The worlds first riddle!

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...

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.

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