Question:medium

A child was born on 13th January 1976 which was a Tuesday. What day of the week will be the child’s birthday in the year 1986?

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Same date across years: add $+1$ for each ordinary year and $+2$ for each leap year between them, then reduce modulo $7$.
Updated On: Jul 15, 2026
  • Sunday
  • Friday
  • Saturday
  • Monday
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The Correct Option is D

Approach Solution - 1

Step 1: Assign a number to each weekday.
Number the days starting from Sunday as \(0\) through Saturday as \(6\), so Tuesday is day code \(2\). This lets us do the whole problem with modular arithmetic instead of a table.

Step 2: Count the leap years in the gap.
From 1976 to 1986 is 10 years. Since the date in question is 13 January, every leap year whose 29 February falls inside this ten-year span adds one extra day of shift. Those leap years are 1976, 1980 and 1984, so there are 3 leap years.

Step 3: Work out the total day shift.
Total shift \( = 10 \text{ years} + 3 \text{ extra leap days} = 13 \) days. Reducing modulo 7 gives \( 13 \bmod 7 = 6 \).

Step 4: Add the shift to the starting code.
New code \( = (2 + 6) \bmod 7 = 8 \bmod 7 = 1 \), and day code \(1\) corresponds to Monday.
\[ \boxed{\text{Monday}} \]
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Approach Solution -2

Another way to approach this is to work out how many total days of shift each option would require from the starting Tuesday, and then check which of those shift values is actually achievable given how many leap years genuinely fall in this ten year window.

  1. Sunday: Landing on Sunday from Tuesday needs a shift of 5 days. Since the ten-year span from 1976 to 1986 must include at least 10 base days of shift plus extra days for each leap year crossed, a total shift as low as 5 is not achievable here.
  2. Friday: Landing on Friday needs a shift of 3 days, which is even smaller and just as unreachable given the same reasoning.
  3. Saturday: Landing on Saturday needs a shift of 4 days, still too small to be consistent with a ten-year span carrying its own leap year adjustments.
  4. Monday: Landing on Monday needs a shift of 6 days. Counting the three leap years that fall within this exact window, 1976, 1980 and 1984, the total shift works out to 10 base days plus 3 extra leap days, which is 13, and 13 reduces to 6 after removing complete weeks. This is exactly the shift Monday requires.

Only a 6-day shift is consistent with the number of leap years genuinely present in this ten-year window, and that shift corresponds to Monday.

So the correct answer is Monday.

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