- Day-of-week with a non-century-boundary year and a direct date: three-step formula, under 60 seconds.
Attempt in second pass (higher setup cost)
- Same-calendar year: requires building the odd-day accumulation table. Budget 90 to 120 seconds.
- Day-count between two dates: requires careful leap-year inclusion for multi-year spans. Budget 90 seconds.
In the AMCAT Quantitative and Mu Sigma MuApt sections, time per question is tight. Spending more than 2 minutes on any single calendar problem usually signals a misidentified type. If a question resists the formula approach after 90 seconds, mark it and move on.
For the campus placement aptitude test format (CRT), most invigilating colleges run a 60-minute section with 20 to 25 questions. Calendar questions are worth the same marks as easier ratio or percentage questions. In a time-pressured section, take the leap-year and day-of-week questions first, then revisit same-calendar problems if time allows.
Calendar questions reward the same habit as time and work problems: recognise the pattern, apply the method, do not rederive. The three-step day-of-week formula reduces a question that looks hard to arithmetic that takes under 40 seconds. That same instinct, spotting the pattern and applying a known method without rewriting from scratch, transfers directly to evaluating language model outputs. At ₹299, TinkerLLM is where that habit meets live model calls, not just aptitude formulas. Practice sets at IndiaBix Calendar are the fastest way to build formula recall before the exam.
Primary sources
Frequently asked questions
How do I identify which type of calendar question I am facing?
Check what is given and what is asked. If you have a full date and need the weekday, apply the day-of-week formula. If the question asks for the next year with the same calendar, count odd-day accumulation. If you are counting days between two dates, subtract and adjust for leap years. If the question only asks whether a year is a leap year, apply the 4/100/400 divisibility rule.
What are odd days in calendar problems?
Odd days are the remainder when you divide a period's total days by 7. An ordinary year has 1 odd day (365 mod 7 equals 1). A leap year has 2 odd days. 100 years has 5 odd days, 200 years has 3, 300 years has 1, and 400 years has 0.
Is the year 1900 a leap year?
No. 1900 is divisible by 100 but not by 400, so it is not a leap year. The standard rule applies to century years only: divisibility by 400 is required. The century code for the 1900s in the day-of-week formula is 0, reflecting this non-leap status.
What century code should I use for years 2000 to 2099?
The century code for years 2000 to 2099 is 6. For 1900 to 1999 it is 0. For 1800 to 1899 it is 2. For 1700 to 1799 it is 4. The pattern cycles in descending order: 6, 4, 2, 0 as you step forward by 100 years.
Which placement tests include calendar questions?
TCS NQT Numerical Ability, AMCAT Quantitative, Mu Sigma MuApt, and most campus CRT aptitude sections include calendar questions. Day-of-week and same-calendar-year are the most common formats.
How long after a non-leap year does the same calendar repeat?
It depends on the year's position in the 4-year leap cycle. Typically 6 years if only one leap year falls in between, and 11 years if two leap years fall in between. The reliable approach is to count odd-day accumulation until the running total hits a multiple of 7, then check that the following year has the same leap or non-leap status as the start year.
A self-paced playground for building with LLMs.
TinkerLLM is FACE Prep's sister property. A guided environment for shipping real LLM applications, the kind of project that earns a paragraph on your resume, not a line.
Try TinkerLLM (₹299 launch)