Each stanza refers to a different meaning of the same word.

Hopefully someone gets this while I'm teaching tonight. Good luck!

YEP!!! That's the answer. Good job, Glycereine.

Yay I have to get back to work though or I'm in trouble :-p.

Wish I'd read this sooner, there were so many replies that I assumed someone had answered it already but I just now finally read through all the answers and realized no one had!

must be even since they add up to 14.

since all five digits = 30, and 1-->9 would equal 45, the four digits not used must be three odd and 1 even or 1 odd and three even.

the three undetermined digits in the pin must be two odds and one even to add up to an even number.(30)

therefore V+III must be 8 and 6, not necessarily in that order.

if i understand the next two clues correctly, the second digit will be the fourth largest number of the five digits.

digit I < (2+ digit II)

digit II cannot be 1 because it is the second-smallest.

digit II cannot be 9 or 7 because there must be 4 digits larger than II in the pin, and I < II.

digit II could be 5 if digit IV were 7, then digit I must be 4.

then digit I must be 4.

so 4,5, 6or 8, 7, 6or 8

4+5+6+7+8=30

but how to determine whether 45876 or 45687? oops, i overlooked the = 10 clue that must be the key if figure II and III equal 10, then figure II cannot be odd. above is wrong! Arrgh! Okay, all the numbers cannot be even because there are only 2468 to use. but, one cannot be odd or the digits could not total 30. At least two digits must be odd. okay, I am going to have to start over, running out of edit time....

Remember that 2 odd numbers added also makes an even number.

Oh, that doesn't mean much. You guys are that young? '92 were just the Rodney King incident, and the streets were rioting all over the country. It was a time of many gangs, drivebys and the like. That's all that means. So, NE is the side of town with more "history."

I'm not that young. I was guessing:

Fireworks

F1

wow. this will be hard considering I wasn't even born in 1992.

!!

Looked it up for you octal numeral system : http://en.wikipedia.org/wiki/Octal

Thank you! Educational and interesting too

Ooh, English is not my native, I initially thought he wanted the number of telephone numbers. octal numeral base is what I wanted to write, but once agian, English is not my native.

It just sort of seemed to me, that both questions were in fact trick questions.

Sorry Randoms I wasn't criticizing your English, I don't know what octal numeral base is either, I was just curious. I'll look it up.

Cheers!

An elderly professor arrives home without his key and realizes he must use the keypad he had installed for this reason.

He is however not only habitually forgetful, but also extremely insecure. As such he has set up that each time he enters his house via the keypad (which he only does when he forgets his key) the combination will be changed before the next use.

The number changes as follows:

The first number is doubled and the 1's place of the double digit becomes the new first number.

The tens place is carried and added to the second number. Again the 1's digit of this number is the new second number.

The tens digit is added to the fourth number and the 1's digit of the this number is the new 4th number.

The third number is increased by 3 (if it goes over 10 the ten's digit is discarded).

Question 1:

A. If the code is 0123, what was it last time the professor forgot his key?

B. How many times has he forgotten his key?

Questions 2:

A. If the code is started with 1111 and it ended in 1 last time but now ends in 2, how many times has the professor forgotten his key.

Sort of? your on the right track(in fact thats part of it) but there's more to it.

This was what I was thinking too, but I can't find anything in the riddle that makes this different than any other like it so I'm stumped

a key?

Sorry but no

2 on the power of a hundred is the number of telephone numbers

2 on the power of a hundred expressed in octal numerical base would be how the last bacteria's number would look like

the chance of of having a six in the number if you are spawned on the 101th day equals the chance of Fred losing his mind and ignoring his own rules on the 101 the day... 0%

EDIT: I said how the number would look like in octal form, since I didn't count 1 as flippable.

Unfortunately the number of phone numbers is pretty easy, but that's not what he's asking . He wants the sum of all the phone numbers, ie 0+1+2+3+4+5+7+8+10+11....

Also forgive my ignorance but what is octal numberical base?

I think you looked at the chance of a 6 the same way I did in my first post, but then I realized I was wrong. Just because it has a 6 or a 9 in it doesn't mean it's a flippable number.

wheel?

Also a good guess!

But no dice, sorry.

I was thinking simply

Engine, or combustion.

Oh that actually fits quite well!!

But alas, not what I'm looking for. When you see the answer you'll know it's the answer (I hope )

I have a hint that won't give it away but I'm saving it for later.

It's a simple math problem. If the 6 and the 9 are thrown away, 2 numbers are thrown away for every ten except for the 60's and the 90's which are all thrown away. That means that by day 100 with 2 numbers assigned a day, there would be 47 numbers thrown away. That means that the factorial of 147 would add to the total of everybody, including the ones that have been thrown away. To factor in the ones that have been thrown away, these must be subtracted from the factorial. The summation of these numbers and the net result is given. The 645 and 945 are the sums of the 60's and 90's respectively.

14 13

147!-E(10x+6)-E(10x+9)-645-945

x=0 x=0

The net sum of all numbers 0-147 that do not include a 6 or a 9 is

5852

1. the only 2 days that have only 2 numbers assigned are day 1 and 2. Remember the polulation is doubling each day not increasing by 2 each day.

2. Remember you cannot use numbers that contain only 1's and 8's (unless completely symetrical) or that contain any 6's or 9's and only 1's or 8's or 0's as other numbers (ie 109, 18, 68, 1968, etc.) unless it cannot be flipped (ie it ends in 0)

Ah, this is quite easy. The top two boxes average the best profit, which are the only two profitable ones. Meaning A1 and B1.

Every box has 10% to contain the riches, meaning on average every box is worth 1.4 dollars. The only boxes worth the risk on prolonged tries are the top ones. Top boxes average a .4 dollar profit each try, so it means, that the top two boxes profit .8 dollars per game on average. That's the best you'll get

Agree with everything but went about it a different way:

open box 1 of each stack each try means 20% chance of winning. In theory 1 in every 5 games should be a winner. 10 dollars spent on 5 games and 14 dollars won means a 4 dollar profit.

4/5 = 80 cents per game profit.

dont u mean 0,0 10,0 and 0,2? or am i missing something

Unless final is correct, the 2nd post is right, which doesn't make a whole lot of sense to me.

yay- 100% first try!

It's actually called the Stroop Effect (Test)- we did it in my intro Psych course.

Here's a couple websites that might be a bit more challenging:

http://faculty.washington.edu/chudler/words.html

http://cognitivefun.net/test/2

http://www.thepersonalbest.com/SelfTests/S...sMainFrame.html

for the first site, click 'run experiment'

for the second, at the end you can continue as 'anonymous' to view your results- they will be at the very top of the list

usually the stroop test measures not how many you get right, but the time difference between two different screens- the one where the color and words match, and the one where they don't. Wikipedia says of its purpose:

"This test is considered to measure concentration effectiveness as well as processing speed;[12] and it is commonly part of executive functions test batteries. It has been used to investigate cognitive aspects of disorders such as brain damage,[12] Attention Deficit Hyperactivity Disorder,[14], dementias, or mental disorders"

sorry if I'm going on and on about it- I just find it really interesting

Fun stuff!!

- Got 100% on the OP's test but was close to running out of time twice

- got 15.43 seconds then 19.24 seconds on the timed stroop test

- Average of 1.1 second on the individual timed on (100% correct)

- 8.86 for the last test with correct colors/names, 13.547 for different names/colors

I said red pickaxe.

Only reason I didn't say hammer was that I have a pickaxe sitting within view. The first reply needs to be spoilered or removed

oil, but now I'm thinking gears.

Nope, but not too far off.

there are 55 omited #s through the first 10 generations. so my guess is 550. lol

i don't think there is a solid stragedy

Unfortunately I'm agreeing with you on this for the moment.

I started bruteforcing it and decided it wasn't worth it lol. Unless I can figure something else out I'll have to wait for a solution.

Without me cars could not go,

I was important ever so.

But now with changes near complete,

I fear I'll become obsolete.

When tides need turning in a rout,

It seems that hope may come about.

For there’s a chance yet of a coup,

Don’t stop now I’m coming through.

Gripping terror, clenching jaws,

I can't escape her deadly claws.

This vicious beast may be my end,

With strength I cannot comprehend.

Keep in mind they also lose 18, 68, etc.

I don't get it. if everyone has a unique number, then if there are 2^100 bacteria there are 2^100 numbers.

I haven't found a definitive pattern yet

the 101st generation should start at 2^100 +1. However they will lose numbers: 6,9,16,19,61,69,91,96,106,109,161,191,601,609,611,901,906,911,619,916,10601,1090

1,11611,11911,etc.