The Maths Topic

I found this one…

uppsidedown L = gamma
gamma(n+1)=n!

read more on en.wikipedia.org/wiki/Gamma_function

I definitely think too hard into stuff. I seriously LOL’d when I saw your answer [com]and what, no cookie? [/com]

Quite right, but we were talking more about a formula to calculate the munber of divisors of any given number

Ok so, 2 problems are still up, feel free to add a 3rd one to the mix, or I will post one tonight.

2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?

Not going to work on this until I’m actually awake, but I’d say it’s somewhere around 5,250. Close?

And fine, here’s your cookie.

Moar liek 4000+.
I’ll give a little hint: 12= 3*4, and divisibility by 3 and 4 are easily studied.

EDIT: While we’re at it, https://en.wikipedia.org/wiki/Mathematical_joke

Man, I totally forgot to add a 3rd problem back.
1b) Geometry! ^^
If the circle has diameter 2, how much’s the coloured area?

If you give some love to math, it will give it back to you, squared

I figured that one out with a little help from my math teacher, i50.tinypic.com/1jvlte.jpg
So the area is… wolframalpha.com/input/?i=hexagon+area
s is the radius, so of the the diameter is 2, s=1
Area=2.59808 cm^2 (or some other unit)

Congratulations! I’ll post the solution for the 2b:

[spoiler]
First, a number is divisible by 3 if the sum of its digits is. EX. 297 is divisible because 2+9+7=18, 4568 is not becuase 4+5+6+8=23.
Secondly, a number is divisible by 4 is the last 2 digits are. IE. 12736 is divisible because 36 is.

Therefore, we can study the problem digit by digit.
So, let’s study the last 2 digits. The possibilities are 00, 04, 08, 12, … 88, 92, 96, and that makes 25 possibilities, without the 00, 04, 08, 20, 40, 60, 80, so that makes 18.
All other digits can range from 1 to 9, so that would make 9*9=81 possibilities for the 3rd and the 4th, then the 5th must be left for the divisibility by 3. You see, having only numbers from 1 to 9 to choose, it is much easier, as they are grouped 3 by 3 for the disision rest by 3: 3,6,9; 1,4,7; 2,5,8 are 3 groups of rest 0, 1, 2. This way, whatever the rest of the division by 3 of the sum of the other digits will be, we will always have the possibility to choose any of 3 digits IE. X3712 has rest 1 so far (like 3+7+1+2=13 has), so we can use either 2, 5 or 8; X8328 has rest 0 (like 8+3+2+8=21), so we can use 3, 6 or 9.

So, all numbers responding to all criteria are 18 * 81 * 3 = 4374 numbers.[/spoiler]

Feel free to add any problems, I won’t solve them and I probably won’t put up any until mid January.

How many consecutive zeroes are at the end of 100! (factorial) ?

You have two string-like fuses which burn in exactly 1 minute. However, the fuses do not burn at a constant rate and may burn quickly at first, then slowly, then quickly again, etc. You have a lighter. How do you measure 45 seconds without using a watch?

Can the mean of two consecutive prime numbers ever be prime?

[spoiler]24 zeros

I figure, ending in 0 gives you a 0 (so 10 of them)
Ending in 5 gives a zero as soon as it’s multipled to an even (another 10)
But, 25 was special, has 2 factors of 5, so 25, 50, 75 and 100 get an extra zero when multiplied to an even, (so another 4)

So, 24 zeros![/spoiler]

well, do decimal numbers count as prime?

(2+3)/2= 2.5

Well, i guess then yeah

Only integer numbers can be prime. It’s a matter of divisibility, and divisibility loses its meaning if you can divide anything by anything without any reminder.
Besides, (solution)[spoiler] if two prime numbers are consecutive, that means there’s none inbetween, and that’s where the mean will always lie, in the middle so it’s pretty much impossible.[/spoiler]

Hey, TosxyChor, how do you do that “Click Here” stuff? Good solution, I like it, makes sense.

After hearing the prime number answer, I thought “outside the box” on this one.

[spoiler]Take one fuse, and light both ends at the same time. It’ll burn up twice as fast, 30 seconds.

Take the other fuse, cut it in half, then light both ends of both halves when the first fuse is entirely burned up. It’ll burn 4 times faster, 15 seconds.

When both pieces of the second fuse are gone, 45 seconds will have gone by.[/spoiler]

Edit: This solution is wrong! But, I think I got it a few posts below this one.

Okay, so here’s a new one:

A death ray is mounted on top of a platform which rotates at a constant speed of one revolution per hour. Every 275 seconds the death ray is fired. Find the smallest value for the total number of times the death ray has been fired when the same spot is hit for a second time.

You can’t do that, because you don’t know where “half the duration” is, the rope is irregular and will burn at different rates.
However, it’s quite the right road, now how to make the second rope last 15 seconds after the first one has burned…

Plus, to make spoilers, just use the [spoiler] tag.

[spoiler]thing to hide[/spoiler]

Cool, thanks for the spoiler tag!

You’re right, my solution didn’t work!

When I said “cut it in half” I just meant, cut it into two pieces. If one piece is a fast burning piece, it’ll be gone quickly. If the other is slow, it’ll take longer. If it’s vice versa, that’s fine. All that matters is that it’ll burn itself up entirely in 15 seconds, you don’t care which piece burns up first, you just care that both pieces are gone. When they’re both gone, 15 seconds have passed.

But, if I think about a constant burning fuse and take off just the tip and cut all four ends, it doesn’t finish in a quarter of the time, you’d need to cut it exactly in half.

But, I think I see the correct solution now:

I had a math moonquest dream once! Click here to see it.

Actually, that’s part of a solution to one of the problems posted here not to long ago.

Yay math! I’ve got a problem
Many numbers can be expressed as the sum of two palindromes. 389 for instance is 383 + 6 as well as 323 + 66.

What three-digit number cannot be expressed as the sum of one or two palindromes? (in other words, requires at least three palindromes to sum to it)

(For the sake of this problem, palindromes with leading zeroes [like 030 and 0550] are not allowed, and neither are negative numbers or non-integers. Single digits are allowed.)

Your spoiler solution is correct, the cutting-idea is wrong since you will need to cut it exactly at the right spot (which you do’nt know). Say, if you cut in in half and one end burns quickly, in 1 sec, so the other end burns in 29 sec and you wait till both ends are burned up, you are waiting 29 seconds instead of 15.