I hope you're good at Math!

Yes, that is what was meant. Laura's answer is correct. Perhaps you should read the replies before making your own.

i'd appreciate it if there was less cheek!!
zex wot the hell did i ever do to you??
u guys r far to sensitive!! lol

Another solution
upper integer value ( square root (22) )

Alon

Here's another : using exactly five 5 digits, plus any number of math. symbols, excluding symbols representing numbers and variables, can you get all numbers from 1 to 10?
How about 10-20?

And for the most fun, try with five 6 digits.

I say no number symbols because one other solution to 2's problem is
2+2+x/x ; x is a natural number.
As a matter of fact, you can get 1 in many ways in math..

Relative

.2 Divided by .2 squared = 5 I don't know where to find the symbols on my keyboard.

[(1+2)-3]divide by 4=0
1*2+3-4=1
1+2+3-4=2
1+2-3+4=4
(1+2)*3-4=5
(1-2)+3+4=6
1+2+3+4=10

Uh, i meant you can use digit 5 , five of them, like

5*5+5+5-5 = 30

Relative

(5+5+5)^(5-5) = 1
((5 + 5 )* 5/5)/ 5 = 2
((5+5)/5)+5/5 = 3
(5 - 5/5)*5/5 = 4
(5*5*5)/(5*5) = 5
((5+5)/(5+5)) + 5 = 6
(sq rt5* sq rt5 )+ ((5+5)/5) = 7
((5+5+5)/5)+5 = 8
(5+5)-((sq rt5*sq rt5)/5) = 9
((sq rt5*sq rt5)/5)*(5+5) = 10

To do it with five 6's, the first three can be arrived at merely by replacing each 5 in this 'fives' solution with a 6.
6-6/6-6/6 = 4
6*6/6 - 6/6 = 5
(6+6)/(6+6) *6 =6
((6/6) *(6/6)) + 6 = 7
(6/6) +(6/6)+6 = 8
((6+6+6)/6)+6 = 9
(6+6)- ((6+6)/6) = 10

Iacomus : that is correct.
Also : 55/55 + 5 =6


There is another, more elegant riddle: do the same using digit 4, and you can use 4 of them.

I don't see what is wrong with your post or why should I disregard it?

Relative

Your original question asked for 'one to ten' and my original post only gave 'one to five'. I have now edited it - after quickly finding the answers I had not included first time - and I think it now is a better response to the question as asked.

1 to 10 using four 4's:

(4/4) * (4/4) =1
(4/4) + (4/4) = 2
(sq rt4*sq rt4)-(4/4) = 3
(sq rt4*sq rt4)*(4/4) = 4
(sq rt4*sq rt4)+(4/4) = 5
((4+4)/4) +4 = 6
(4+4)-(4/4) = 7
(4+4)*(4/4) = 8
(4+4)+(4/4) = 9
(4! /4)+(sq rt4 * sq rt 4) = 10

I didn't want to use numbers such as '44' because I felt that this was just a shade dishonest in that there is a hidden '4*10 + 4' in there and it is no longer using '4'. The same would be true of using a decimal point as this is a hidden 'divide by ten'. IMHO of course.

Very good, Iacomus!
The original author uses

0 = 44-44
1 = 44/44
2 = 4/4+4/4
3=(4+4+4)/4
4=4+(4-4)/4
5=(4*4+4)/4
6=(4+4)/4+4
7=44/4-4
8=4+4+4-4
9=4+4+4/4
10=(44-4)/4

It is possible go get even more numbers

Relative

I'm sure that you are right and that many more numbers could be arrived at. However, this type of problem is a devourer of time and I will save it for the nights when I can't sleep, or when I am stuck in an airport without reading material (or in solitary confinement maybe! )

Uh, I hope not!

A tough problem, an oldie
I am going to repeat an old problem that first appeared in a weekly newspaper, where riddles were posted by no other than the great M. Gardner. A prize of 100$ was later promised for a solution.

"Five men find themselves shipwrecked on an island, with
nothing edible in sight but coconuts, plenty of these, and
a monkey . They agree to split the coconuts into five equal
integer lots, any remainder going to the monkey.
Man 1 suddenly feels hungry in the middle of the night
and decides to take his share of coconuts at that very
moment. He finds the remainder to be one after division
by five, so he gives this remaining coconut to the monkey
and takes his fifth of the rest, lumping the coconuts that
remain back together. A while later, Man 2 wakes up
hungry too, and does exactly the same - takes a fifth of the
coconuts, gives the monkey the remainder, which is again
one, and leaves the rest behind. So do men 3, 4, and 5.
In the morning they all get up, and no one mentions anything
about his coconut-affair the previous night. So they share
the remaining lot in five equal parts finding, once again, a
remainder of one left for the monkey. Find the initial number
of coconuts.
"

The problem is HARD, so don't think there's something wrong with you if you can't solve it. The first solution was hard labour. Allegedly, it was Paul Dirac himself who provided the clever solution which is in fact very elegant and simple. Also allegedly, this had to do with his then current work on anti-electrons. Now there's a clue in there..

Relative

Then it is now established for certain that I am no Dirac but I am one of the monsters (I was hoping I'd be somewhere in the middle :-( )

So far I have this monstrosity.

Where x is the number they started with and y is how many coconuts each man received in the final share out:

4^5x - 5^6y = A

And A is where the monstrosity creeps in, because:

A = 5 + 4^1 * 5 + 4^2 * 5 + 4^3 * 5 + 4^4 * 5 + 4^5 = 11529
[To save writing in the parenthesis - the '+' and '*' operations are carried out in sequence starting from the left]

If it wasn't for that darn monkey getting one coconut each time this would be quite easy! But this has just ceased to be relaxation. This is WORKING!!

Hm, I spelled it out on paper and it seems to me that my monster is the same


All we have to do now is run this on a decent excel , khm..uh I mean paper sheet.

But, you still have a chance at Dirac, while I don't !

Relative

After walking away from this problem, and with the help of a few beers, I now suspect that I am doing this the hard way.

Clue: The monkey - I never did like that monkey anyway! - falls into the water and is eaten by a shark. As is the way of belief on that island he returns as four ghost coconuts, which snuggle with the 'real' coconuts in the original pile.

I will have another look at this after my usual weekend of riotous and ribald excesses (I wish!)

I could not wait a moment longer.

Let initially there be ?'x' coconuts in the pile.
Now, we proceed step by step as follows:

Number of coconuts before the First division = x
Number of coconuts removed in the First division = [{(x - 1)/5} + 1]
Remaining number of coconuts after the First division = [4(x - 1)/5] = A

Number of coconuts before the Second division = A
Number of coconuts removed in the Second division = [{(A - 1)/5} + 1]
Remaining number of coconuts after the Second division=[4(A - 1)/5] = B

Number of coconuts before the Third division = B
Number of coconuts removed in the Third division = [{(B - 1)/5} + 1]
Remaining number of coconuts after the Third division = [4(B - 1)/5] = C

Number of coconuts before the Fourth division = C
Number of coconuts removed in the Fourth division = [{(C - 1)/5} + 1]
Remaining number of coconuts after the Fourth division=[4(C - 1)/5] = D

Number of coconuts before the Fifth division = D
Number of coconuts removed in the Fifth division = [{(D - 1)/5} + 1]
Remaining number of coconuts after the Fifth division = [4(D - 1)/5] = E

Now, for the first part of the problem, the number of coconuts remaining (after the fifth man finished dealing with the pile) is one more than a multiple of 5, i.e., (a) 5y + 1 (say) whereas for the second part, i.e., (b) the number of coconuts remaining is an exact multiple of 5, i.e., 5y (say) where ?'y' is the number of coconuts each man receives in the final division.

On simplification (getting the final expression in terms of x and y), we get for the two questions (the two parts of the problem) the following equations:

1024x - 15625y = 11529 for (a) and
1024x - 15625y = 8404 for (b).

From these equations, we obtain: x (minimum) = 15621 for (a) and
x (minimum) = 3121 for (b).

Thus the number of coconuts left after each of the men finished dealing with the pile are:

15621, 12496, 9996, 7996, 6396, 5116 for (a) and
3121, 2496, 1996, 1596, 1276, 1020 for (b).

However I am still thinking about the ?'Rubber Band' problem.

I am sorry to have been a bit slow over last two days.

Iacomus : your thinking is very interesting!

Tryagain : I just wonder how you solved
1024x - 15625y=11529 ?
And congrats on the correct answer

However, there is still a competition for the generalized problem:

We have N men on the island, blah blah, and the monkey that gets one coconut on every iteration!