The worlds first riddle!

Two trains, A and B, leave Pickleminster for Quickville at the same time as two trains, C and D, leave Quickville for Pickleminster. A passes C 288 miles from Pickleminster and D 315 miles from Pickleminster. B passes C 280 miles from Quickville and D half way between Pickleminster and Quickville. Now, what is the distance from Pickleminster to Quickville? Every train runs uniformly at an ordinary rate.


"Three men," said Crackham, "Atkins, Brown, and Cranby, had to go a journey of twenty miles. Atkins could walk two miles an hour, Brown could walk three miles an hour, and Cranby could go in his donkey cart at six miles an hour. Cranby drove Atkins a certain distance, and dropping him to walk the remainder, drove back to meet Brown on the way and carried him to their destination, where they all arrived at the same time.
"How long did the journey take? Of course each went at a uniform rate throughout."


The Crackhams made their first stop at Bugleminster, where they were to spend the night at a friend's house. This friend was to leave home at the same time an ride to London to put up at the Crackham's house. They took the same route, and each car went at its own uniform speed. They kept a look-out for one another, and met thirty miles from Bugleminster. George that evening worked out the following puzzle:
"I find that if, on our respective arrivals, we had each at once proceeded on the return journey at the same speeds we should meet at thirty-six miles from London."
If this were so, what is the distance from London to Bugleminster

Mark, I read the book. Is this riddle from the same man?


Two ferry boats start at the same instant on opposite sides of the river. One is faster than the other. They cross at a point 720 yards from the left shore on their way to their respective destinations, where each one spends 10 minutes to change passengers before the return trip. They meet again at a point 400 yards from the right shore.

How wide is the river


Made I laugh. Keep it clean folks!

What can you hold in your left hand but not in your right hand

well, since I'm on the role with such tasks...
They will get to their destination in exactly five hours.
Cranby takes Atkins exactly to 15th mile (two and half hours), comes back 5 miles to meet Brown on 10th mile (50 minutes, or 3 hours and 20 minutes of walking for Brown and for total time), and then drives him to destination in another 1 hour and 40 minutes - from the time he left Atkins, he drove exactly 15 miles or 2 and half hours, which is exactly time needed that Atkins come to destionation from 15th mile mark.
Huh...

@Try - right hand?

MyO, At last we agree. Five hours it is or at least better be!

"@Try - right hand?"


London to Bugleminster; 66miles

Try: I'm taking Dudeney problems and changing the numbers. My copy gives answers, but not methods. If that's the case for all copies, then folks can't just look up the methods and plug the new numbers into them.

FERRY BOATS
1 mile

MOU:
"They will get to their destination in exactly five hours."

Try:
"London to Bugleminster; 66miles"
I think you added when you should have subtracted.

Mark

FERRY BOATS
1 mile


Let u and v be the speeds of the boats and w the width of the river. Let a = 720 yd and b = 400 yd be the distances from the left and right shores given in the question. What we are told about the crossing points means that:

a / u = (w-a) / v
(w+b) / u = (2w-b) / v

Divide each side of the first equality by the corresponding side of the second equality and you obtain an equality where the speeds of the boats no longer appear, namely a/(w+b) = (w-a)/(2w-b), or rather a(2w-b) = (w-a)(w+b), which boils down to w = 3a-b (after ruling out w = 0 and dividing by w). The width of the river is therefore 3a-b = 1760 yd, or exactly 1 mile.

Or, the simple way:

The first time the boats meet, they have travelled a combined length equal to the width of the river. The second time, they have travelled a combined length equal to three times the river's width. The boat which had travelled 720 yd at the first meeting has therefore travelled three times that (2160 yd) at the second meeting. As this boat has then travelled the width of the river plus 400 yd, the river is thus shown to be 1760 yd wide.


"I'm taking Dudeney problems and changing the numbers. My copy gives answers, but not methods. If that's the case for all copies, then folks can't just look up the methods and plug the new numbers into them."

And you called me sneaky. :wink:

Henry Ernest Dudeny, 536 PUZZLES & Curious Problems.

1. CONCERNING A CHECK
A man went into a bank to cash a check. In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He pocketed the money without examining it, and spent a nickel on his way home. He then found that he possessed exactly twice the amount of the check. He had no money in his pocket before going to the bank. What was the exact amount of that check


2. DOLLARS AND CENTS
A man entered a store and spent one-half of the money that was in his pocket. When he came out, he found that he had just as many cents as he had dollars when he went in and half as many dollars as he had cents when he went in. How much money did he have on him when he entered


3. LOOSE CASH
What is the largest sum of money - all in current coins and no silver dollars - that I could have in my pocket without being able to give change for a dollar, half dollar, quarter, dime, or nickel


4. GENEROUS GIFTS
A generous man set aside a certain sum of money for equal distribution weekly to the needy of his acquaintance. One day he remarked, "If there are five fewer applicants next week, you will each receive two dollars more." Unfortunately, instead of there being fewer there were actually four more persons applying for the gift.
"This means," he pointed out, "that you will each receive one dollar less."
How much did each person receive at that last distribution

Mark cannot play, unless he keeps the book closed.

Boy, those 100 year old puzzles sure are hard. I return to riddles, the answers to which can be found in crackers.


One side of a very accurate ruler is to be marked in units of 1/24 inch and in units of 1/9 inch. Including the two end points. How many different marks must be made from the 1-inch mark to the 2-inch mark


The Ohio license plates have two letters, followed by two numbers, followed bytwo letters. How many different license plates are possible


When George turned 14 years old, his father was 42 years old, which was three times George's age at the time. Today is George's birthday and George's father is now twice as old as George is.

How old is George



Whim wants to draw an equilateral triangle (all sides the same length), and wants to do it as neatly as possible. To that end, he is using a large piece of graph paper made up of squares, and will have each edge go between two points on the graph paper.

Can such a neat triangle be drawn

RULER
31 counting the 1" and 2" marks.

LICENSE PLATES
45,697,600

GEORGE
28

TRIANGLE
No. This is equivalent to asking if an equilateral triangle can be constructed with integer coordinates.

Let (0,0) and (X,Y) be two of the vertices. The third vertex can be found by rotating the point (X,Y) 60 degrees about (0,0).
The third vertex is (Xcos(60)-Ysin(60),Xsin(60)+Ycos(60))
or (X/2-Ysqrt(3)/2,Xsqrt(3)/2+Y/2)
Since X and Y are integers, neither of the coordinates of the third vertex are integers.

Mark:

RULER
31 counting the 1" and 2" marks.

LICENSE PLATES
45,697,600

GEORGE
28

Mark could have just said, ?'Yes' or ?'No' but chose instead to enlighten.

TRIANGLE
No. This is equivalent to asking if an equilateral triangle can be constructed with integer coordinates.


No. One way of seeing this is to note that by either Pick's Theorem or the determinant formula for the area of a triangle the area of an equilateral triangle with every vertex a lattice point must have the form n/2, for some integer n. Since the area of an equilateral triangle is also equal to sqrt(3)/4 * side^2, where side^2 is an integer if the vertices are lattice points, it would follow that if such a triangle existed then sqrt(3) would be rational, which is nonsense.

Alternately, we can assume that one of the vertices is at (0,0). Let the other two vertices be (a,b) and (c,d), and it then follows that a^2 + b^2 = c^2 + d^2 = 2(a*c + b*d). We may assume that at least one of a,b,c,d is odd (otherwise, keep dividing by 2 until this is the case). Assume that a is odd, then b must also be odd since a^2 + b^2 is even, and so a^2 + b^2 must be of the form 8k+2 for some integer k. Finally, it then follows from this that c and d must also be odd, and so 2(a*c + b*d) must be divisible by 4, contradicting the fact that it's of the form 8k+2.


Bill learned to count in a base different from base 10, so that, for instance, instead of writing 136, he writes 253.

What base does Bill use



Al, Bob, Chris, and Dan play with a special deck of 32 cards. Dan deals them out unequally, then says, "If we want to all have the same number of cards, Al should divide half his cards between Bob and Chris, then Bob should do the same with Chris and Al, then Chris should divide half his cards between Al and Bob".

How did Dan initially distribute the cards


Phyllis had made six of seventeen free-throw attempts. How many consecutive free throws must she make to raise her percentage of free-throws made to exactly 50 percent


Donald Knuth, one of the most famous computer scientists in the world (and who was first published as a kid in Mad Magazine) believes that it's possible to make any positive integer by starting with a single 3 and then using some combination of the operations of factorial !, square-root sqrt(), and greatest integer []. Note that n! = 1*2*...*n (e.g. 6!=720), and that [x] is the greatest integer less than or equal to x (e.g. [3.14]=3).

As an example, we can make 26 by [sqrt((3!)!)], since 3!=6, 6!=720, sqrt(720)=26.8, and [26.8]=26.

Can you show how that it's possible to make 10

BILL
7

PHYLLIS
5

KNUTH
[sqrt([sqrt(sqrt((3!)!))]!)]

CARDS
A: 4
B: 7
C: 13
D: 8

Mark yet again amazes. Have a Christmas on me.

BILL
7

PHYLLIS
5

CARDS
A: 4
B: 7
C: 13
D: 8

KNUTH
[sqrt([sqrt(sqrt((3!)!))]!)]

I half gave the answer. Since [sqrt(26)] = 5 and [sqrt(5)!] = [sqrt(120)] = 10, we get one solution just by stringing everything together just so:
10 = [sqrt([sqrt([sqrt((3!)!)])]!)]
I note that there is an extra [] operation in there we don't really need.


Nine different two-digit numbers can be formed with the digits 1, 3, and 7.
(A digit may be used twice in a number). How many of these numbers are prime


A jar has eight red, nine blue, and ten green marbles. What is the least number of marbles that you can remove from the jar and be certain that you have three of the same color


Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, obtaining an answer of 43.

What would her answer have been if she had worked the problem correctly


Here's one from the USA Mathematical Talent Search; it's not easy, but it's certainly doable. (So they tell me!)

For a positive integer n, let P(n) be the product of the nonzero base 10 digits of n. Call n "prodigitious" if P(n) divides n. What is the maximum number of consecutive prodigititious positive integer's n

Hint: the answer is not 12; meditate on numbers ending in a 3, 6, or 9

PRIMES
7 (all but 33 and 77)

MARBLES
7

CINDY
15

Mark at 8.54 am. Yes! 8.54am in the morning. It ain't not decent, at that time I don't even got the horse outta the barn.

PRIMES
7 (all but 33 and 77)

MARBLES
7

CINDY
15


A rectangular floor is composed of square tiles of the same size, eighty-one along one side, and sixty-three along the other. If a straight line is drawn diagonally across the floor from corner to corner, how many tiles will it cross


A knife weighs as much as two spoons; three spoons weigh as much as a knife and a fork; a plate weighs as much as a knife and a spoon. If a fork weighs four ounces, how much does each of the other utensils weigh


Adrian (no relation) drove to work on Monday at 40 mph and arrived one minute late. He left at the same time on Tuesday, drove at 45 mph, and arrived one minute early.

How far does Adrian drive to work

fork=4
spoon=4
knife=8
plate=12



andrian drives 720m

usamashaker-

Welcome to A2K.

As for how far Adrian drives to work....if you take your answer and multiply it by 26.821833 you will be on the right track. :wink:

135 tiles

hi andrian

i don't know how???? 19311.71976