The worlds first riddle!

[size=8]PASSWORD
93102[/size]

Mark:

PASSWORD
93102


I am sure Stormy will show her appreciation in a tangible way!



The sum of the first two digits is exactly twice the sum of the last four. The smallest number that is the total of four different digits is 6
(0 + 1 + 2 + 3).

The sum of the first two digits, therefore, must be at least 12. The largest possible total for two different digits is 17 (8 + 9). Since the total of the first two digits is twice the sum of the last four, this total must be an even number.

So the possible totals for the first two digits are 12, 14, and 16.
But 16 (7 + 9) is impossible, because the total of the last four digits would have to be 8, and four different digits can't total 8 if one of them is 7 or 9.

Likewise, the total of the first two digits can't be 14 (5 + 9 or 6 + 8) because four different digits can't total 7 if one of them is 5, 9, 6, or 8.

The sum of the first two digits must therefore be 12. Since the sum of the last four digits is 6, none of the digits can be greater than 3. So the first two digits must be 9 and 3, in that order. The remaining digits must be 0, 1, and 2, in some order. The only order in which the sum of the middle three digits is twice the sum of the last two digits is 93102.





At the A2K Sports Store, Ping Pong Balls come packaged in boxes of 6, 15, and 20.

What is the largest number of ping pong balls that you can not purchase

By getting two boxes of 6, you have 12 ping pong balls. But you can not get 13 ping pong balls since no combination of 6, 15, and 20 adds up to 13.

So, in other words, what is the greatest number of ping pong balls that can NOT be made from 6, 15, and 20?




Due to my flight schedule, I regret I will not be around until Friday at the earliest.

<pauses>


Therefore here is Thursday's challenge:


I give you the answer; yes, give!

The answer to the problem is 45678.


You arrive at this by subtracting one number from another. The two numbers must contain the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. You must use each of these digits once, but only once.

E.g. nnnnn - nnnnn = 45678

What two numbers do you subtract to arrive at this answer

The pause was for the cheering to subside, but it would not print above

[size=8]45678

Take your pick:
57984 - 12306
58047 - 12369
58074 - 12396
59406 - 13728
59460 - 13782
63582 - 17904
65382 - 19704
69783 - 24105
70359 - 24681
70539 - 24861
75138 - 29460
75318 - 29640
75894 - 30216
80295 - 34617
82095 - 36417
86217 - 40539
86271 - 40593
87603 - 41925
87630 - 41952
87693 - 42015
97026 - 51348
97062 - 51384[/size]

[size=8]PING PONG BALLS
49[/size]

Hiya all, yes, I'm backkkkkkkkk!


Mark:

45678

Take your pick:
57984 - 12306
58047 - 12369
58074 - 12396
59406 - 13728
59460 - 13782
63582 - 17904
65382 - 19704
69783 - 24105
70359 - 24681
70539 - 24861
75138 - 29460
75318 - 29640
75894 - 30216
80295 - 34617
82095 - 36417
86217 - 40539
86271 - 40593
87603 - 41925
87630 - 41952
87693 - 42015
97026 - 51348
97062 - 51384


That is truly amazing.

However, I was somewhat surprised to note you missed out:


48,615 - 02,937 = 45,678
48,651 - 02,973 = 45,678



Then, a whole seven minutes passed before you came up with:

PING PONG BALLS
49


Beginning with 50, every number following can be made up of combinations of 6, 15, and 20.

# of ping pong balls # of packs of 6 # of packs of 15 # of packs of 20
50 = ………………….. 0 ……………….. 2 …………………1
51 = …………………..1 ……………….. 3 …………………0
52 = …………………..2 ……………….. 0 …………………2
53 = …………………..3 ……………….. 1 …………………1
54 = …………………..9 ……………….. 0 …………………0
55 = …………………..0 ……………….. 1 …………………2

You can get all numbers greater than 55 by adding 6 or multiples of 6 to the numbers above. So this shows that all numbers above 49 can be made from combinations of 6, 15, and 20. All that we had to show was the first instance where six numbers in a row could be made (since 6 was the smallest box).

The following shows the number of ping pong balls which can be made from packages of 6, 15, and 20:

6, 12, 15, 18, 20, 21, 24, 26, 27, 30, 32, 33, 35, 36, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 50, and all numbers after it. - Strange but true.



If consonants are free, determine the cost of the five vowels A, E, I, O, and U from the following clues:

Audacious costs $260
Equivocation costs $340
Inimitable costs $255
Onomatopoeia costs $435
Unambiguous costs $225

Facetious costs Er bai wu

so heres something i read on yahoo news:

the us census bureau estimates population by assuming 1 person is born every 7 seconds, one person dies every 13 seconds, and one person enters the country as an immigrant every 31 seconds. so the population increases by one person every how many seconds?

277/2821 or 0.0981921

Adrian

Facetious costs: Er bai wu


For the non Chinese speakers; if a person speaks phrases such as "er bai wu" (two hundred fifty)


a = $60
e = $75
i = $40
o = $50
u = $25

To solve it you could set up 5 simultaneous linear equations:
(Brackets are used to indicate each equation)



You could also solve this with Cramer's Rule


As for the question raised by Thoh, I like Adrian's answer.




Given the set of letters below, by starting at the center of the rhombus and moving through the letters, each move from center must be to the right, left, up or down (in other words, you may not move diagonally).

How many ways can you find to spell the word PASCAL

thoh13's population question:

The answer is the inverse of Adrian's answer.

Adrian provide the population increase per second.

It is too.

Pascal's rhombus is a dead give away. Having said that it still took me almost half an hour to realise what the solution would be. Eventually the light went on, a rhombus is four triangles...

I knew this guy years ago that loved the Fiat twin cam 2.0L engine. He had an old 124 with a turbo 2.0L in it, huge intercooler, damn thing ran high 10's for the quarter. Still looked like crap though.

population riddle

x/7 - x/13 + x/31 = 1
x = roughly 10.2
so population increases by 1 person every 10.2 seconds

Adrian:
….He had an old 124….


"…ran high 10's for the quarter…" Man, that is so slow, he would be better off taking the bus.



The answer is 124 ways!

You could use "Pascal's Triangle" to help solve it. I have replaced the letters of PASCAL by numbers representing how many different ways you could reach that particular letter. Then add up all the numbers where the letter L is (all the numbers around the outside of the figure) to get your final answer.









Sequence #1: What is the missing number in this sequence:
43, 41, 37, 31, 29, ___, 19, 17



Sequence #2: What letter comes next in the following sequence:
A, S, D, F, G, H, J, ___

[size=8]SEQUENCE #1
That is a prime example of a monotonically decreasing sequence. Michael Jordan would be proud of you.

SEQUENCE #2
The key to this problem is under your right middle finger.[/size]

Mark:

SEQUENCE #1
That is a prime example of a monotonically decreasing sequence. Michael Jordan would be proud of you.


For those who don't follow basketball, 23 was his shirt number.


The answer is 23 or 25.

To get 23, it is a list of consecutive prime numbers in descending order (23 is missing).

To get 25, begin with 43, then subtract 2 to get the next term, subtract 4 to get the next term, then subtract 6 to get the next term, then repeat subtracting 2, then 4, then 6, then 2, ...



SEQUENCE #2
The key to this problem is under your right middle finger.


To get K, look at the second row of your computer keyboard




At the local swimming pool, a certain pipe can fill it in 2 hours. Another pipe can fill it in 5 hours. A third pipe can empty the pool in 6 hours.

With all three pipes turned on exactly at the same time, and starting with an empty pool, how long will it take to fill the pool

[size=8]SWIMMING POOL
T/2 + T/5 - T/6 = 1
T = 15/8 hours[/size]

Shoot! If you go any faster, I will have to match questions to your answers.


In an effort to slow y'all down, take a board that is marked with squares numbered 1 through 36. You bet by placing chips on these numbers.
Then a player rolls a pair of standard six-sided dice, and the winning number is the product of the values on the dice. For example, if the dice show 3 and 5, the winning number is 15.

Players who bet on the winning number win $10 for every $1 they wager; the others lose.

An enterprising A2K member decides to play the game.
On which number or numbers should he/she bet

And in the long run, should he/she expect to win or lose money, in other words, what is the expected payoff

[size=8]DICE GAME
6 and 12 average one win per nine tries.

Expected payoff is $1/9 per game (about 11 cents).[/size]

Mark:

SWIMMING POOL
T/2 + T/5 - T/6 = 1
T = 15/8 hours



In 1 hour, the first pipe fills 1/2 the pool, the second pipe fills 1/5, and the third pipe empties 1/6. That is, in 1 hour the pool fills:
1/2 + 1/5 - 1/6 = 15/30 + 6/30 - 5/30
= 16/30 = 8/15 of the pool.

For the whole pool to fill, it would take 15/8 or 1 7/8 hours.



DICE GAME
6 and 12 average one win per nine tries.

Expected payoff is $1/9 per game (about 11 cents).



The number 6 and the number 12 can each be produced in four different ways. Since these are the numbers which occur the most, you should bet on one of them. If you place a bet, your chances of winning are 4/36





From enquires I have made, there is a thread on A2k that consists of 58% girls and 42% boys. There are 6 more girls than boys in the discussions.

How many TOTAL participants are there