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TS
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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 05 May 2005, 21:01
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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 ?

Approach: 6 leap years * 2 = 12 days
20 *1 = 20 days

32/7 remainder 4 i.e should it be wed+4 i.e sunday ?



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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 05 May 2005, 21:12
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What is the quickest way to solve this problem ?
Qn) If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 ?
Approach: 6 leap years * 2 = 12 days
20 *1 = 20 days
32/7 remainder 4 i.e should it be wed+4 i.e sunday ?
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how do you know that there were 6 leap years? cuz there could be 7 leap years during the 25 years period (but 26 different years), if 1978 was a leaap year.
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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 06 May 2005, 20:47
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What is the quickest way to solve this problem ?

Qn) If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 ?

Approach: 6 leap years * 2 = 12 days
20 *1 = 20 days

32/7 remainder 4 i.e should it be wed+4 i.e sunday ?
Show more


it shifts one day once on the same date of the next year. If it bumps into leap year, then it shifts one day more.

For example: Aug 2nd 1978 is Wednesday, then Aug 2nd 1979 is Thursday unless there is a leap year.

Now how many leap year during this period? given 6 times, I get Saturday.
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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 06 May 2005, 21:50
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What is the quickest way to solve this problem ?
Qn) If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 ?
Approach: 6 leap years * 2 = 12 days
20 *1 = 20 days
32/7 remainder 4 i.e should it be wed+4 i.e sunday ?

how do you know that there were 6 leap years? cuz there could be 7 leap years during the 25 years period (but 26 different years), if 1978 was a leaap year.
Show more


1978 is not a leap year. Only years divisible by 4 (by 400 in case of years with 00) are leap years.

1978 - 2003 = 25 Years => 25 days
1978 - 2003 => 6 leap years => 6 days

31/7, remainder 3, so it must be a Saturday.
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If Aug 2nd 1978 is a wednesday what day was Aug 2nd 2003 06 May 2005, 22:31
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amt
1978 is not a leap year. Only years divisible by 4 (by 400 in case of years with 00) are leap years.
1978 - 2003 = 25 Years => 25 days
1978 - 2003 => 6 leap years => 6 days
31/7, remainder 3, so it must be a Saturday.
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how do you konw that 1978 is not a leap year? does the question support this generalization? you are not supposed to use outside information. your answer should be justified by the informatio provided in the question.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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