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# November 16, 2001, was a Friday. If each of the years 2004, 2008, and

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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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parkhydel
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
PS91151.02

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

Cheers,
Brent
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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parkhydel
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

PS91151.02
Straight to point:

Though this requires a lot of math as a rule of thumb:
if its Friday this day this year it will be Saturday next year (if this year isn't a leap year(366days)
& if this year is a leap year next year it will be +1 day i.e., Friday in leap>> next year Saturday+1=Sunday

So doing some goofy math counting write down:
2001-fri,
2-sat,
3-sun,
4-mon
5-tue+1=wed
...........14>>>Sunday

or

u can also do it like: count normally.. fri,sat,sun,mon,tues-2005(+1)......2009(+1)...2013(+1)...... 2014->Thurs+3=sunday

or

If it's a long gap then find the no.of years between: here 13 years n 3(+1)>>> 16 days from friday: friday+16=firday+2=Sunday(16/7 gives remainder=2)

As simple as this

OA:A

HOPE this helps.
THANKS
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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parkhydel
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

PS91151.02

This is essentially a remainder problem. First, there are 13 years. Think of how old you'd be on Nov 16, 2014 if you were born on Nov 16, 2001.

Three of the years have 366 days, and the other ten years have 365 days.

It helps to break the calculations down to bite-size pieces. Find the remainders for 366 and 365, then use the additive property of remainders described below, rather than calculating the total number of days and finding the remainder for that big number.

To find the remainder when 366 is divided by 7, look for an easy-to-find multiple of 7 near 366... like 350.

366 is 16 more than 350, so the remainder when 366 is divided by 7 is the same as the remainder when 16 is divided by 7... that is 2.

Likewise, 365 is 15 more than 350, so the remainder when 365 is divided by 7 is the same as the remainder when 15 is divided by 7... that is 1.

In total, here's what we've got, using abbreviated notation.

3 x 366 DIV 7 = 3 x R2 = R6
10 x 365 DIV 7 = 10 x R1 = R10
_______________________________
R6 + R10 = R16 = R2

To finish it off, FRIDAY + R2 = FRIDAY + 2 DAYS = SUNDAY

For this approach, it helps to understand the additive property of remainders. For example,

remainder when dividing 7 by 4 = 3
remainder when dividing 14 by 4 = 2

Adding these two remainders 3 + 2 = 5 and then dividing 5 by the divisor of 4 leaves a remainder of 1.

If you add 7 + 14 = 21 and then divide by 4, the remainder is also 1.

Likewise, if you add several of the same remainders together, you can multiply the remainders and then divide by the divisor to get the remainder.

For example, 7 x 3 = 21. The remainder when you divide 7 by 4 is 3. Multiply that remainder of 3 by 3 = 9 and then divide by 4 to get a remainder of 1, which is the same remainder you get when you divide 21 by 4.

So for this problem:

3 years with 366 days, so 3 x remainder of 2 = 6
10 years with 365 days, so 10 x remainder of 1 = 10

Add 6 + 10 = 16 then divide by 7 to get a remainder of 2.

It takes a lot of words to write this out, so instead I suggest using notation like 7 DIV 4 = R3 to mean "7 divided by 4 yields a remainder of 3"

Or if you're familiar with MOD notation, it's the same thing... 7 MOD 4 = 3
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November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
Using the whiteboard, i did it as follows:
In year 2012, the day would come out to be friday. This is because the sum of (number of days in each year remainder by 7) is divisible by 7.
2,3,4,5...14 denote the years
The number after '-' denote number of days in that year remainder 7
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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I know this may not be the best approach, but as someone who gets excited to see the day her birthday falls on every year, I know for a fact that if my birthday falls on Friday this year (2020) then it is bound to fall on Saturday next year (2021). And similarly if it is a leap year, it will fall 2 days after the current day. (This is also logical in the sense that adding every 365 days to the current date is as good as 365 days +1 which will be the day after today in the case of a normal year.) So therefore, by doing my calculation like this, I really didn't take more than a minute. Sometimes real life logic works great.

2001 - Friday
2002 - Saturday
2003 - Sunday
2004 - Tuesday (Leap year so +2)
.
.
.
.
2014 - Sunday

Try it with your birthdays to enjoy this trick to the most! Hit kudos if you enjoyed this solution.
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November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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parkhydel
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

PS91151.02

Between 2001 and 2014 are 10 non-leap years and 3 leap years.
Each non-leap year is composed of 365 days; each leap year is composed of 366 days.
Thus, from November 16, 2001 to November 16, 2014, the total number of days = 10(365) + 3(366).

Since November 16, 2001 is a Friday, every passing of 7 days will yield another Friday.
364 is a multiple of 7.
Rephrase the total number of days in terms of 364:
10(365) + 3(366) = 10(364+1) + 3(364+2) = 13(364) + 16 = 13(364) + 14 + 2 = (multiple of 7) + (multiple of 7) + 2 = Friday + 2 days = Sunday

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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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parkhydel
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

PS91151.02

We see that there is a 13-year span from November 16, 2001 to November 6, 2014. If each year were 365 days, the total number of days in this 13-year span would be 365 x 13 = 4,745 days. However, since 3 of the years during this span each has 1 extra day, the actual number of days in this span is 4,745 + 3 = 4,748 days. Since 4,748/7 = 678 R 2, we know that there were 678 weeks and 2 days from Friday, November 16, 2001 until November 16, 2014. The remainder of 2 tells us that November 16, 2014 was 2 days after Friday, so it was Sunday.

this is the best solution
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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Solution:

Total number of years from November 16, 2001 to November 6, 2014 = 13 years

=13 * 365 = 4745 days

Since, 2004, 2008, and 2012 had 366 days we have total number of days = 4745 + 3

= 4748 days

= 4748/7 = 678 weeks and 2 extra days (as remainder)

Thus, November 16, 2014 was 2 days after Friday

=> Sunday (option a)

Hope this helps
Devmitra Sen(GMAT Quant Expert)
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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13 years + 3 days
13*365 + 3

remainder for each respective term when divided by 7
6*1 + 3 = 9

Remainder for 9 divided by 7 = 2

So two days after friday
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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$$\frac{365}{7}=reminder(\frac{350+14+1}{7})=1$$
each normal year we go on 1 day, in 2004,2008 and 2012 we go on 2 days
2001 +0
2002 +1
2003 +1
2004 +2
2005 +1
2006 +1
2007 +1
2008 +2
2009 +1
2010 +1
2011 +1
2112 +2
2013 +1
2014 +1

$$reminder(16/7)=2 \to Sunday$$

A
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
Since each day repeats itself after 7 days, the remainder of the total number of days will give us the shift from the present day.

From 2001-2014, there were 3 leap years,i.e., 2004, 2008, and 2012.
So, the normal years will be 13-3 = 10.

Thus, total number of days = 365*10 + 366*3
(Total number of days)%7 = (365*10)%7 + (366*3)%10 = 3 + 6 = 9

Again, as each days repeats itself after 7 days, shift = 9 - 7 = 2

So since November 16, 2001, was a Friday, November 16, 2014, will be Sunday.
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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COVID19butGMAT22
I know this may not be the best approach, but as someone who gets excited to see the day her birthday falls on every year, I know for a fact that if my birthday falls on Friday this year (2020) then it is bound to fall on Saturday next year (2021). And similarly if it is a leap year, it will fall 2 days after the current day. (This is also logical in the sense that adding every 365 days to the current date is as good as 365 days +1 which will be the day after today in the case of a normal year.) So therefore, by doing my calculation like this, I really didn't take more than a minute. Sometimes real life logic works great.

2001 - Friday
2002 - Saturday
2003 - Sunday
2004 - Tuesday (Leap year so +2)
.
.
.
.
2014 - Sunday

Try it with your birthdays to enjoy this trick to the most! Hit kudos if you enjoyed this solution.

Great solution, this is indeed the fastest approach, especially if you are familiar with this real life logic.

10 non-leap years * 1 day forward each = 10 days forward
3 leap years * 2 days forward each = 6 days forward
10 + 6 = 16 days forward = 2 weeks and 2 days extra
2 weeks forward takes us to Friday, and the 2 extra days takes us to Sunday.

(If we didn't know in advance that a birthday advances 1 day forward each non-leap year, then we could divide: 365/7 = 52 weeks + 1 day extra)
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Re: November 16, 2001, was a Friday. If each of the years 2004, 2008, and [#permalink]
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