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June 25, 1982, fell on a Friday. On which day of the week d

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June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 14 Feb 2011, 16:19
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June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note: 1984 was a leap year.)

A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 15 Feb 2011, 10:15
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One year = 365 days except for leap years = 366 days
52 weeks in a year 7 days per week = 364 days

1982 25th a Friday, if we move 365 days in the next year we end up on the 25th, but the weekday will be Saturday since we have moved 52 weeks + 1 day into the future.
Hence,

1982 25th Friday
1983 25th Saturday (+1 day)
1984 25th Monday (+1 day and +1 extra day for leap year)
1985 25th Tuesday (+1 day)
1986 25th Wednesday (+1 day)
1987 25th = Thursday

If the system had been 364 days per year, every year we would have end up on the same day every year.
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 15 Feb 2011, 10:59
1
1982 -Friday
1983 -Fd+1=Sat
84+1=Sun
85+1=mon
86+1=tue
87+1=wed
1987 wed+1=Thursday
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 15 Feb 2011, 11:25
June 25, 1987

The following is a formula/mathematical approach to solving this type of question. It is kind of cumbersome; but once acquainted, you can easily solve these types of questions.

year 1600-> 0 odd days
In every 100 years, there are 5 odd days
+300-> 5*3 = 15 odd days = 15%7= 1 odd day

Upto 1900-> 1 odd day

Upto 1986:
86/4 = 21 leap years
86-21=65 ordinary years

Every leap year: 2 odd days,
ordinary year: 1 odd day
So; 21*2+65 = 42+65=107 odd days = 107%7 = 2 odd days

So; from 1600-1986 end = 1+2 = 3 odd days

in 1987(non-leap year)
Jan=31; Feb=28; March=31; April: 30 May:31 June: 25
31+28+31+30+31+25 = 176%7 = 1 odd day

Total: 3+1=4 odd days

odd days-day
0-Sunday
1-Monday
2-Tuesday
3-Wednesday
4-Thursday
5-Friday
6-Saturday

Ans: Thursday

*****************************
June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall?

june 25,1983: Saturday
june 25,1984: Monday (Leap year: Feb would be 29 days- so days 2+)
june 25,1985: Tuesday
june 25,1986: Wednesday
june 25,1987: Thursday
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 03 Jun 2014, 07:25
banksy wrote:
June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note: 1984 was a leap year.)

A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday


Similar question to practice: the-19th-of-september-1987-was-a-saturday-if-1988-was-a-126468.html
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 09 Feb 2017, 04:00
good question :-

remainder theorem can be used.

since 84 is a leap it contains one day extra..so totally 1826 days.

now every 7 days gives next friday..so 1826/7 will give remainder 6 indicating the last set of division alone is not complete to give :thursday !!
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 04 Nov 2017, 12:41
I set this up like a remainder problem. There are 5 years between 6/25/1982 and 6/25/1987...

365 * 5 = 1825
Add one day for leap year = 1825+1 = 1826
Divide 1826 days by 7 = 7/18 = 2, 7/42 = 6, 7/6 0 remainder 6 days.

So since the remainder is 6 days, the answer is one day less than Friday, which is Thursday.
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Re: June 25, 1982, fell on a Friday. On which day of the week d  [#permalink]

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New post 24 Nov 2018, 12:17
banksy wrote:
June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note: 1984 was a leap year.)

A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday


Total years = 5
Total days = 5 *365 + 1 (for leap year) = 1826
1826/7 gives a remainder of 6.

Therefore, Friday + 6 days = Thursday
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Re: June 25, 1982, fell on a Friday. On which day of the week d &nbs [#permalink] 24 Nov 2018, 12:17
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