Determine all possible value(s) of a real number N, such that N satisfies this equation:

(N) = N*<N>

Note: <x> is the least integer ≥ x, and, (x) = <x> - x.

N = 0:

Basically the left side becomes <0> - 0...which is 0, and the right side is 0 * 0...which is also 0...so N = 0 works

N = 0.5:

The left side becomes 1 - 0.5...which is 0.5 and the right side is 0.5 * 1...which is also 0.5...so N = 0.5 works

...and I'm just now looking at this, but I'm thinking those may be only answers...Non-zero integers do not work since they would cause the left side to be zero always, but the right side to be a positive number...still checking and working on a proof...

2(ELEVEN) + TEN+ ONE = THIRTY + THREE, where ELEVEN is divisible by 11.

First of all, I agree with huofo's answer, but since he didn't show how it was derived, I figured I would put my solution down. Also, note, that while my methodology would work with not too much effort, it was sort of luck that I came across the answer as quickly as I did. There may be a faster way, but this way works...

Wow, ok...let's start with some observations and use these as axioms:

Axiom 1. E, T, and O can't be 0

Axiom 2. N + N + N + E = Y + E...which means N + N + N = Y

Axiom 3. By Axiom 2 we know that if N is odd, Y is odd...and if N is even, Y is even and also N and Y can't be 0 or 5...

N = 1, Y = 3

N = 2, Y = 6

N = 3, Y = 9

N = 4, Y = 6

N = 6, Y = 8

N = 7, Y = 1

N = 8, Y = 4

N = 9, Y = 7

Axiom 4. E + E + E + N + X = T + E (where X <= 3), so that means E + E + N + X = #T (where X <= 3)

Axiom 5. Since ELEVEN is divisible by 11, we know that (EL + EV + EN) is also divisible by eleven.

Axiom 6. Since ELEVEN is divisible by 11, we know that (E + E + E) - (L + V + N) is a multiple of eleven...so that means 3E - L - V - N is a multiple of eleven...

Let's say E >= 5:

Then the left side becomes a 7 digit number. If this is the case, then T MUST be equal to 9 (in order to get the right side to be a 7 digit number)...But if E > 5, then the left side becomes too large to allow the right side to equal it. So we now know that 0 < E <= 5 and we know that T > E (by at least a factor of 2):

E = 1, T >= 2

E = 2, T >= 4

E = 3, T >= 6

E = 4, T >= 8

E = 5, T = 9

So T cannot be 1

If E = 1:

T = 2 or 3

Using Axiom 4, we can say that 2 + N + X = 9 (where X <= 3), so N must be 7 or 8 (so by axiom 3, Y must be 1 or 4)...but Y can't be 1, so N can't be 7...

So that means, T must be 2 or 3, and N must be 8, Y must be 4

if T = 3, when we add what we have so far, we end up with ####35 = ####45...so obviously that doesn't work...so T must be 2.

We now have 2(1L1V18) + 218+ O81 = 2HIR24 + 2HR11

We then have (by Axiom 6) 3 - (L + V + 8) is a multiple of eleven...so L + V must be 6...so the only combination of remaining numbers that allows for this is:

L = 0, V = 6 or L = 6, V = 0

We also know (by plugging the numbers we have) that 2L = H + 2 + X (where X <= 1), so 2L = H + 2 OR 2L = H + 3...so we have:

if L = 0, H = 7, but we can see right away that there is no way this will work, since 2 * 101### can never equal 27####...so L can't be 0.

so then L must be 6, and H must then be 9:

We are left with 2(161018) + 218+ O81 = 29IR24 + 29R11

Let's simplify this a little bit:

322254 + O81 = 29IR24 + 29R11

Going with that, we can then say that 2 + O + 1 = R + R, so O + 3 = 2R:

if O = 3, R = 3...which doesn't work

if O = 5, R = 6...which doesn't work

if O = 7, R = 5...it works, so we go with that.

322254 + 781 = 29I524 + 29511

It is now trivial to solve for I to see if our final solution satisifies the original:

323035 = 29I524 + 29511

293524 = 29I524...so I must be 3...and that does work.

So, we have our solution already, without needing to check if E >= 2. There MAY be other solutions out there that satisfy this, but this was the first one we came across...

so, we know that NOD must be between 317 and 999 (inclusive) in order to get a 6 digit number

We also know that D must be either 0, 1, 5, or 6 so that D * D = a number ending in D...

However, if D were 0, then the last 2 digits of the result would be D...because ##0 * ##0 = ###00...so D can't be 0.

So given that, we know:

N must be 3, 4, 5, 6, 7, 8, or 9

D must be 1, 5, or 6

and O cannot be 0.

The basic pattern now is simply to try the values of N, and what you find is that when you choose a specific value for N, there are only 2 or 3 numbers you have to try out. Let me show by example:

If N was 3, then O must be 1 because 396 (highest number allowed with N=3) squared is 156816...But that means the only two numbers it could be are 315 and 316...which we've already shown are too low for it. So N cannot be 3.

If N was 4, then O must be 1 or 2...and if O is 2, then 426 (lowest number allowed with N=4, O=2) squared results in a number beginning with 1...so O can't be 2, O must be 1...but once again, we run into the problem that 415 squared (smallest number allowed with this pattern) is too high to result in a 14#### number. So N can't be 4.

If N was 5, then O must be 2 or 3...if O is 2, our lowest number is 521, which when squared gives a number higher than 25####, so O can't be 2...so O must be 3...but, 536 squared ends up still only yielding 2#####, so O can't be 3, and N can't be 5...

If N was 6, then O must be 3 or 4 (catch the pattern here so far)? If O is 3, then 631 is the only number to yield a value 3#####, but unfortunately it doesn't fit the pattern, so O can't be 3. O must be 4...but again, we find that 641 and 645 do not fit...so N can't be 6

If N was 7, then O must be 5 or 6. If O is 5, then we check 751 and 756 as our possible values...and lo and behold, we find that 756 works:

(NOD)² = onLOAD

(756)² = 571536

So

N = 7

O = 5

D = 6

L = 1

A = 3

Keep 'em coming K, whenever I don't have work to be doing, these are a great way to spend my time

You have a tap that can rotate and has the water going straight downwards out of it(like this) and a cup that is fixed to the bottom of the sink by it's base just out of reach of the water.

With out using any other items how can you fill the cup?

Simply spin the tap at different speeds...the angular momentum will carry the water outwards...find the speed that allows the water to reach the cup and just keep spinning it at that speed (if the tap doesn't spin 360 degrees, you need to spin it back and forth at that speed).

888 + 88 + 8 + 8 + 8 = 1000

Aww...bobpi beat me by about 5 seconds...nicely done

pickett ur right.

any other possiblities.

If you assume the following about the OP:

1. Each letter is a distinct number (meaning two letters can NOT have the same numerical value)

2. The numbers cannot start with 0

Then, I would say "no" there are no other possibilities. Here's the way to solve this and essentially the proof that there aren't any others:

Given: ABCDE*4=EDCBA

A must be even and not 0

ABCDE must be less than 25000

So, therefore A must be 2

2BCDE * 4 = EDCB2

BA must be a multiple of 4

B must be 1, 3, 5, 7, or 9

E must be 3 or 8 to result in number ending in 2..but 3 doesn't work...so E must be 8

2BCD8 * 4 = 8DCB2

B must be 0 or 1 so that answer is less than 90000, but can't be 0, since that would result in an answer that is not a multiple of 4...so B = 1

21CD8 * 4 = 8DC12

D must be either 7 or 2 so that 4 * D + 3 = a number ending in 1...(so 4*D must end in 8...aka, D must be 7 or 2)

But D can't be 2, since A is 2...so D = 7:

21C78 * 4 = 87C12

Finally, that leaves C being either 0, 3, 4, 5, 6, or 9

well, 4 * C + 3 = number ending in C...

so C can not be 0, 3, 4, 5, or 6...which leaves C being 9

Therefore the equation becomes:

21978 * 4 = 87912

I can give an explanation and steps to solving if people want...but it's pretty self-explanatory...

21978 * 4 = 87912

So, ABCDE = 21978

Boy, this is getting much tougher...You are just evil K Sengupta

. Let me first explain why this one is a lot harder. Obviously dealing with CUBE roots is no fun for anyone. Also, there are only 7 of the 10 digits used in this problem. Finally, when initially working it out, ITS and ANY can be interchangable...it isn't until you try to plug them into the SIGN part that you have to distinguish which 3-digit number goes with which word...so this one is VERY difficult. The most difficult part lies with the following:

We know that at least one of ³√(ITS) or ³√(ANY) can not be evaluated to whole numbers, since:

5^3=125

6^3=256

7^3=343

8^3=512

9^3=729

And we know that neither of them can be 7^3, since that has a 3 repeated in it (and neither of them have repeated numbers)...BUT that leaves us with all numbers that a "2" in them. Given that they do not share any common numbers, we know that at least one of the values must not be a whole number.

Finally, let me preface this with the statement that this is very difficult to explain well. My method does work, and isn't as much trial and error as it may sound in the below description...

Ok, so we know that the range for either ³√(ITS) or ³√(ANY) must be between 4.64 and 10 (exclusive)...which means that the ³√(SIGN) must be between 9.999... and 20 (exclusive).

Therefore, S cannot be 8, 9, or 0 because the highest number we can get from that range would be just under 8000.

Assuming you're not just REALLY mean, I would have to assume that the answer to this will be an equation that basically satisfies: x³√(a) + y³√(a) = z³√(a) which would leave us with the equation x + y = z...otherwise, there really wouldn't be much of a way to satisfy it.

If this is the case, we must find 3 digit numbers that contain perfect cubes as factors so that we can factor them out:

1³=1

2³=8

3³=27

4³=64

5³=125

6³=256

7³=343

8³=512

9³=729

x or y can't be 8 or 9, because anything multiplied by them would yield at least a 4 digit number.

So, we know that z must be between 3 and 13 (inclusive).

If either x or y is 1, we know that a must be a 3-digit number...so a would have to be between 100 and 125. If this were the case, then we would know that one of the 3-digit numbers would start with either 10 or 12. Which in turn makes it so that the other starts with 8 or 9...

So, we get these possibilites for the 3 digit number pairs:

102/816, 103/824, 104/832, 105/840, 106/848, 107/856, 123/984, 124/992

But we can rule out any that have duplicate numbers between them, so we are left with:

103/824, 104/832, 107/856, 123/984

That leaves us with only 4 possibilities:

³√103 + 2*³√103 = 3*³√103: ³√(103) + ³√(824) = ³√(2781) which doesn't work

³√104 + 2*³√104 = 3*³√104: ³√(104) + ³√(832) = ³√(2808) which doesn't work

³√107 + 2*³√107 = 3*³√107: ³√(107) + ³√(856) = ³√(2889) which doesn't work

³√123 + 2*³√123 = 3*³√123: ³√(123) + ³√(984) = ³√(3321) which doesn't work

So, we know that neither x or y is 1...which makes our life a bit more interesting, but we now know that z cannot be 3 or 4!

Let's start on the other side:

If x or y is 7, we know that a MUST be 2:

BUT that would mean one of the 3 digit numbers would be 686, which doesn't match, so x and y are not 7...

If x or y is 6, then we know that a MUST be 2 or 3, so we get these possibilities:

6*³√2 + 4*³√2 = 10*³√2: ³√(512) + ³√(128) = ³√(2000) which doesn't work

6*³√2 + 5*³√2 = 11*³√2: ³√(512) + ³√(250) = ³√(2662) which doesn't work

6*³√3 + 4*³√3 = 10*³√3: ³√(768) + ³√(192) = ³√(3000) which doesn't work

6*³√3 + 5*³√3 = 11*³√3: ³√(768) + ³√(375) = ³√(3993) which doesn't work

So, again, x or y is not 6...so x or y must be 2, 3, 4, or 5

If 5, the other must be 4 and then a must be 2, 3, 4, 5, 6, or 7

5*³√2 + 4*³√2 = 9*³√2: ³√(250) + ³√(128) = ³√(1458) which doesn't work

5*³√3 + 4*³√3 = 9*³√3: ³√(375) + ³√(192) = ³√(2187)...

HOLY COW, we have one that works!!

³√192 + ³√375 = ³√2187...which simplifies to 5*³√3 + 4*³√3 = 9*³√3

So we have:

I=1

T=9

S=2

A=3

N=7

Y=5

G=8

Now, maybe I can get some work done...Great puzzle...very difficult...

By the way, if the original words maintain their association with their sorted versions, what might this worm-infected dictionary be useful for?

Don't know if it's what you were thinking, but anagrams...you simply take a word and alphabetize the letters, then look it up in this dictionary and find all the other words with the same alphabetized list. and voila you have all of the anagrams for a given word.

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

SCENE + (IT)*(IS) = NEATLY

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

Unfortunately I have to go to a meeting, so I cannot finish this right now...but I'll give you my first thoughts...This one's a little harder, since there must be 1 digit that isn't used at all...

IT * IS can only be a 3 or 4-digit number (100 up to 9801)...so, we are adding a 5 digit and a 3 or 4 digit to get a 6 digit number...so we know that N must be 1 (since the 6 digit number must start with 1)...and S must be 9 (because adding a 3 or 4 digit number to a 5 digit number to get a 6 digit number means that the 5 digit number must be between 90199 and 99900)...which using the pattern of letters, limits us to numbers between 92313 and 98717

We can also deduce that E must be 0, because if E = 2, then (IT)*(IS) must be a 5 digit number (123456 - 98212 = 25244)...and that can't be. and E can't be 1, because N is 1.

So we now have:

9C010 + (IT)*(I9) = 10ATLY

We now know that the smallest number we can get from (IT)*(IS) is 4335 (because 102345 - 98010 = 4335).

The largest number we can get from (IT)*(IS) is 87*89 = 7743

So our range of values for (IT)*(IS) = 4335 to 7743

Our only possible numbers for SCENE are 92010, 93010, 94010, 95010, 96010, 97010, and 98010.

And our range of values for NEATLY = 102345 to 108765

We can then say that I must be 6, 7, or 8 because anything less would yeild a (IT)*(IS) result less than 4335.

if I=8, then C = 3,4,5,6,7

if I=7, then C = 5,6,8

if I=6, then C = 7,8

Like I said, that's all I've got for now...if no one has solved this when my meeting's over, I'll work on it some more

Interesting since a Roman Legion was about 6000 men... how'd they count them all?

So, there is actually a notation that isn't as well known with Roman Numerals. The Romans would typically write out all of the M's that were needed to do higher order numbers, however, when following the rules that were stated above, and when getting to VERY high numbers they actually had a notation that they used.

1000=M

2000=MM

3000=MMM

Now, you actually put a bar over the numbers, to represent multiplication by 1,000:

4000 is actually written as:

__

IV

So, then 5000:

_

V

5150:

_

VCL

And 100,000:

_

C

and 1,000,000:

_

M

And to go higher than that, you simply put 2 bars over the numbers to represent multiplying by 1,000,000.

1,000,000,000 would then be:

=

M

So, to represent their 6000 men, would be

__

VI or if they weren't being lazy and decided to disregard their own rules: MMMMMM

Hope this helps.

Complete the below sequence by filling in the missing letters..

M, V, E, _, J, S, U, _, P

I would assume it's supposed to be M, N (unless there is some crazy twist to this one)...

However, I think P should no longer be part of this sequence as it has been kicked out of the family...

Well, because it's a 5-digit number squared resulting in a 9 digit number... we know that the 5-digit number can only be between 10000 and 31622...

So, that means that E must be either a 1 or 2.

We also know that I cannot be 1 (because then G would be 1), and I cannot be 5 or 6 for the same reason...

So, we know

if I = 2, G = 4

if I = 3, G = 9

if I = 4, G = 6

if I = 7, G = 9

if I = 8, G = 4

if I = 9, G = 1

We also know that if E = 1, then I cannot be 9 (because G would have to be 1)

This greatly limits the number of possible 5-digit number we can have that match that pattern (about 60):

11223, 11227, 11228

11332, 11334, 11337, 11338

11443, 11447

11552, 11553, 11554, 11557, 11558

11662, 11663, 11667, 11668

11772, 11773, 11774, 11778

11882, 11883, 11887

11992, 11993, 11994, 11997, 11998

22113, 22114, 22117, 22118

22334, 22337, 22338, 22339

22443, 22447, 22449

22553, 22554, 22557, 22558, 22559

22663, 22667, 22668, 22669

22773, 22774, 22778, 22779

22883, 22884, 22887, 22889

22993, 22994, 22997, 22998

Which, you COULD just trial and error with those results, and it wouldn't take long to find that the only answer is: (22887)² = 523814769...I'll leave it to someone else to narrow it down more if they like...

Going by the results I have listed above, then if E=1, then that would make R=1 (because 11998^2 = 143952004 and 11223^2 = 125955729...)

So we then know that E MUST be 2...so we are down to about 30 possible numbers:

22113, 22114, 22117, 22118

22334, 22337, 22338, 22339

22443, 22447, 22449

22553, 22554, 22557, 22558, 22559

22663, 22667, 22668, 22669

22773, 22774, 22778, 22779

22883, 22884, 22887, 22889

22993, 22994, 22997, 22998

We can then determine that R MUST be 4 or 5 because 22113 * 22113 = 488984769 and 22998 * 22998 = 528908004

But it can't be 4, because since E = 2, we know that 488984769 > 42???????...so R=5

Now, it should be noted that we now know that the result begins with 52...and the UPPER bound on our results is 528908004...so we know that the number that works must be close to the upper bound (specifically within 22883 and 22998)...and really, we can get rid of 22998, since the result has zeroes in it, but we'll ignore that for now...so we are left with only these 8 possible numbers:

22883, 22884, 22887, 22889

22993, 22994, 22997, 22998

So we know that U can only be 8 or 9, and I can only be 3, 4, 7, 8 or 9...

Now, let's say that G=4...then we know that I=2 or I=8...but I can't equal 2 since we know that E=2...so I=8:

(22UU8)^2 = 52SULT8N4

which makes U=9:

(22998)^2 = 52S9LT8N4..but we rule this one out right away, because 22998 was our upper bound of 528908004, which obviously doesn't work...so G cannot be 4, and in turn, I cannot be 8...and we also find that U cannot be 9

Since U cannot be 9, we know that U must be 8:

(2288I)^2 = 52S8LTING

So, now we have 4 possible answers:

22883, 22884, 22887, and 22889...

We can then find out that S MUST equal 3 (since for all 4 of them, they have the same value for S)...Which means that I cannot be 3 anymore::

(2288I)^2 = 5238LTING

So we only have 22884, 22887, and 22889 as our possible numbers now...We can really just start eliminating them. First assume I=4, then you have 22884*22884 which is 523677456...but we know that it must be 5238?????...so that doesn't work...

Now try I=9, and you get 523906321...which again does not match 5238?????...so that only leaves I=7:

(22887)^2 = 523814769

(EEUUI)^2 = RESULTING

And as you can see, the result works out to be:

R=5

E=2

S=3

U=8

L=1

T=4

I=7

N=6

G=9

Whew...

can someone explain what the hell is going on

i thought the answer was a formula at first mathematical but then i thought it was the periodic table (but when i looked at it for some reason i was thinking code instead of pattern) but how can a series of numbers be the answer to a pattern (shouldnt it be a formula or at least where the pattern comes from)?

i think im missing something

You are correct in that it is the periodic table. Each of the numbers is the atomic number of an element. Plug in the element's symbol and you get an sentence. The answer I gave of: 16 8 92 90 95 68 53 20 translates into "South America"...which is my answer to his question "Where can you find peru? Argentina? Honduras?"

Hope that helps.

Well, because it's a 5-digit number squared resulting in a 9 digit number... we know that the 5-digit number can only be between 10000 and 31622...

So, that means that E must be either a 1 or 2.

We also know that I cannot be 1 (because then G would be 1), and I cannot be 5 or 6 for the same reason...

So, we know

if I = 2, G = 4

if I = 3, G = 9

if I = 4, G = 6

if I = 7, G = 9

if I = 8, G = 4

if I = 9, G = 1

We also know that if E = 1, then I cannot be 9 (because G would have to be 1)

This greatly limits the number of possible 5-digit number we can have that match that pattern (about 60):

11223, 11227, 11228

11332, 11334, 11337, 11338

11443, 11447

11552, 11553, 11554, 11557, 11558

11662, 11663, 11667, 11668

11772, 11773, 11774, 11778

11882, 11883, 11887

11992, 11993, 11994, 11997, 11998

22113, 22114, 22117, 22118

22334, 22337, 22338, 22339

22443, 22447, 22449

22553, 22554, 22557, 22558, 22559

22663, 22667, 22668, 22669

22773, 22774, 22778, 22779

22883, 22884, 22887, 22889

22993, 22994, 22997, 22998

Which, you COULD just trial and error with those results, and it wouldn't take long to find that the only answer is: (22887)² = 523814769...I'll leave it to someone else to narrow it down more if they like...

ANSWER:

16 8 92 90 95 68 53 20

Not sure this one is correct:

49 1 8 74 26 74 92 16 16 73 52 16 20 7 53 4?

Because the words don't seem to make sense...specifically the 92 16 16 73 52 16 part....

OK all you programmers, do the exhaustive search now.

You asked me to do it:

I wrote some code to solve this one...and the only answer that works is this...

224
--------
716
358
358
--------
40096

     179

Sorry for anyone that has spent a lot of time on trying to solve this manually...but this only took about 3 minutes to write the code for and get the answer :c) Probably not the most efficient code in the world, but I went with the quick and dirty method...here's the code (Java) in case you are wondering:

for (int i = 100; i < 1000; i++) {
for (int j = 100; j < 1000; j++) {
int twoOnes = Integer.parseInt(new Integer(j).toString().substring(2));
int twoTens = Integer.parseInt(new Integer(j).toString().substring(1, 2));
int twoHuns = Integer.parseInt(new Integer(j).toString().substring(0, 1));
boolean validValue = true;
if (i * j >= 10000 && i * j < 100000) {
if (twoOnes * i > 100 && twoOnes * i < 1000) {
if (twoTens * i > 100 && twoTens * i < 1000) {
if (twoHuns * i > 100 && twoHuns * i < 1000) {
String s = "" + i + j + (twoOnes * i) + (twoTens * i) + (twoHuns * i) + (i * j);
int[] counts = new int[10];
for (int k = 0; k < s.length(); k++) {
int val = Integer.parseInt("" + s.charAt(k));
counts[val]++;
}
for (int k = 0; k < counts.length; k++) {
if (counts[k] != 2) {
validValue = false;
break;
}
}
} else {
validValue = false;
}
} else {
validValue = false;
}
} else {
validValue = false;
}
} else {
validValue = false;
}
if (validValue) {
System.out.println(" " + i);
System.out.println(" " + j);
System.out.println("--------");
System.out.println(" " + (twoOnes * i));
System.out.println(" " + (twoTens * i) + " ");
System.out.println(" " + (twoHuns * i) + " ");
System.out.println("--------");
System.out.println(" " + (i * j));
System.out.println();
}
}
}
}

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

Ok, changed my mind...the starting over definitely play a part in it...in fact, my new answer is:

infinite sum of (chance of winning * chance of starting over raised to the nth power)

sum from n=0 to infinity of (11/63 * (73/126)ⁿ)

And I don't know what that converges to...but I'm going to guess somewhere between 1/3 and 1/2...maybe 40%?

Ok, it's been a while since I've done conditional probabilities...

First of all, my answer (for now) is you have a 11/63 chance of WINNING (approximately 17% chance). Here a few thoughts which led me to this:

IF you pick urn 1:

you have a 1/3 chance of winning, 5/18 chance of losing, and the 7/18 chance of starting over.

IF you pick urn 2

you have a 1/63 chance of winning, 8/63 chance of losing, and 8/9 chance of starting over.

Since you play until a terminal state, the starting over parts really have no impact, because you then are starting your odds all over again from scratch.

You also have a 1/2 chance of picking each urn...SO...

your chances of winning by urn 1 are (1/2 * 1/3) =1/6

your chances of winning by urn 2 are (1/2 * 1/63) = 1/126...

so, 1/6 + 1/126 = 11/63 which is your total chance of winning.

Like I said, been a while since I did probabilities, but that's my answer for now.

oh boy...this thread will be locked very shortly, I assume...classic Honestants/Swindlecants problem that has been posted on this site more than any other riddle...it's a great one, but been done before...

*Said in Robotic monotone* "I come from France!"

Well played, sir. Great reference...And I also agree with this answer...

A man is 6 feet tall. His wife is 5 feet tall and his two children are 4 feet and 3 feet. He wants to buy a flat mirror to place on a wall so that he and his family can stand in front of it and everyone can see themselves (their whole body) and everyone else at the same time. What is the minimum height of the mirror he should buy?

(Assume a persons eyes are at the top of their body)

Do we have any indication as to how far away they all will be standing? or is that up to us to decide the optimum for this problem? And along those lines, do they have to be standing together? or can they all be standing at different distances from the mirror?

I would have to say the second word is "power"

My second, like an extension, -Extension cord supplies power

or a mechanical device. -Mechanical could mean "powered"

Battle focused attention, -Many battles are fought for power

or Authorities long vice. -Authorities often get greedy with power

And the first one kind of makes me think of "Will"...which means that the answer would be "will power" which also fits the first 2 lines of the third stanza

Absolutely correct - loved that show as a kid!

Please do explain you methods!

I'd like to provoke an answer. Ask a Question (or a Q) = Q

In the late 30's, early 40's these water bound vessels were considered dangerous, WWII saw the use of German U-Boats = U

especially when stuck in the eye of a hurricane. "EYE" of a hurricane = I

there certainly was help along the way, help = aid

Mr. Freud would have been upset he was left out of this case. Freud came up with the concept of the superego, ego, and the ID...so, AID - ID = A

The time now is 3 o'clock. The hands on the clock at 3:00 = L

A discreet location for personal messages I was thinking an E-Mail address...which has @ in it = A

mountains the shape of mountains = "M"

finish the story the end of a story = ending

with out the ding of the gong ENDING - DING = "en" = N

So we get the letters QUIALAMN...BUT the riddle says "Before the storm..." so that would move the A before the I...and it also says "After mountains"...so move the A after the M...and you get:

QUAILMAN which is one of Doug's alter egos.

Excellent riddle...I really enjoy these that you write. Very well done.

I'm going to say your name is Doug...and your pet is Porkchop...your best friend is Skeeter...am I right? If so, I can explain...but I'll let people work on it more...