Roundabout Math 2 Trivia Quiz

Answer: Monday

The only time the 7th will fall EXACTLY 14 weeks after the 30th when the following year is a leap year, is if the 30th is in November, and the 7th is in a March that follows a February during a leap year. Otherwise, it won't work out exactly. The 30th MUST be a Friday, because two days before today is yesterday's yesterday, and three days after the day after that is four days after that. Since that day is four days after a Wednesday, yesterday's yesterday MUST be a Wednesday, making TODAY a Friday.
The first US Presidents elected in a century are:
18th - George Washington
19th - Thomas Jefferson
20th - Theodore Roosevelt
21st - George W. Bush
The Presidents of the United States during the first occurrence in a century of a Friday, November 30th appearing in the year before a leap year are:
18th - John Adams - 1799
19th - John Quincy Adams - 1827
20th - Calvin Coolidge - 1923
21st - George W. Bush - 2007
Since the US President TODAY is the same man as the first US President elected this century, this year is 2007. Therefore, today is Friday, November 30th, 2007. Last year, Christmas fell on MONDAY, December 25, 2006.

Answer: 4

The number of letters in each word of the three sentences is as follows:
3-7-2-7-2-3-5-2-5-5-9-4-9-2-2-4-3-6-5
3-7-2-7-2-3-5-4-3-3-7-2-7-2-3-5-2-5-5-9-4-9-2-2-4-3-6-5
3-6-2-4-8-2-1-6-6-5-2-3-4-6-2-3-6-5

The number of letters in each word that is the number of letters in each word of the three sentences (i.e.- 'one' has three letters in it, 'three' has five letters in it, 'five' has four letters in it, etc...) is as follows:
5-5-3-5-3-5-4-3-4-4-4-4-4-3-3-4-5-3-4
5-5-3-5-3-5-4-4-5-5-5-3-5-3-5-4-3-4-4-4-4-4-3-3-4-5-3-4
5-3-3-4-5-3-3-3-3-4-3-5-4-3-3-5-3-4

The number of three letter number words that give the number of letters in the three sentences are:
First sentence - 6 (there are six words in the first sentence that have an amount of letters in them such that the amount of letters can be written as a three-letter word - i.e.- one, six, ten. Of, in, in, to, do, series - five two-letter words and a six-letter word, 'two' and 'six' each possessing 3 letters.)
Second sentence - 8 (there are eight words in the second sentence that have an amount of letters in them such that the amount of letters can be written as a three-letter word - i.e.- one, six, ten. Of, in, of, in, in, to, do, series - seven two-letter words and a six-letter word, 'two' and 'six' each possessing 3 letters.)
Third sentence - 10 (there are ten words in the first sentence that have an amount of letters in them such that the amount of letters can be written as a three-letter word - i.e.- one, six, ten. Answer, to, is, a, single, number, is, number, in, series - one one-letter word, four two-letter words and five six-letter words, 'one', 'two' and 'six' each possessing 3 letters.)

This explains the first four terms in the sequence - "3-6-8-10". The '3' represents numbers which, when written out, have three letters in their names.

The next 4 numbers in the sequence, "4-8-10-4" represent the same thing.

The last four numbers, "5-5-10- ?" represent the quantity of words in the three sentences with an amount of letters that, as numbers, when written out have five letters in their names.
First sentence - 5 (there are five words in the first sentence that have an amount of letters in them such that the amount of letters can be written as a five-letter word - i.e.- three, seven, eight. The, numbers, letters, the, the - three three-letter words and two seven-letter words, 'three' and 'seven' each possessing 5 letters.)
Second sentence - 10 (there are ten words in the first sentence that have an amount of letters in them such that the amount of letters can be written as a five-letter word - i.e.- three, seven, eight. The, numbers, letters, the, are, the, numbers, letters, the, the - six three-letter words and four seven-letter words, 'three' and 'seven' each possessing 5 letters.)
Third sentence - 4 (there are four words in the first sentence that have an amount of letters in them such that the amount of letters can be written as a five-letter word - i.e.- three, seven, eight. The, question, the, the - three three-letter words and an eight-letter word, 'three' and 'eight' each possessing 5 letters.)

Therefore, to complete the sequence as described, the '?' should be replaced with a '4'.

Answer: 1173

Every combination of any two of its digits is a prime number. Since there are no two-digit even prime number, none of the digits in the number can be even, since in some of the combinations of two of its digits, that even number would end up as the ones digit, making it an even two-digit number. The same can be said for the number 5, since there are no two-digit prime numbers ending in a '5'. This leaves 1, 3, 7 and 9. While many two-digit combinations of these numbers are prime, 39, 91 and 93 are not. This means that 1, 3 and 7 are the only digits that can be represented in the four-digit number, and at least one of these digits must be repeated at least once. Since, of 11, 33 and 77, only 11 is prime, the '1' must be the digit that gets repeated.

If the '1' is repeated four times, making the four-digit number '1111', there is only one combination of three of the digits, '111', which is not prime, thus contradicting the second condition. Similarly, if the '1' is repeated only three times, we will get one of the following sets of combinations:
111, 113, 131 and 311, of which 75% are prime, OR
111, 117, 171 and 711, none of which are prime.
Neither of these two possibilities fulfills the second condition.

Now, if we check the combinations where the '1' is repeated only once and each of the other digits, '3' and '7' appear (of course), then 113, 131, 137, 173, 311 and 317 are prime, while 117, 171, 371, 711, 713 and 731 are not. Exactly half of these numbers are prime, which fulfills the second condition.

So, the digits of the four-digit number MUST be 1, 1, 3 and 7.

And, coincidently, ALL four-digit combinations of 1, 1, 3 and 7 happen to have a prime number of prime factors (either two or three, both of which are prime). But only one, 1173, has a prime factorization for which the sum of its factors (3, 17 and 23) is also a prime number (43). It also happens, that if you consider ALL of its factors (1 and 1173 included) they add up to another prime number (1217).

Therefore, the answer is 1173.

Answer: $8.50

Draw a square and let it represent the 200 people in the room. Now, draw a vertical and horizontal line dividing the square in quarters. Let the top half of the square represent "half of the people in each of the groups", and the left half of the square represent "each of the people in half of the groups". Now, in each quarter of the square, write a negative dollar amount in it corresponding to the amount of money collected from each of the three sets of people: In the top two quarters, write "-$2". In the left two quarters, write "-$2", and in all four of the quarters, write "-$1", representing each of the people in each of the groups (or, all of the people). Now, you will see that, assuming even distribution, one quarter of the people have given $1, one quarter have given $5, and the remaining half have each given $3. This means that the least amount of money any of the people could conceivably have left after this step is $0, and the maximum is $4.
Now, draw another square and divide it, as above, into four quarters. Again, the top half will represent "half of the people in each of the groups", and the left half will represent "each of the people in half of the groups". The second part of the question states that I redistributed half of the accumulated pool of money among "each of the people in half of the groups" (which is the left half of the second square we have drawn). The accumulated pool of money is equal to $600 (10 people in 10 of the groups = 100 people x $2 = $200, 20 people in 5 of the groups = 100 people x $2 = $200, and 20 people in 10 groups = 200 people x $1 = $200. $200 + $200 + $200 = $600), half of which is $300. Since we are dividing it among half of the total amount of people, or 100 people, "each of the people in half of the groups" will get $3. We would then write +$3 in the left two quarters of the second square we have drawn. Then, we are dividing the remaining accumulated pool of money in half again (into $150 and $150), and distributing half of it to the top two quarters of the second square we have drawn, at $1.50 each, so we would write +$1.50 in the top two squares. Then since we are dividing the remaining $150 among all 200 of the people (at $0.75 each), we would write "+$0.75" in each of the four quarters. We can see from this that one quarter of the people were paid back $5.25, one quarter were paid back $4.50, one quarter were paid back $3.75, and the remaining quarter were paid back $0.75. Therefore, the maximum amount any one of the people could be paid back was $5.25, and the minimum was $0.75.
Since it is impossible to tell whether any of the people in the first square are in their corresponding quarters in the second square, there are no restrictions on the greatest or least possible amounts any of the people could have. Therefore, the maximum amount any of the people could have at the end would be one of the people who gave the least, but were paid back the most ($4 + $5.25 = $9.25). The minimum would be any of the people who gave all of their money, but were only paid back $0.75. The difference between these two amounts is $8.50.

Answer: 40

The river is traveling downstream at 2mph. This means that if I can generate a constant velocity in my canoe (lets call that constant velocity 'V'), I will be traveling 4mph FASTER going downstream than I will when I travel upstream. The only referent that we have is the flashlight. Since it had been in the water for 60 minutes traveling at 2mph (floating with the river's current) it must have traveled 2 miles. Now, I turned around at 2/3 of the distance between the point I put my canoe in the water and the campsite (let's call that the 'intended distance', and use 'D'), and traveled at V + 2mph ahead of the flashlight and then turned around at some point and traveled at V - 2mph back upstream until I caught it. Since I traveled twice D, I must have gone the equivalent of 1/2 of my intended distance downstream in search of my lunch and then 1/2 of my intended distance again back upstream. This means that I would have traveled 1 1/2 * D upstream at V - 2mph, and 1/2 * D downstream at V + 2mph. Since I took only 1 2/3 the amount of time in doing so, we can write the equation:
(3D/2) / (V - 2) + (D/2) / (V + 2) = 5/3 * (D / V - 2).
Solving, we get:
3/2(D/(V - 2)) + 1/2(D/(V + 2)) = 5/3(D/(V - 2))
1/2(D/(V + 2)) = 1/6(D/(V - 2))
D/(2V + 4) = D/(6V - 12)
2V + 4 = 6V - 12
V = 4
Therefore, I was creating a constant velocity of 4mph, meaning I was traveling at 2mph upstream and 6mph downstream (accounting for current).
If I traveled 1/2 of D downstream in search of my lunch, I turned back around after catching it while I was only 1/6 of the way along D. Then, after turning around, I caught up with the flashlight when I was 2/3 of the way along D, meaning that I had then traveled upstream for 1/6 of D. This means, that if I dropped the flashlight at some random point along my travel downstream, the flashlight would have traveled a fraction of D downstream with the current (let us call that fraction 'x'), and it took me the same time to travel xD (the random fraction of D that the flashlight traveled) PLUS 1/6D before I turned around, plus another 1/6D after I caught up with my lunch. Since I was traveling at 6mph for xD + 1/6D, and 2mph for another 1/6D, and the flashlight was traveling at 2mph for xD, we can now write the following formula:
xD / 2 = ((x + 1/6)D) / 6 + (1/6D) / 2
3xD / 6 = ((x + 1/6)D) / 6 + 3*(1/6D) / 6
3xD = (x + 1/6 + 1/2)D
3x = x + 2/3
2x = 2/3
x = 1/3
Therefore, the flashlight (which we have determined has traveled 2 miles) traveled 1/3 of the intended distance, which we now know was 6 miles. So, if the entire trip is 6 miles, and the rock formation is only 1/3 of the way along D (or rather, 2/3 D away from the camp), we can easily determine the amount of time the trip will take. Since having my friend in my canoe won't affect my time in any way, we will travel at 6mph for 2/3 of a 6 mile trip, which is 4 miles. Traveling 4 miles at 6mph will take 40 minutes.

Answer: 6

Since you are only making turns in increments of 60 degrees, and each of the legs of your journey are only 100 feet, if you randomly choose to turn left and right through infinity, you will create tracks that form a multitude of equilateral triangles for which adjacent triangles share a side. When enough of these triangles populate an area, they will form hexagons, each containing six such triangles due to the manner in which they are formed. If a circle is circumscribed about one of these hexagons, it will share the center of the hexagon and all 6 vertices of the hexagon will rest on the circumference of the circle.

If you walk, turn, walk, turn, walk turn, etc. you will arrive at many points. Occasionally you will arrive back at a previous turning point. Each point has the possibility of being 100' away from up to 6 other points. Let us say that if you can walk 100' from one point and end up at another point, those two points are adjacent. Adjacent points are listed in the following account of your journey (we will not list the origin point, as you will never return to it and there is not a red 'X' sprayed there):

BEGIN - walk 100' and spray an X at point A,
TURN LEFT at 60 degrees, and walk 100' to point B (adjacent to point A) and spray an X,
TURN LEFT at 120 degrees, and walk 100' to point C (adjacent to points A and B) and spray an X,
TURN RIGHT at 60 degrees, and walk 100' to point D (adjacent to point C) and spray an X,
TURN RIGHT at 120 degrees, and walk 100' to point E (adjacent to points B, C and D) and spray an X,
TURN LEFT at 180 degrees (essentially turning around and heading back), and walk 100' to point D (adjacent to points C and E) and spray an X,
TURN LEFT at 240 degrees, and walk 100' to point F (adjacent to points D and E) and spray an X,
TURN LEFT at 300 degrees, and walk 100' to point G (adjacent to points E and F) and spray an X,
TURN RIGHT at 180 degrees (essentially turning around and heading back), and walk 100' to point F (adjacent to points D, E and G) and spray an X,
TURN LEFT at 360 degrees (essentially going straight ahead), and walk 100' to point H (adjacent to points D and F) and spray an X,
TURN RIGHT at 240 degrees, and walk 100' to point D (adjacent to points C, E, F and H) and spray an X, and tie the rope to the post.
TURN RIGHT at 300 degrees, and walk 100' to point E (adjacent to points C, D, F and G) and spray an X.

If you are to draw a circle with its center at the post at point D, with a radius of 100', AND if you can walk 100' from one point and end up at another point, those two points are adjacent, and the points adjacent to point D are points C, E, F and H, then you will pass by every red 'X' that you sprayed at points C, E, F and H. If you look back at the account of the journey, you will see that you painted one X at point C, two at point E, two at point F, and one at point H, for a total of 6.

Answer: 80

First, since we are dealing with animals, none of the operations involved in this question will deal with fractional quantities. That said, the first sentence states that my friend took 1/3 of twice the amount I started with. This can be represented by 2x/3, leaving me with 4x/3 penguins, since 2x/3 + 4x/3 = 2x, which is twice as many as I began with. This means that the amount I started with (and by extension the amount my friend borrowed) is an even amount, but the amount I started with is also divisible by 3. Also, the amount that my friend took from me (which is 2/3 of the amount I started with) is greater than 16, since he gave 15 of them back to me in the second sentence, and an kept additional amount greater than one (since he kept that one as a pet) all but one of which he gave back to me in the fourth sentence. Now, he could not have borrowed more than 20 penguins from me, because in that case, when I bought twice as many to make up for the ones he took, I would have 4/3 of my original amount plus an additional 4/3 of my original amount, or 8/3 of my original amount, plus the fifteen that he gave back to me. If the number of penguins he took is greater than 20 (and since it has to be an even number, the least of those is 22), the minimum amount of penguins I would have after the second sentence would be 4 x 22 (since he had 2/3 of my original amount, which is 1/4 of 8/3 of my original amount, we will multiply that amount by 4), or 88 + 15, which is 103. Since I never had an amount of penguins greater than 100, this can not be the case.
Now, since my friend had to have taken MORE than 16 penguins from me, I must have an amount of penguins GREATER than 24 (and an even amount that is also a multiple of 3, as stated earlier), or else he would be unable to take at least 16 penguins from me. This amount MUST be 30, since anything more than that would mean that he borrowed more than 20 penguins from me, which was impossible.
Since I started with 30 penguins:

I started with 30, and by the time my friend borrowed 1/3 of them, I had twice as many, or 60. If he took 1/3, I was left with 2/3 of 60, or 40 penguins. I bought twice as many as he took to make up for the ones he had taken, which since he took 20, is an additional 40 penguins. Then I had 80 penguins. He gave me back 15, so then I had 95, which was 5 times the amount I ended up with, or 19. My friend still had 5 of my penguins (he gave me back 15 out of 20), and he gave me back 4 of them, keeping the last one as a pet. Then I had 99 penguins. Since my pen only holds 100 penguins, I decided to sell some of them, and got rid of 2/3 of them, or 66, leaving me with 33. This was 7 less than 40 penguins, which is twice what my friend borrowed from me. The wildlife refuge wanted to buy a more even number that 66, so they bought an additional 14 penguins, reducing my 33 penguins down to 19, and raising the total amount of penguins that they bought from me to 80.

Answer: four

1) If this sentence is false, then the following sentence is also false.
THIS SENTENCE IS FALSE, BECAUSE FOR IT TO BE TRUE, BOTH SENTENCES REFERRED TO BY IT MUST BE FALSE. THEREFORE, IF IT WERE TRUE, IT WOULD BE FALSE. SINCE THIS CAN NOT BE, THEN THIS SENTENCE MUST BE FALSE, MEANING THAT THE NEXT SENTENCE IS TRUE.
2) If the answer to this question is two, then the correct thing to write in the blank is "four".
THIS SENTENCE IS TRUE, PER THE PREVIOUS SENTENCE.
3) If more of these sentences are true than false, the answer to this question is three.
THIS SENTENCE IS FALSE, SINCE THERE ARE MORE TRUE SENTENCES THAN FALSE ONES, AND THE ANSWER IS NOT THREE.
4) If the second sentence is true, then the answer to this question is two.
THIS SENTENCE IS TRUE, BECAUSE THE TENTH SENTENCE STATES THAT THE ANSWER TO THE QUESTION IS THE NUMBER OF TRUE SENTENCES CONTAINING THE WORD "FALSE". SINCE THE SECOND SENTENCE IS TRUE AND THERE ARE TWO SENTENCES FULFILLING THE CONDITIONS IN THE TENTH SENTENCE, THIS SENTENCE IS TRUE.
5) More than half of the previous sentences are not true.
THIS SENTENCE IS FALSE, BECAUSE TWO OF THE PREVIOUS SENTENCES ARE TRUE AND TWO ARE FALSE, SO THE NOT TRUE SENTENCES DO NOT MAKE UP "MORE THAN HALF" OF THE PREVIOUS ONES.
6) More than half of the previous sentences are not true.
THIS SENTENCE IS TRUE, BECAUSE THERE ARE NOW THREE FALSE SENTENCES AND TWO TRUE SENTENCES MAKING UP THE PREVIOUS FIVE SENTENCES.
7) If the blank says "four", then four is the answer to this question.
THIS SENTENCE IS FALSE, PER THE TWELFTH SENTENCE.
8) None of the sentences containing the word "false" are true.
THIS SENTENCE IS FALSE, BECAUSE THE TENTH SENTENCE AND THE LAST SENTENCE ARE BOTH TRUE AND CONTAIN THE WORD "FALSE".
9) The next sentence is false.
THIS IS FALSE BECAUSE EITHER IT IS FALSE AND THE NEXT SENTENCE IS TRUE, OR IT IS TRUE AND THE NEXT SENTENCE IS FALSE. EITHER WAY, THE ANSWER TO THE QUESTION (WHICH ACCORDING TO THE NEXT SENTENCE DEPENDS ON THE NUMBER OF TRUE SENTENCES CONTAINING THE WORD "FALSE") STAYS THE SAME, BECAUSE ONLY ONE OF THESE SENTENCES IS TRUE, BUT BOTH CONTAIN THE WORD "FALSE". IF THE NEXT SENTENCE WERE FALSE, THEN THE FOURTH SENTENCE WOULD ALSO BE FALSE, MAKING THE FIFTH SENTENCE ALSO FALSE, WHICH MAKES THE ELEVENTH SENTENCE TRUE, WHICH CREATES A CONTRADICTION WHEN THE THIRD SENTENCE IS CONSIDERED, GIVEN THAT THE THIRD SENTENCE WOULD BECOME TRUE. BUT SINCE IT CONTAINS THE WORD "FALSE", THERE WOULD NOW BE FOUR TRUE SENTENCES CONTAINING THE WORD "FALSE".
10) The answer to this question is the number of true sentences containing the word "false".
PER THE NINTH SENTENCE, THIS SENTENCE IS TRUE.
11) The fifth and sixth sentences are both false.
THIS SENTENCE IS FALSE PER THE FIFTH AND SIXTH SENTENCES.
12) The word "five" should be written in the blank if the answer to this question is four.
THIS SENTENCE IS TRUE (THOUGH TOPICALLY IRRELEVANT) PER THE NEXT SENTENCE.
13) This sentence and the previous sentence are true if most of these sentences are false, and false if most of these sentences are true.
SINCE UP TO THIS POINT, SEVEN OF THE SENTENCES ARE FALSE, THIS AND THE PREVIOUS SENTENCE MUST BE TRUE. THIS BECOMES THE SECOND TRUE SENTENCE CONTAINING THE WORD "FALSE". (ALSO, IF IT WERE FALSE, THEN LOGICALLY, IT WOULD BE TRUE, MEANING IT CAN NOT BE FALSE).

Therefore, the answer to the question, per the tenth sentence is the number of true sentences containing the word 'false', which is two. Per the second sentence, if the answer is two, then you should have typed 'four' in the blank.

Answer: 33

Without using a graphing calculator and getting into a lot of calculus, we can use logical deduction to figure this out, even though there are no constants given in the question.

First, let's set some variables:
Let M = my age today
Let B = my brother's age today
Let D = M - B = the difference between my age and my brother's age (and also the number of years between when my father met my mother and when I was born)
Let F = my father's age when he met my mother
Let W = M + 1 = which wedding anniversary my parents are celebrating this year.

This year's wedding anniversary is one more than my age since they got married the year before I was born.
Now, we are going to need to create two equations so that we can solve this problem. The first is relatively easy. My father's age TODAY can be represented by F + D + M, so since the sum of our ages last year is the same as our father's age today:
(M-1) + (B-1) = F + D + M
AND:
M + B -2 = F + D + M
M - 2 = F + D + (M - B)
M - 2 = F + 2D
M = F + 2D + 2

Now the tricky one:
If I am four times as old as my brother was when I was two-thirds as old as he was when the average of our ages was twice the age that our father was when he met our mother, we must represent my brother's age in terms of mine, since none of the times his age are mentioned refer to his age today.
I am four times as old as my brother (M/4 is my brother's age at this point) was when I was (to convert his age to my age in the same year, we must add D, the difference between our ages, or M/4 + D) 2/3 as old as (if I am 2/3 as old as something else, that thing is 3/2 as old as me at that same point, so 3/2*(M/4 + D) would represent this) he was when the average of our ages (if he was 3/2*(M/4 + D), then I would be (3/2*(M/4 + D)) + D years old at the same time) was twice the age that our father was when he met our mother (set the average of our ages equal to 2F).

So, the second equation would look like this:
((3/2*(M/4 + D)) + ((3/2*(M/4 + D)) + D)) / 2 = 2F
(2*(3/2*(M/4 + D)) + D)) / 2 = 2F
(3/2*(M/4 + D)) + (D/2) = 2F
(3M/8 + 3D/2) + (D/2) = 2F
3M/8 + 2D = 2F
3M/16 + D = F

If we insert this second equation into the first, we get:
M = (3M/16 + D) + 2D + 2
13M/16 = 3D + 2
13M/16 - 2 = 3D

Now, from this third equation, we can deduce some things. First (and most important) is that my age today MUST be a multiple of 16. Therefore, I am 16, 32, 48, 64, 72, 80, 96, etc... And, that my age MUST be such that the result of the left side of the third equation produces a multiple of 3. For this to work, my age must be one of every third multiple of sixteen, beginning with 32 (or 32, 80, 128, 176, etc...). Now, since all multiples of 16 are even numbers, and only an even number added to an even number will give an even result, the value of D generated by the third equation MUST be an even number, or else the average of mine and my brother's ages (in any given year) will provide a fractional result. Therefore, only EVERY OTHER third multiple of 16 beginning with 32 may be considered. Only 32 will work, because in each other case using a number larger than 32, the result of calculating the first equation (without simplification) will result in taking an average in which my age today is less than some age in my past, which is, of course, impossible. Therefore, my age today is 32. If I am 32, and my parents were wed the year before I was born, then this year would mark their 33rd wedding anniversary.

Answer: 11:12

If the clock loses one second for every minute that goes by, then the clock will show 59 seconds have gone by at the end of every minute. This means that after 60 minutes of real time, the clock will show that 3,540 seconds have gone by, or 59 minutes. Except that in addition to losing one second every minute, it loses an additional minute every hour, meaning that each hour of real time shows as 58 minutes on the minute hand of the clock. This means that after 24 hours, the clock will have lost 48 minutes.

Now, since the clock loses one hour every second, the hour hand must be spinning backward very quickly! But, after 24 hours, or 1,440 minutes, or 86,400 seconds, the clock will have lost 86,400 hours. Since it will be in the same place it began if this amount is a multiple of 12 (it is, since 7,200 * 12 = 86,400), the hour hand will be up at the 12, except that since THE CLOCK loses the aforementioned amounts of time (not just each individual hand), the hour hand will actually have lagged back four minute marks to be between the 11 and the 12. The minute hand will be at two minute marks past the 2, and the second hand will be straight up at the 12, having lost an even 60 seconds every 60 minutes, and having passed through 59 cycles of 60 every 60 minutes. Therefore, the clock will read 11:12 (and if it read AM or PM, it actually would read AM, since the parity changes 3,600 times - an even number - leaving the AM or PM value unchanged).