The worlds first riddle!

Oh boy, how I hate getting dressed in the dark and having to find my way home!

Mark: CHESS "My logic was correct."

Fairy Nuff just a typo, I have made one or two myself.


Mark:

TICKETS
Although implied, but not explicitly stated, the tickets must have been consecutively numbered. Otherwise, there are multiple solutions for each of the four response combinations.

The responses were both negative. The ticket numbers are 1299 and 1300.

Neat answer!

What Molly realized was that the tickets were consecutively numbered. If the tickets were numbered abcd and abc(d+1) and my answer to Spike's question had been yes, the only conclusion Molly could have reached would have been that a+b+c+d=12, and regardless of my answer to Waldo's question there would not have been a unique solution.

So my answer to Spike's question must have been no, and it follows that the tickets could not have been numbered in this manner. If the numbers were abc9 and ab(c+1)0 we'd have 2a+2b+2c+10=25, and so 2(a+b+c)=15, which is impossible. If the numbers were ab99 and a(b+1)00 we'd have 2a+2b+19=25 or a+b=3, leading to the four possibilities 0399 and 0400, 1299 and 1300, 2199 and 2200, 3099 and 3100.

Of these, three of them would have had me answer "yes" to Waldo's question, and only the pair 1299 and 1300 would have had me answer "no". It follows that these were my ticket numbers as Mark correctly deduced.




When Gus turned x years old he noticed to his surprise that between them; x^2 and x^3 included all the digits from 0 to 9, with none repeated.

How old was he


Yep! That's it, no more no less.

[size=8]GUS
69[/size]

Mark:

GUS
69



How about a round of applause for Mark, because Gus was 69; 69^2 = 4761 and 69^3 = 328509. The number of ages to check can be reduced by noticing that if x is Mortimer's age, x^2 must have 4 digits and x^3 must have 6, and so we have 10^5 ≤ x^3 < 10^6, giving that 47 ≤ x < 100.
Furthermore, x can't end in a 0, 1, 5, or 6, as then x^2 and x^3 would have the same last digit. The number of cases to check can be reduced still further, but there are only 33 cases left so there isn't much of a point.



When Mark (no relation) recently did a conversion of a positive integral Celsius temperature c = 275 to its Fahrenheit equivalent f (which turned out to be 527), he noticed to his amazement that he could have simply moved the last digit of c to the front to obtain f.

Doing some intense calculations he failed to discover the next largest such example.

Does one exist, and if so, what is it

[size=8]To avoid a non-integer conversion, the last digit must be zero or five. Moving a zero to the front will reduce the number. Therefore, the last digit must be 5.

Call the solution 10x+5

Solve:
(10x + 5) * 9/5 + 32 = x + 5 * 10^n
to get:
x = (5 * 10^n - 41) / 17

There is a solution when (5 * 10^n - 41) mod 17 = 0.

There are an infinite number of solutions of the form:

[(5 * 10^n - 410) / 17] + 5

where n = 3, 19, 35, ..., 16m+3, ...

The number you seek is 2,941,176,470,588,235,275[/size]

Mark:

To avoid a non-integer conversion, the last digit must be zero or five. Moving a zero to the front will reduce the number. Therefore, the last digit must be 5.

Call the solution 10x+5

Solve:
(10x + 5) * 9/5 + 32 = x + 5 * 10^n
to get:
x = (5 * 10^n - 41) / 17

There is a solution when (5 * 10^n - 41) mod 17 = 0.

There are an infinite number of solutions of the form:

[(5 * 10^n - 410) / 17] + 5

where n = 3, 19, 35, ..., 16m+3, ...

The number you seek is 2,941,176,470,588,235,275




Credit where credit is due, that is one heck of an answer!

Let c = x_{n}*10^{n-1} + ... + x_{1}*10^1 + x_{0} with x_{n} > 0, then f = x_{0}*10^{n-1} + (c-x_{0})/10. We also have that f = (9/5)*c + 32. Notice that in order for f to be integral c must be divisible by 5; this implies that x_0=5 since it cannot equal 0 (since as a number f>c). Our equation then becomes (9/5)*c + 32 = 5*10^{n-1} + (c-5)/10 implies c = 5*(10^n - 65)/17.

Now it turns out that 10 is a primitive root modulo 17 (don't worry dear reader about what this means), and so it follows that c is integral if and only if n is of the form 16*m + 3. When m=0 we get c=275; when m=1 we get the next highest such temperature, which is 5*(10^{19}-65)/17 = 2941176470588235275.

(Not a lot of people know that. However, thanks to Mark if the subject comes up in polite conversation, please feel free to regale them with the answer)




Billy Bob recently had another birthday. When someone mentioned (I am not naming names) that he was getting up there in years, he replied that he was actually quite young. Indeed, he pointed out, he is the youngest age such that the sum of the divisors of his age, not including the age itself, exceeded his age, yet the sum of no subset of these divisors equaled his age.

How old had Billy Bob just turned

I think it's

ע

[size=8]BILLY BOB
70[/size]

Adrian:

I think it's

ע

Well buddy, I think you are €


Mark:

BILLY BOB
70


Yes! That's more like it, I can just about understand that.

Clearly this number cannot be prime, nor can it be a product of two primes. To see this latter case, let these primes be p and q; if p=q the fact that the sum of factors cannot exceed the number is easy to see. If p and q are distinct, with p>q, we have that p(q-1) is greater than or equal to p ==> pq is greater than or equal to p+p is greater than or equal to p+q+1 and the same conclusion follows.

Looking at the first few examples we see that there does exist such a number with three distinct prime factors, 70 = 2*5*7, since the sum of its factors not including itself is 1+2+5+7+10+14+35=74, but no subset of these factors sums to 70. We still need to check those numbers smaller than 70 that have 3 or more prime factors (not necessarily distinct), but that's easily done.

Numbers that have these properties are known as weird

[size=8]BORIS
618033988750

After failing to spot a pattern, I wrote a program to calculate the first 100 values. I noticed that f(n)/n seemed to be converging. Then, I recognized the beginning of the famous golden ratio.[/size]

Mark:

BORIS
618033988750

After failing to spot a pattern, I wrote a program to calculate the first 100 values. I noticed that f(n)/n seemed to be converging. Then, I recognized the beginning of the famous golden ratio.


Wow! That answer goes straight into the Hall of Fame

Clearly f(n) is always less than n, and checking the first few values by hand it appears to be increasing, with f(n+1) equaling either f(n) or f(n)+1. A reasonable guess is that f(n) is approximately c*n, for some constant c. To find what c is if this is the case, consider f(n)/n = 1 - f(f(n-1)))/n. In the limit as n goes to infinity, the left side becomes c, and the right side becomes 1 - c^2. Since one must equal the other c=1-c^2, and rejecting the negative root we have that c is (-1+sqrt(5))/2. Examining the values of c*n rapidly leads to the conclusion that f(n)= [c*(n+1)], where [x] is the smallest integer less than or equal to x.

To prove this, note that if x>0 is not an integer, then [-x] = -[x] - 1. Now n - [c*[c*n] + c] = n + [-c*[c*n] - c] + 1 = n + 1 + [ -c^2*n - c*e - c ], where e = c*n - [c*n]. Since c^2 = 1-c, this equals n + 1 + [-n - c*n - c*e - c] = 1 + [c*n - c*e - c] = [c*n + c] + [f - c*e - 2c + 1], where f = c*n + c - [c*n + c]. Now if e+c<1, f=e+c; if e+c is greater than or equal to 1, f=e+c-1. It follows easily from this that [f - c*e - 2c + 1] is always 0. Since f(0) = 1 = [c*0 + c], the truth of our hypothesis follows, and it also follows that f(1000000000000) = 618033988750.



I was down at the gym yesterday looking to loose 70 pounds to bring me back to a fighting fit 260, when the electrics on the cycle machine gave up the unequal struggle and burnt out. What I want to know is:

If my bicycle wheel is 30 inches in diameter, and the wheel turns at a constant rate of 3 revolutions per second, what is the linear speed in miles per hour of a point on the tire

Please express any answers in miles per hour to save me massive foreign calculations.
I joke not; a Hong Kong foot is exactly 0.371475 metres. You wanna do the conversion???

a point on the tire will move 90pi inches/second or 16miles/hr......but since the bike is stationary, the tire is not going anywhere xcept in a circle.

Thoh:

A point on the tire will move 90pi inches/second or 16miles/hr ......but since the bike is stationary, the tire is not going anywhere xcept in a circle.




The same goes for me because, I had been exercising for several seconds before I fell to the floor exhausted. On looking up I got to thinking:

The mean distance of Earth from the Sun is 9.29 x 10^7 miles. Assuming that the orbit of earth around the sun is circular, and that one revolution takes 365 days.

What the heck is the linear speed of Earth



Many moons ago I asked:

"Using each of digits 1, 2, 3 once, and only basic arithmetic operations, can you write a formula for 19"


Perhaps not surprisingly there were no takers. I have received a number of missives doubting there is an answer. Therefore, to answer my critics:

19 = sqrt(((3!)!)/2 + 1)

I don't know that this amusing formula has any deeper significance. I do have a heuristic explanation that suggests that there are more such formulas than one might guess at first.

The number ((3!)!)/2=360 clearly has many factors, and so can be written as A•B in many ways. Often A,B are of the same parity, so can be written as x+y and x-y for some integers x,y, whence 360=x^2-y^2.
Since 2y=A-B, we know that y will be small if A and B are close to each other. In particular, 360=18•20 yields 360=19^2-1^2, which is equivalent to our formula above.

This kind of thing happens a few more times; for instance, 71=sqrt(7!+1), and we can use three 9's to write 603 as the square root of 9!+9sqrt(9).

Earth
2pi(9.29E7) / (365x24)= 66633mi/hr or 18.5mi/sec

ok...along the same lines. what is the linear speed of the solar system (in km/s) as it rotates around the galactic center 28 thousand light years away? Each revolution takes 220 million years.

The Earth speed is approximately 29.76 km/s. Its speed of flight in the Solar System would be of 29.76x sqrt2 = 42,1 km. /s.

Whoops, almost forgot.


Thoh:

Earth
2pi(9.29E7) / (365x24)= 66633mi/hr or 18.5mi/sec



Or proving there is more than one way to skin a dragon:
2<pi>(9.29 * 10^7)/365 days = 1599199.767.. mi/day • 1 day/24 hr = 66633.323... mi/hr .



(No dragons were harmed in concocting this question)

Wizard Thoh went to fight a 3-headed, 3-tailed dragon. He has a magic sword that can in one stroke, chop off either; one head, two heads, one tail, or two tails.

This dragon is of a type related to the hydra; if one head is chopped off, a new head grows. In place of one tail, two new tails grow; in place of two tails, one new head grows; if two heads are chopped off, nothing grows.

What is the smallest number of strokes required to chop off all the dragon's heads and tails, thus killing it



BTW For those of you concerned at Mark's absence, may I assure you he should post bail today after a whip round down at the local cat house.

not the answer i had in mind.....where does the sqrt(2) come from?
...pretend the solar system is just a blob w/o any subdivisions and its flying thru the galaxy

DRAGON
I get 9. Minimizing the total number of heads and tails during the battle, here's the chopping strategy. One dragon was harmed (killed) formulating this solution.

Speed II

[size=7]Approximately 400 km/s[/size]




As some of you might know, Mark collects cheese labels. Yes! I kid you not.

I'm thinking of a word that means someone who collects cheese labels.

What is it

solar system:

400 must be a VERY approximate estimate....

distance: 2pi(28000 lyr)(9.46E12 km/lyr)=1.664E18 km
time: (220E6 yr)(365 d/yr)(24 hr/d)(3600 s/hr)=6.938E15 s

speed = 1.664E18 km / 6.938E15 s = 240 km/s

[size=8]CHEESE LABELS
I should probably not be allowed to answer this question since it is about me. However, I can't resist.

Someone who collects cheese labels is a fromologist. However, what you probably don't realize is that I collect only Camembert cheese labels. Therefore, I am a tyrosemiophilist.[/size]

Try loves women. That makes him a philogynist. Turns out, he's particularly fond of wives (including those that aren't his own - possibly yours). What does that make him?

We've all heard of "leg men" and "boob men" (men who are attracted to the legs and boobs of women). Try is a butt man. Yep, he's attracted to women's cabooses. What does that make him?