To find the day of the week on 17th June 1998, we can use Zeller's Congruence formula or simply check a reliable calendar.
Using Zeller's Congruence:
For the Gregorian calendar, Zeller's formula for the day of the week is:
h=(q+⌊13(m+1)5⌋+K+⌊K4⌋+⌊J4⌋+5J)mod 7h=\left(q+\left\lfloor\frac{13(m+1)}{5}\right\rfloor +K+\left\lfloor\frac{K}{4}\right\rfloor +\left\lfloor\frac{J}{4}\right\rfloor +5J\right)\mod 7h=(q+⌊513(m+1)⌋+K+⌊4K⌋+⌊4J⌋+5J)mod7
Where:
- hhh = day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, ..., 6 = Friday)
- qqq = day of the month
- mmm = month (3 = March, 4 = April, ..., 12 = December; January and February are counted as months 13 and 14 of the previous year)
- KKK = year of the century (year % 100)
- JJJ = zero-based century (year / 100)
For 17 June 1998:
- q=17q=17q=17
- m=6m=6m=6
- Year = 1998, so K=98K=98K=98, J=19J=19J=19
Calculate:
h=(17+⌊13(6+1)5⌋+98+⌊984⌋+⌊194⌋+5×19)mod 7h=\left(17+\left\lfloor\frac{13(6+1)}{5}\right\rfloor +98+\left\lfloor\frac{98}{4}\right\rfloor +\left\lfloor\frac{19}{4}\right\rfloor +5\times 19\right)\mod 7h=(17+⌊513(6+1)⌋+98+⌊498⌋+⌊419⌋+5×19)mod7
=(17+⌊13×75⌋+98+24+4+95)mod 7=\left(17+\left\lfloor\frac{13\times 7}{5}\right\rfloor +98+24+4+95\right)\mod 7=(17+⌊513×7⌋+98+24+4+95)mod7
=(17+18+98+24+4+95)mod 7=256mod 7=4=(17+18+98+24+4+95)\mod 7=256\mod 7=4=(17+18+98+24+4+95)mod7=256mod7=4
Day 4 corresponds to Wednesday.
Final answer:
17th June 1998 was a Wednesday.
