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November 16, 2001, was a Friday. If each of the years 2004,
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24 Aug 2020, 09:52
24 Aug 2020, 09:52
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58% (02:10) correct
41% (01:43) wrong based on 24 sessions
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November 16, 2001, was a Friday. If each of the years 2004, 2008, and 2012 had 366 days, and the remaining years from 2001 through 2014 had 365 days, what day of the week was November 16, 2014 ?
A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday
A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday
Re: November 16, 2001, was a Friday. If each of the years 2004,
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24 Aug 2020, 09:53
24 Aug 2020, 09:53
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
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Re: November 16, 2001, was a Friday. If each of the years 2004,
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24 Aug 2020, 10:05
24 Aug 2020, 10:05
2
Carcass wrote:
November 16, 2001, was a Friday. If each of the years 2004, 2008, and 2012 had 366 days, and the remaining years from 2001 through 2014 had 365 days, what day of the week was November 16, 2014 ?
A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday
A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday
The period from November 16, 2001 to November 16, 2014 is 13 years.
If we ignore the leap years, the total number of days during that period = (13)(365)
Since 3 of the years during that period had 1 extra day, the TOTAL number of days during the given period = (13)(365) + 3
Important: Although we COULD evaluate (13)(365) + 3 to get 4749, we can save ourselves a bit of time by making a few observations
Notice that 364 is a multiple of 7. In fact 364 = (7)(52)
So, (13)(365) + 3 = (13)(364 + 1) + 3
= (13)(364) + (13)(1) + 3
= (13)(364) + 16
= (13)(364) + 14 + 2
= 7[(13)(52) + 2] + 2
Since 7[(13)(52) + 2] is a multiple of 7, we know that after 7[(13)(52) + 2] days, the day will be FRIDAY
So, after 7[(13)(52) + 2] + 1 days, the day will be SATURDAY
And after 7[(13)(52) + 2] + 2 days, the day will be SUNDAY
Answer: A
Cheers,
Brent
