spectacles

eclipse

Nope to both, however, your second guess is correct for my FIRST WAI:

Alright, so it has been a while since I've posted on BD...and it's been even longer since I came up with a word puzzle...so I thought I would try another one.

Through rings I can be seen

And by my rings I am known

Through thousands of years

My legend has grown

My death is historic

In more ways than one

I fell in controversy

And now my time is done

But my name will live on

For being a hero to man

If you're old enough to see me

Next year is when you can

Who am I?

[Another oldie - can you get them all?]

360 degrees in a circle

12 days of christmas

5 sides on a pentagon

What about:

40 W in a P

4 L on a D

5 F on a G

6 S on a D

4 Legs on a Dog

5 Fingers on a Glove

6 Sides on a Die

Not sure of the 40 W in a P...

Some of my favorites:

Monopoly (mono = one, poly = more than one)

Everyone

Pianoforte (musical term literally meaning soft-loud)

Sophomore (sopho = wise, more = foolish)

Mythological

Seashore

Nevermore

Butthead :c)

I said that,"that 'that' that that man wrote should have been underlined."

"MI"

(2) M --> M[iI]

(2) M[iI] --> M[iIII]

(1) [MIII]I --> [MIII]IU

(3) [MI]III --> [MI]U

(2) M[iUU] --> M[iUUIUU]

(4) [MIIUUI]UU[] --> [MIIUUI][]

(4) [MII]UU --> [MII]

(3) [M]III[] --> [M]U[]

"MU"

This is not a correct answer. Very good try.

"MI"

(2) M --> M[iI]

(2) M[iI] --> M[iIII]

(1) [MIII]I --> [MIII]IU

(3) [MI]III --> [MI]U

(2) M[iUU] --> M[iUUIUU]

(4) [MIIUUI]UU[] --> [MIIUUI][] - You added an extra "I" after the "M" between these two steps.

(4) [MII]UU --> [MII]

(3) [M]III[] --> [M]U[]

"MU"

First off, sorry if this has been posted. I haven't seen it on here, and searching for this one was difficult (since "MI" and "MU" are both under 3 letters, and the search doesn't like that...). So again, I apologize if this is a duplicate.

Can you use only the following 4 rules to convert MI to MU?

1) xI --> xIU

2) Mx --> Mxx

3) xIIIy --> xUy

4) xUUy --> xy

Please note: "x" and "y" are placeholders for any string of characters. For example, MUUI would match rule 1 (x=MUU), rule 2 (x=UUI), or rule 4 (x=M, y=I)

Also note: the unidirectional nature of the rules...meaning you CANNOT go from MUU to MIIIU using rule 3...you can only go from MIIIU to MUU.

Also: Before anyone asks, you can use the rules as many or as few times as you like, but you can ONLY use those 4 rules.

I like to use the convention of XXX --#--> YYY where "XXX" is the start string, "#" is the rule number used, "YYY" is the resulting string. So here's a quick example showing a valid conversion:

Convert MUI to MIU

MUI --1--> MUIU --2--> MUIUUIU --4--> MUIIU --1--> MUIIUUIIU --4--> MUIIIIU --3--> MUUIU --4--> MIU

Anyways, this is a classic puzzle. Hope this makes sense! Good luck!

4 & 3 ?

Yep, 4 and 3 are the two missing numbers...

What are the next two numbers in the following number series?

101, 301, 701, 901, 311, 721, 131, 731, 931, ?, ?

941, 151, 751, 361, 761...could keep going...it is infinite.

Also important to note, this actually isn't the beginning of the sequence...101 is actually the 26th position in the sequence.

I'll leave it at that to see if others can figure out what the sequence is...

picture puzzles 5 to 7

This would be talking about the RMS Titanic.

Ah sorry. That should be:

1,2,3,13,16,19,22,179,201,223,245,267,289,311,56163,52518... They continue to increment in 355s to at least 100,000 (I have used a computer to generate the sequence). However, this does not describe the sequence. It so happens that the sequence groups into arithmetic progressions but what actually is it?

Still seems like you mis-typed the last number or two...because that still shows it decrementing. Is it supposed to be:

56163, 56518? or 52163, 52518?

Also, I just wanted to verify that it increments by 355 after that? not 311?

ok...

so this lady goes to work everyday... one fine sunday she goes out shopping and buys new shoes.. the next day she gets killed at her work

guess the situation...

u can ask me ques..

have fun

She works somewhere where there is a low clearance (stuff flying overhead...) and she bought new high heels...when she wore them to work, she got hit in the head or decapitated or something along those lines...

Or she works where the floor is electric, and she didn't buy rubber soled shoes...

1/= ^/0(_) ( @|\| /93@|) +#1$ +#3|\| ^/0(_) $|^3|\||) \/\/@^/ +00 |\/|(_)( # +1|\/|3 0|\| ^/0(_)/9 ( 0|\/||^(_)+3/9. +#3 $0!(_)+10|\| +0 +#1$ /91|)|)!3 1$: “ @ $3( /93+ ”.

...You take that back...I do not spend too much time on my computer...and I didn't think the answer was too much of "a secret"...

To be exact, here is the full message:

IF YOU CAN READ THIS THEN YOU SPEND WAY TOO MUCH TIME ON YOUR COMPUTER. THE SOLUTION TO THIS RIDDLE IS: " A SEC RET ".

There are only 2 more numbers in the sequence:

10000, 1111111111111111

Basically the sequence is just the number 16 in each of the different bases from 16 to 1:

16 in base 16: 10

16 in base 15: 11

16 in base 14: 12

etc...

16 in base 3: 121

16 in base 2: 10000

16 in base 1: 1111111111111111

I think maybe.

Wouldn't the leangth of the line to be measured be equal to the circumfrence of the wheel if point "A" made a complete rotation?

Not quite...the distance along the X-Axis would be the circumference of the wheel...because that's how far point "A" travelled along the X-Axis...the question is asking the distance in the x-y plane, which is not quite the same thing...

neither is right, you win in 1/3 of cases

according to my calculation p doesn't change. p=(t-1)/(2t-1) while t is the number of cards those are in unremoved color.

So, I agree with these answers...but I decided to put the math and proof down for this:

#1 is just a specific implementation of the original case (not removing a card)...so let's say that n is the number of red and black cards in the deck...which means that the total number of cards in the deck is 2n...

After you pick a card, you have a ^(n-1)/_(2n-1) chance of picking the same color card (so p = ^(n-1)/_(2n-1))

Now, if we take one card away at the beginning (let's say red for the sake of argument...but it could be either) we are left with 2n-1 cards in the deck: n-1 red, n black.

Now let's say you pick a black card first:

you now have a 50% chance of matching the colors...^(n-1)/_(2n-2)

if you choose red card first:

you have a ^(n-2)/_(2n-2) probability...

So, your total probability for this scenario is:

p = ^(n-1)/_(2n-1)*^(n-2)/_(2n-2) + ⁿ/_(2n-1)*^(n-1)/_(2n-2)

Now it's just simple algebra:

^{(n^2-3n+2)}/_(4n^2-6n+2) + ^{(n^2-n)}/_(4n^2-6n+2)

^{(2n^2-4n+2)}/_(4n^2-6n+2)

^(2n-2)(n-1)/_(4n-2)(n-1)

^(2n-2)/_(4n-2)

^(n-1)/_(2n-1)

Which is the exact same probability that we had in case #1...and it matches what nobody had as his answer. Nice problem.

Unfortunately, I couldn't get this narrowed down very far, so I broke down and wrote a program to figure it out for me. Also, unfortunately, It appears there are a number of valid solutions:

1249758 = 9 * 138430 + 1926 + 1962

1378692 = 9 * 152756 + 1926 + 1962

1537290 = 9 * 170378 + 1926 + 1962

2149758 = 9 * 238430 + 1926 + 1962

2684907 = 9 * 297891 + 1926 + 1962

3178692 = 9 * 352756 + 1926 + 1962

3584907 = 9 * 397891 + 1926 + 1962

4078692 = 9 * 452756 + 1926 + 1962

5137290 = 9 * 570378 + 1926 + 1962

5384907 = 9 * 597891 + 1926 + 1962

6284907 = 9 * 697891 + 1926 + 1962

I haven't verified all of them by hand (to make sure my program worked...) but I did verify the first one it came up with. With the first one the answers are:

1249758 = 9 * 138430 + 1926 + 1962

0=E, 1=M, 2=A, 3=O, 4=R, 5=I, 7=L, 8=N, 9=Y

ROLE = 4370

LONELY = 738079

Best answer. One of those lateral solutions that makes more sense and fits the original parameters even more than the intended solution.

I would agree, except for the part of the OP that says "...and then have a wonerful night together..." Because that implies that they both had a wonderful night together...and I'm going to go out on a limb and say that if she actually did those things to him, he did NOT have a wonderful night (unless he's a serious masochist and somehow survived all of the above) :c) But I agree that it's a great answer.

So...there are 2 possible solutions to this problem. Each with 8 different configurations/rotations...that gives a total possible 16 "answers" to this problem...but only 2 distinct solutions. The first can be found if you don't use the number "2" in the solution (so use numbers 013456789). The second can be found if you don't use the number "7".

The sums for those two are as follows:

for the solution that doesn't use the 2, the sums of the rows/columns/diagonals are 11, 12, 13, 14, 15, 16, 17, 18...and for the solution that doesn't use the 7, the sums are 9, 10, 11, 12, 13, 14, 15, 16.

1 8 6

9 7 0

4 3 5

and

3 1 8

9 2 0

4 6 5

I took the OP to mean if b is rs, then the original equation was

x² - ax + rs = 0

...but he wrote it down as

x² + ax - sr = 0

(flipped the digits in b as well as the + and - sign)...

PART 1

So, we know that E must be either 1, 2, or 3...because any higher digit and there would be no way that 4 single digits could add up to 9*E.

If E = 1...we need A + B + D + E = 9...or A + B + D = 8... where A, B, and D cannot be 1 (since E is 1)...and there are no possible combinations that work...so E does not equal 1.

Let's move to E = 3...then we need A + B + D + E = 27...or A + B + D = 24...and there is only one possibility for that 7 + 8 + 9...but that would only give us one of the 4 2x2 squares...so we know that E must be 2.

E = 2

A + B + D + E = 18

A + B + D = 16

the combinations that work for this equation are:

1 7 8

1 6 9

3 4 9

3 6 7

4 5 7

So, What we need to find out of these 5 combinations are 4 of them such that 4 of the numbers are used only once, and 4 of them are used twice...

if we remove the 3 6 7 combination, we can then see that 1, 4, 7, and 9 are used twice...3, 5, 6, and 8 are used only once...so, now you just have to arrange accordingly...and our answer is

8 1 6

7 2 9

5 4 3

PART 2

We can then use the same sort of logic to determine that if they add to 7*E...E must be 3...and we get combinations of

A + B + D = 18

1 8 9

2 7 9

4 5 9

4 6 8

5 6 7

Which if we remove the 4 5 9, we see that 1 2 4 5 are used once, 6 7 8 9 are used twice...so our answer is

1 8 4

9 3 6

2 7 5

So, you may be interested in noting that the OP stated that the digits are between 1 and 9...not 0 and 9...so, hopefully that will help you narrow down your possible answers a bit :c)

At the north pole, the sun always rises and sets in the south...at the south pole, the sun always rises and sets in the north...

You can actually argue that the sun rarely ever rises directly east. It only does that once or twice a year if you are in the tropics...otherwise, it always rises slightly north or south of east...

ok, well I agree with all of the posts that say it should be trivial to print out the even numbers to any number...and I also agree that 10 minutes is a LONG time to write that program...here it is in Java (took about 20 seconds to write, literally...this assumes you will pass in the value upon execution...):

for (int i = 0; i <= Integer.parseInt(args[0]); i++) {
if (i % 2 == 0) {
System.out.println(i);
}
}
}

public static void main(final String[] args) {

So, I believe it would be something like this in C (been while since I've done C...):

for (int i = 0; i <= atoi(argv[0]); i++) {
if (i % 2 == 0) {
printf("%d\n", i);
}
}
return 0;
}

int main(int argc, char *argv[]) {

You could also replace the Integer.parseInt(args[0]) and atoi(argv[0]) with a constant value (500 or 10000 in this case), and in order to change the upper bound, simply change that value...that's it...So, I say that it will take him 0 minutes to write the second program, since if he did the first one correctly, he will not have to do any more writing for the second.

I will also note that this doesn't do any error handling or checking to make sure the user isn't stupid...adding those checks in adds about 1 more minute of programming (to wrap stuff in try-catch blocks and gracefully exit...so, that's about it)...

Ok, so, let's divide up the real numbers into 5 groups and prove values in each group:

N < -1

If N was less than -1, then the right hand side of the equation would yield a positive number greater than 1 ALWAYS. The left hand side of the equation would always be a negative number somewhere between 0 and -1...therefore there are no solutions for N < -1.

-1 <= N < 0

If N was in this range, we get the right hand side of the equation to be 0 (since <N> becomes 0), but the left hand side would have to be 0 - N...so the solution would have to satisfy 0 - N = 0...which does not exist in this range.

0

From the above proof, we see that 0 - N = 0 is a solution...so when N = 0, we have a solution.

0 < N <= 1

IF N was in this range, the right hand side is N (because <N> becomes 1), and the left hand side would be 1-N...so the solution would have to satisfy 1 - N = N...therefore, the only solution in this range is 0.5

N > 1

This is the same argument as the N < -1...RHS becomes positive number greater than 1, LHS always is less than 1...so no solutions exist in this range.

Therefore, N = 0 and N = 0.5 are the only solutions that satisfy this equation. QED