The worlds first riddle!

DrewDad - (are you sure you don't drive a ?'stick shift?)


Glass balls: Two drops, if you're lucky enough to guess right....


Mark:
GLASS BALLS
I think I've seen this answered here. I know I've seen it at another site. If I remember right, the correct answer is 14.



The answer is 14. ( Or, : Two drops, if you're lucky enough to guess right.... ) The strategy is to drop the first ball from the K-th story; if it breaks, you know that the answer is between 1 and K and you have to use at most K-1 drops to find it out, thus K drops will be used. It the first ball does not break when dropped from the K-th floor, you drop it again from the (K+K-1)-th floor, then, if it breaks, you find the critical floor between K+1 and K+K-1 in K-2 drops, i.e., again, the total number of drops is K.

Continue until you get above the top floor or you drop the first ball K times. Therefore, you have to choose K so that the total number of floors covered in K steps, which is K(k+1)/2, is greater that 100 (the total size of the building). 13*14/2=91 -- too small. 14*15/2=105 -- enough.
Obviously, the only possible strategy is to drop the first ball with some "large" intervals and then drop the last ball with interval 1 inside the "large" interval set by the two last drops of the first ball. I

f you claim that you can finish in 13 drops, you cannot drop the first ball for the first time from a floor above 13, since then you won't be able to detect the critical floor 13. The next cannot be above 25 etc.


DD
Boys/Girls: None. The father lost his balls in a freak riddle accident.


Mark wrote, "That's the best answer I've ever seen in this forum!"

Drew, tell your dad to frame that reply, it is unlikely you will see its like again.

Mark:
BOY/GIRL
I get a number that is indistinguishable from 3.


The correct answer is indeed more than 2 but less than 4.

Mark continues in a most jovial tone, "LOCKERS
I know this has been done here. These lockers aren't square, are they?"



The answer is 31. The locker number i is accessed as many times as it has divisors. E.g., locker number 6 is accessed 4 times - by the first, second, third and sixth kids. Thus the lockers which will be open are those whose number has an odd number of divisors, i.e., full squares

(since we have an involution on the set of divisors of i, namely, j --> i/j whose fixed point, if any, is the square root of i). The number of full squares less than 1000 is (isqrt 1000) which is 31.



Paula, correctly points out domestic animals did not invent scissors.


Domesticated animals. False Probably.

Waiting at red lights. True Who has ever counted?

Eyes open. True..(it's true for me)


We then seem to have a conversation from some X rated mag, about animal fetishes.

"I think you're just twitterpated over that belly-baring paulaj.


Twitterpated, what is that?.....
Tryagains eyes rove paulajs avatar?
Thats exciting dear.


< with a smug and cavalier attitude --> <-- paulaj leans back in puter chair, puts hands behind head,...and then wonders why she is derailing this thread again

Haven't you ever seen Bambi?
In the Spring, the animals start "falling in love."
The owl says they're twitterpated.

"I promised myself I wasn't going to do this, again!"

Apparently that was a piecrust promise.
(surely you've seen Mary Poppins).


Come on you two, get a room. This is a family forum.
I should explain at this point I am viewing this on a small laptop and therefore not privy to such images. <makes note to get 26inch screen>



Last, but not least a warm welcome to our newest member Turtlette.

May I apologise for my earlier correspondence. I was getting excited, or was it confused between, Ninja Turtles and Majorettes. No, it was defiantly, cheerleaders. :wink:

In the last 4,000 years. False (see above)

Red lights. True (see above)

Sneeze. True
It's impossible to sneeze with your eyes open.

FALSE- which means, it must be true.
Leonardo Da Vinci invented the scissors.

Leonardo. False

TRUE- also the parachute, machine gun and what is believed to be a primitive precursor to our modern-day helicopter.


I hope you enjoy the forum, and the whole site. Something for everyone.


Hold the front page - late breaking news…

Horrendous???

Mark, you keep me on my toe's, or is that knees? :wink:

"I have only just returned home, where did that night go?"
You weren't north of Boston, were you?


Not with your auto parked outside, ?'cuddly bumkins' well, that is what it sounded like.

Or, as Turtlette said, "Sounds like an illusion."





Place six three digit numbers of 100 plus at the end of 685 so that six numbers of six digits are produced [like 685123]. When each number is divided by 111 six whole numbers can be found



How many three digit numbers are divisible by 17




We have a number (integer) with 6 as the last (right-most) digit (936 for example). If we erase the 6 and put it on the left end of the number (693 in our example), then we have a number four times our original number. We see that 936 doesn't work.

What is the smallest number that does fit the above conditions

Thats it for now, I have to go out :wink:

"Not with your auto parked outside, ?'cuddly bumkins' well, that is what it sounded like. "

You sure twisted that around, you sly devil.

6-DIGIT NUMBERS
685203 + N*111 (N=0 to 7)
That's eight numbers by the way.

DIVISIBLE BY 17
53

6 ON THE RIGHT
153846

53

What Adrian said :wink:

Mark:
6-DIGIT NUMBERS
685203 + N*111 (N=0 to 7)
That's eight numbers by the way.


Of course there are eight such numbers. The first is greater than or equal to 685100. Dividing that by 111 gives us 6172.072. . . So the first number of our answer should be 111x6173 or 685203. The next answer is 111 more than that (111x6174=685314). Then 111 more than that, etc.

The answers:
685203, 685314, 685425, 685536, 685647, 685758, 685869, 685980.


Mark:

6 ON THE RIGHT
153846


6 on the right, or left?
Mark responds, "You sure twisted that around, you sly devil."

You humans really crack me up.


We can solve this by trial and error. But first we can deduce that the left-most digit must be a 1 (6.../4 = 1...). Then a little trial and error shows that the answer must have at least four digits. There are a lot of numbers to try, with four or more digits. So trial and error is beginning to seem a little too time-consuming.

First solution: Maybe we can solve it algebraically. If our number is x, then 4x=6m+(x-6)/10, where m is some power of 10. 40x=6n+x-6, where n is the next power of 10. 39x=6n-6 or 13x=2(n-1) or x=2(n-1)/13. So n-1 (one less than a power of 10, in other words, a string of nines) would seem to be a multiple of 13.

We can now test the first few strings or nines, to see if any of them is a multiple of 13. 999999 turns out to be a multiple of 13 (13 x 76923). So x=2 times 76923 = 153846. And multiplying this by 4, to see if it solves our problem, we get 615384. So it is the solution.
We can do this first solution by writing out x as a sum of powers of ten (10^(s-1)a+10^(s-2)b+...+10p+6). The algebra is similar to the above.

Second solution: Here is a clever way to solve it. 1...6x4=61... So the last digit of the product must be 4: 1...46x4=61...4. So the next to last digit of the solution must be 8 (do you see why?): 1...846x4=61...84. So the third to last digit must be 3: 1...3846x4=61...384. So the next digit must be a 5: 1...53846x4=61...5384. The next digit must be a 1, which is what we were looking for. The smallest such number is 153846, which should work. But we must multiply it out to make sure we haven't made a mistake: 153846x4=615384. It works.

Third solution: We can do something similar to the second solution, but work from the left end. 1...6x4=61... 61.../4 is between 1525... and 154999... So our second digit is 5: 15...6x4=615... 615.../4 is between 15375... and 153999... So our third digit is 3: 153...6x4=6153... 6153.../4 is between 153825... and 15384999... So our fourth digit is 8: 1538...6x4=61538... 61538.../4 is between 153845... and 1538474999... So the fifth digit is 4. We've been looking for a 4, as the right-most digit of our product must be a 4. So 153846x4 should be 615384. Multiplying it out shows that it is our answer.


Mark:
DIVISIBLE BY 17
53


Well, the first one is greater than 100. 100/17=5.88235. . . So the first is 6x17=102. We don't need to list these three digit numbers. The last one is less than 1000. 1000/17=58.8235. . . So our three digit numbers are 6x17, 7x17, 8x17, . . ., 58x17. There are 53 such numbers.


Friends, Romans, countrymen, please welcome Adrian upon his return to the forum. For those who have not had the pleasure:

I first met Adrian on Sun Dec 28, 2003 8:42 pm. When he said, "Tryagain are you taking the piss or what?"

His astute observation and keen wit has been appreciated ever since.
As proof he, with the minimum use of bandwidth writes, "53" and agrees with Mark. - safe bet.


I knew it! Turtlette has been sent by the Pinkertons.

Li'l Turtle wrote, "She's a pill!"
What! Do you mean; she is hard to swallow?

"Btw, where is she? Did you scare her away?" I have a lot of questions for a newbie."
Yesterday, a hatchling, today a Newbie, tomorrow???

"I hope I'm not getting on your nerves."
Hope no more, swing your leg over.


Tryagain wrote"

May I apologise for my earlier correspondence. (YES! )I was getting excited, (What a surprise) or was it confused (No, you were excited, you got it right) between, Ninja Turtles and Majorettes. No, it was defiantly, cheerleaders. (cheerleaders?...are you talkin' bout...paulaj' ...pom-poms?)

Li'l Turtle. I will get my people to talk to your people. I can't be dealing with all this talk of pom-poms.

BTW It's not the size that matters, so they say. Who the hell are they anyway? Early to bed, early to rise, that's my motto.



Here are all of the divisors of 1008: 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 36, 42, 48, 56, 72, 84, 112, 126, 144, 168, 252, 336, 504, 1008.

Actually, I think that one divisor is missing.


Is there a fast way to figure out what it is
Ok! What the hell is it



Every letter stands for a different digit:
EVE/DID=.TALKTALKTALKTALKTALKTALK



Here is a classic problem. Mark and Paula were 10 km apart. They rode their bicycles at a constant 10 km/hr toward each other. As they started off, a fly took off from one bicycle and flew to the other bicycle at 20 km/hr, then it flew back to the other bicycle, and then back and forth until it was crushed between the two bicycles when the two bicycles collided.

No person was hurt in the making of this riddle. The same can not be said of the fly.

How far did the fly, fly

1008
By pairing the divisors from the ends toward the middle and checking the last digit of their products, you find that 16 doesn't have a mate. Since 8 was paired with 126, 16 must be paired with 126/2=63.

FLY
Left out of the problem is the fact that Try was 4.9 km behind Paula pedalling furiously at 20 km/hr (he's quick when he's jealous) to intercept her. Upon reaching her, he grabbed her and reversed direction at the same speed with her mounted on the handlebars.

Therefore, the answer is that the fly's trip was infinitely long as the bicycles never met. Or, the fly could have pooped out after 10 km.

EVE DID TALK
242/303 = .7986...

Woops, overslept.


Mark:
1008
By pairing the divisors from the ends toward the middle and checking the last digit of their products, you find that 16 doesn't have a mate. Since 8 was paired with 126, 16 must be paired with 126/2=63.


Divisors occur in pairs. Look at the list. 1x1008=1008, 2x504=1008, 3x336=1008, 4x252=1008, etc. If the original number (1008) is not a square, then there should be an even number of pairs. If it is a square, then its square root (a factor) is paired with itself, and there are an odd number of pairs. 1008 is not a square, and I have listed an odd number (29) of divisors. So I was right; there is at least one missing divisor (assuming that the numbers that I listed are actually divisors). And we can start from both ends and pair up divisors and see which one does not have a mate. If we do that (I started to do that above) we find that 16 is not paired with a larger number. 1008/16=63. So 63 is the missing divisor.




The excellent book, Self-Working Number Magic (subtitled "101 Foolproof Tricks") by Karl Fulves contains the following cute puzzle (much like Word Arithmetic), with an incomplete solution.
Every letter stands for a different digit:
EVE/DID=.TALKTALKTALKTALKTALKTALK...

The book's solution:
There are a number of ways (including Geometric Series) to show that 0.TALKTALKTALKTALKTALKTALK...=TALK/9999. So EVE/DID=TALK/9999. The author then leaps to the conclusion that DID must be a factor of 9999. 9999=3x3x11x101. The factors of 9999 of the form DID are 101, 303, and 909. Also we know that E<D. A great deal of experimenting with the other numbers shows that there is only one solution of this type:
242/303=.7986798679867986...

Marks solution:
EVE DID TALK
242/303 = .7986...

No, I doubt he ever read that book.


My solution:
Of course DID need not be a factor of 9999. It must be a factor of 9999xEVE. Much more experimenting with all of the numbers shows that there is a second solution:
212/606=.3498349834983498...



Mark:
FLY
"Left out of the problem is the fact that Try was 4.9 km behind Paula pedalling furiously at 20 km/hr (he's quick when he's jealous) to intercept her. Upon reaching her, he grabbed her and reversed direction at the same speed with her mounted on the handlebars."


I would like to place on record that, I have never, ever, mounted Paula, on the handlebars.
<Flicks thru ?'Mountain Bike' mag>

"Therefore, the answer is that the fly's trip was infinitely long as the bicycles never met. Or, the fly could have pooped out after 10 km."


There is more than one way to solve this. First, let's look at the easy way. The two bicycles were 10 km apart. They crash halfway between them in half an hour. The fly travels 10 km in that half an hour. That's the answer.

How about the hard way? The fly goes twice as fast as the either bicycle. So the fly initially covers 20/3 km while the bicycle A covers 10/3 km. The fly turns around. Bicycle B is now 10/3 km away. The fly returns to bicycle B, going twice as fast as bicycle B. So the fly goes 20/9 km while bicycle B goes 10/9 km. Then the fly turns around and goes 20/27 km, while bicycle A goes 10/27 km. After an infinite number of flights, the fly has gone 20/3+20/9+20/27+20/81+... That appears to be a Geometric Series. The sum is S=a/(1-r), where a is the first term 20/3, and r is the ratio of successive terms 1/3. So the sum is 10 km.





One famous puzzle is to make a certain positive integer out of four fours (in base 10, with standard mathematical operations, but with no other numbers besides the four fours). For example, you can make 1 with 44/44. The rules seem to vary. Some say that you can also use three or fewer fours; others say that you can use ".4~" for a repeated "0.44444..."

Some say you can use sines, logs, gamma function, etc. Let's say that you must use four fours, not fewer. It turns out that you can make any number from 1 to 100 (and others) out of four fours and these operations

• addition
• subtraction
• multiplication (*)
• division
• square root (sqrt())
• fourth powers (^4)
• factorial (!)
• .4~ for a repeated 0.44444...

We won't allow logarithms (or percent or int()).

Can you make 1-10



Can you divide by 0
Why, or why not


What is 13 raised to the 0 power

What is 0 raised to the 0 power



If you add 1/2 + 1/4 + 1/8 + 1/16 + ... will you ever get to 1

Try: "Of course DID need not be a factor of 9999. It must be a factor of 9999xEVE. Much more experimenting with all of the numbers shows that there is a second solution:
212/606=.3498349834983498... "
Nice job!

DIVIDE BY 0
No. It is undefined. Assume A/0=B and A!=0. Then B*0=A, but B*0=0.

POWERS
13^0=1
0^0 is undefined

INFINITE SUM
No, but you can get as close as you wish.

FOUR FOURS
1 = 4/4 + 4 - 4
2 = 4/4 + 4/4
3 = (4*4 - 4)/4
4 = 4 + (4 - 4)*4
5 = (4*4 + 4)/4
6 = 4 + (4 + 4)/4
7 = 4 + 4 - 4/4
8 = 4 + 4 + 4 - 4
9 = 4 + 4 + 4/4
10 = (44 - 4)/4

I leapt to the same conclusion as the author. However, it did not require a great deal of experimenting to find the solution. As you pointed out, E<D; so 101 is ruled out. 909 is also ruled out because that leads to EVE*11=TALK, but EVE*11 must end with E, not K. Therefore, you're left with DID=303, and EVE*33=TALK. E can't be 1 or EVE*33 would end in 3, not K. Therefore, E=2.

We now have 2V2*33=TALK; so 2V2*33=TAL6. We only need to ensure that our choice of V yields unique values for T, A, and L. Since 2, 3, and 6 are already used, we only need to check 0, 1, 4, 5, 7, 8, and 9. So, after only seven multiplications, we have our solution.

Mark:
FOUR FOURS

1 = 44/44
2 = 4/4 + 4/4
3 = 4/sqrt(4) + 4/4
4 = 4 + 4 - sqrt(4) - sqrt(4)
5 = 4 + 4/(sqrt(4) + sqrt(4))
6 = (4 + 4 + 4)/sqrt(4)
7 = 4 + 4 - 4/4
8 = 4 + 4 + 4 - 4
9 = 4 + 4 + 4/4
10 = 4 + sqrt(4) + sqrt(4) + sqrt(4)


Some of these are not optimal (fewest operators):
44/44
4/4+4/4
(4+4+4)/4
4*(4-4)+4
4!/4-4/4 (not optimal)
(4+4)/4+4
44/4-4
4+4+4-4
4+4+4/4
4+4+4-sqrt(4) (not optimal)
4!/sqrt(4)-4/4 (not optimal)
4!-4-4-4 (not optimal)
4!/sqrt(4)+4/4 (not optimal)
4+4+4+sqrt(4)
4*4-4/4 (not optimal)
4+4+4+4
4*4+4/4
4*4+4/sqrt(4)
4!-4-4/4
4!+4-4-4

Mark:
DIVIDE BY 0
No. It is undefined. Assume A/0=B and A!=0. Then B*0=A, but B*0=0.

POWERS
13^0=1
0^0 is undefined

?'The question about 0 raised to the 0 power leads us into hideous tangles along one route, but along another can we agree to assign this symbol the value 1.?


INFINITE SUM
No, but you can get as close as you wish.



Mark writes, "Try: "Of course DID need not be a factor of 9999. It must be a factor of 9999xEVE. Much more experimenting with all of the numbers shows that there is a second solution:
212/606=.3498349834983498... "
Nice job!"


Thank you, but that is nothing compared to:

"I leapt to the same conclusion as the author. However, it did not require a great deal of experimenting to find the solution. As you pointed out, E<D; so 101 is ruled out. 909 is also ruled out because that leads to EVE*11=TALK, but EVE*11 must end with E, not K. Therefore, you're left with DID=303, and EVE*33=TALK. E can't be 1 or EVE*33 would end in 3, not K. Therefore, E=2.

We now have 2V2*33=TALK; so 2V2*33=TAL6. We only need to ensure that our choice of V yields unique values for T, A, and L. Since 2, 3, and 6 are already used, we only need to check 0, 1, 4, 5, 7, 8, and 9. So, after only seven multiplications, we have our solution. "

Now, that is super
Have you thought of writing your own book?


Li'l Turtle writes, "What Adrian said" (which is what Mark said, 53)
Last edited by turtlette on Thu Apr 07, 2005 1:30 pm; edited 1 time in total

You catch on fast. Just copy what they write and you cant go wrong.



The good, bad and downright ugly!
I don't know where they came from, I have never seen them before.



1. 9.2 + 2.9 equals

(A) 11.11 (B) 12.1 (C) 18.4 (D) 9.49 (E) 11.1


2. Given that 3.84 ×2.75 = 10.56, the value of 1.056 /0.00275 is

(A) 0.384 (B) 3.84 (C) 38.4 (D) 384 (E) 3840


3. If your heart pumps about 80 millilitres of blood each second, then the volume of blood, in litres, that it pumps in one day is about

(A) 7 (B) 70 (C) 500 (D) 5000 (E) 7000


4. In the diagram below, the sum of the numbers represented by letters in each row and in each column, is given by the adjacent number, e.g. the sum of the letters in the top row is w + w + w + z + y = 416. What number should replace the **



w............. W.............w..............z..............y..............416
w..............x..............x..............y..............x...............**
x...............y..............y..............w.............x..............498
y...............z..............y..............z..............z..............118
456..........306...........271..........193..........308

Tryagain wrote:

"You catch on fast." (thanks, it's a gift! I did it to unclog your thread.)

GOOD, BAD, AND UGLY
1) B
2) D
3) E
4) 502

Li'l Turtle writes, "I did it to unclog your thread."

Careful! That there thread is the only thang that is keeping my pants up!


Mark:
GOOD, BAD, AND UGLY (But not necessarily in that order.)

1) B
2) D
3) E
4) 502



GBU cont/-

5. A mathematics test consists of 10 questions. Ten points are given for each correct answer and three points are deducted for each incorrect answer. If a student answered all questions and scored 61, how many correct answers did he have

(A) 7 (B) 5 (C) 9 (D) 8 (E) 6



6. The houses along the street in which I live are numbered in order by consecutive odd numbers, starting with 1, on one side of the street, and by even numbers on the other side.

My house is numbered 137. If the numbering had commenced at the other end of the street, my house would have been numbered 85. The number of houses on my side of the street is

(A) 112 (B) 222 (C) 111 (D) 220 (E) 110



7. In the figure angles XYZ and WZY are 900. XZ and YW intersect at point T and TS is perpendicular to YZ. If WZ = w units, XY = x units, and YZ = y units, what is the length of TS

(A) [(xw + xy + yw)/x]
(B) [ yw/(y+w)]
(C) [ xw/(x+w)]
(D) [(xy + xw + yw)/y]
(E) [(x2 + y2 + w2)/(x+y+w)]



8. Letters from branches of a bank reach the head office in Harare and get placed in particular trays for the attention of the manager.

Those from the Masvingo branch are placed in tray M, those from the Gweru branch in tray G, those from the Kwe Kwe branch in tray K and those from the Bulawayo branch in tray B.

One day the secretary in Harare receives one letter from each of the Masvingo, Gweru, Kwe Kwe and Bulawayo branches.

In how many ways can she put these letters in the trays M, G, K and B, one in each tray, in such a way that no letter is in the correct tray

GBU cont/-

(A) 7


The number of houses on my side of the street is

(C) 111


Letters:

9 ways

Tryagain wrote: -->

Careful! That there thread is the only thang that is keeping my pants up!

Sorry, I didn't mean to cause stress...I apologize :-) No hard feeling's?

7) C (where was the figure?)

It's interesting that the height of the intersection is independent of how separated the parallel segments are.

I have taken Tryagain's riddle and wondered what would happen if you have four balls.


You have FOUR identical glass balls and a 100-storey building. If you drop such a ball from the a certain floor, it might break (obviously, if it breaks when dropped from the ith floor, it will also break when dropped from all floors above it.

You have to find such a floor that the balls do not break when dropped from the floors below it, but do break when dropped from it and floors above it. You may expend all four balls.

What is the minimum number of drops you will have to make?

Li'l Turtle write, "Sorry, I didn't mean to cause stress...I apologize. No hard feeling's?"

No! It's ok, they have subsided.



GBU cont/-
Whim (Good to see you back)

(A) 7
5. (A) Let X denote the number of correct answers recorded by Blessing. Then
10X3(10X) = 61, and solving gives X = 7.
Whim

The number of houses on my side of the street is

(C) 111

(C) On one side of my house there are [(1371)/2] = 68 houses whereas on the other side there are [(851)/2] = 42 houses. Thus altogether there must be 111 houses on my side of the street.

Whim
Letters:

9 ways

(B) One can write out all possible permutations of the letters, and eliminate those with one or more letters in the correct tray(s), but if there were more than 4 trays this would be very tedious and open to error. A more sophisticated approach, which can be used for any number of trays, is using n!(1-[1/1!]+[1/2!]-[1/3!]+-+[((-1)n)/n!]) as the total number of arrangements in which every letter is incorrectly placed. With n = 4 for this example we then have 24(1-1+[1/2]-[1/6]+[1/24]) = 24([(12-4+1)/24]) = 9.

Note that this formula can easily be proved by induction.

Mark:
7) C (where was the figure?)
The (my) question is rubbish, and so is the (my) answer.

(B) In order of magnitude we have 28, 27, 27-2, 2+26, 26, and thus the middle number is 27-2.


"It's interesting that the height of the intersection is independent of how separated the parallel segments are."

Mark, only you, well, possibly Whim, could notice that fact. Good one.


Hello, Whim is up to his old tricks again I see. This angle will take some thinking about.



11. An arithmetic progression is made up of natural numbers with common difference 1. Let the progression be a, b, c, d, e. We know that a + b+ c+ d+ e = f3 while b + c + d = g2, where f and g are also natural numbers. What is the least possible value of c

(A) 1125 (B) 675 (C) 25 (D) 1024 (E) 175



12. Given that 6x+y = 36 and 6x+5y = 216, then x is equal to

(A) [1/4] (B) [3/4] (C) [5/4] (D) [3/2] (E) [7/4]



15. Three men are gambling in a bar. They start with money in the ratio 7 : 6 : 5 and finish with sums of money in the ratio 6 : 5 : 4 (in the same order as before). One of them won $12. How many dollars did he start with

(A) 420 (B) 1080 (C) 432 (D) 120 (E) 90



16. A set of four positive numbers has a sum of 12. What is the maximum value of their product


(A) 96 (B) 81 (C) 256 (D) 144 (E) 60