- Step 1: Calculate the day for 20 October 1986.
From 20 Oct 1981 to 20 Oct 1986, there are 5 full years (1982, 1983, 1984, 1985, 1986). Among these, 1984 is a leap year. Normal years contribute 1 odd day, leap years contribute 2 odd days.
Total odd days = $(4 \times 1) + (1 \times 2) = 4 + 2 = 6$ odd days.
Since 20 October 1981 was Tuesday, 20 October 1986 will be Tuesday + 6 days = Monday.
- Step 2: Calculate odd days from 20 October 1986 to 10 June 1987.
Days remaining in 1986 from Oct 20: Oct ($31-20=11$), Nov (30), Dec (31).
Odd days in 1986 = $(11 \pmod 7) + (30 \pmod 7) + (31 \pmod 7) = 4 + 2 + 3 = 9 \pmod 7 = 2$ odd days.
Days in 1987 till June 10 (1987 is a normal year): Jan (31), Feb (28), Mar (31), Apr (30), May (31), Jun (10).
Odd days in 1987 = $(31 \pmod 7) + (28 \pmod 7) + (31 \pmod 7) + (30 \pmod 7) + (31 \pmod 7) + (10 \pmod 7) = 3 + 0 + 3 + 2 + 3 + 3 = 14 \pmod 7 = 0$ odd days.
Total odd days from 20 Oct 1986 to 10 Jun 1987 = $2 + 0 = 2$ odd days.
- Step 3: Final day calculation.
The day on 10 June 1987 will be Monday + 2 days = Wednesday.