haven't asked for assistance before from you mob
but i think there may be suitably equipped mathematicians out there that will consider this some sort of breeze..
using only the numerals 2008, and in that order !! , create a set of mathematical equations using those numerals to develop the set of numbers from 1 - 20 ( or even further)
as much of an old fart as i am, i succeeded in getting all but 15, 17, 19
by way of example and quite possibly the most elegant
log (2) (008) = 3 (unfortunately not mine - one of the kids in the class)
and others
2 ** (0) + 0 + 8 = 10
(2! + 0!) * 0! + 8 = 11
2! + 0! + 0! + 8 = 12
those bloody primes ... ! ( and odd numbers... )
anyway, glad for any assistance, ciao 4 now
PS _ any maths symbol at all, but no extra digits, and in order of 2008
20 - (0!+cube root 8) = 17
20 + 0! - cube root 8 = 19
(2+0!)! + 0! + 8 = 15
a few further ones...
20 - (0!-cube root 8) = 21
20 + 0 + cube root 8 = 22
20 + 0! + cube root 8 = 23
(2+0+0!)x8 = 24
(2+0) x (0!/8% ) = 25
we can probably get quite far if you're allowing every mathematical symbol... I'll leave it to other users to carry on past 25.
Originally posted by doodinthemoodthe fact that you have multiple ones twos and zeroes available (via 0!, cube root 8, 2, and 0) it makes it much more reasonable to get the "in between" numbers than something like the "four fours" problem which is similar. the tough part it seems with 2008 is finding an initial number from which to go +-1 or +-2. nice job dood
20 - (0!+cube root 8) = 17
20 + 0! - cube root 8 = 19
(2+0!)! + 0! + 8 = 15
a few further ones...
20 - (0!-cube root 8) = 21
20 + 0 + cube root 8 = 22
20 + 0! + cube root 8 = 23
(2+0+0!)x8 = 24
(2+0) x (0!/8% ) = 25
we can probably get quite far if you're allowing every mathematical symbol... I'll leave it to other users to carry on past 25.
for 25 why not use: 200/8 π
wow - and so fast
we did go as far as 21,24,25,27,28 & 29
but with so many gaps i was just trying to get the initial 20, and did manage to get the 200/8 as 25, which is one of the neater ones in the group
but am i being too fussy if the numeral 3 in the cube root function, (distinguishing it from 2, 4 ...,) has introduced an extra digit ?
i appreciate it is a symbol in its own right now, perhaps i'm just too fussy.
i did try the integral from 2 - 0 of (0+8), but of course i have discounted the introduced 'c' to zero, and (2! +0!)*(-0!+8) = 21,
i am still surprised at how many there were, sqrt(2+0)*sqrt(0+8)=4 etc
thanks again, i needed that burst to get past my first hurdle..
Cube-roots are allowed? How about that extra three you need to write the cube roots, or doesn't that one count as "extra" number?
Numbers obtained above: 3 10, 11, 12, 15, 17, 19
Once you have 25, you can simply take the root to obtain 5.
- (2 + 0!)! - 0! + 8 = 1
- ( 2 + 0! + 0)! + 8 = 2
- ( 2 + 0!) ! + 0! + 8 = 3
- ( 2 + 0! + 0! ) + 8 = 4
- ( 2 + 0! + 0 ) + 8 = 5
- ( 2 + 0 + 0 ) + 8 = 6
- ( 2 - 0! + 0 ) + 8 = 7
2 - 0! - 0! + 8 = 8
2 - 0! + 0 + 8 = 9
(2 + 0!)! - 0! + 8 = 13
( 2 + 0!)! + 0 + 8 = 14
(2 + 0! + 0! )! - 8 = 16
(2 + 0) * (0! + 8) = 18
2 * (0! + 0! + 8) = 20
That covers 1 - 20.
I couldn't find anything for 26 so fast. You can get huge numbers by repeating the ! of course.
(2 + 0! ) * ( 0! + 8 ) = 27
Originally posted by doodinthemoodIncorrect, the function would be the third root of the base (in this case 8) or 8 to the power of one-third. In both cases, an extra digit is required in order to complete the expression, otherwise it is meaningless. Square-root, although technically has a two, can also be written without any numbers.
As far as I considered it, cube root is a function, not a number, because 3 isn't involved anywhere in the process. If it isn't allowed you should also not allow percentages.
Percentage is a tricky one, but has a widely accepted symbol, however that symbol is also used for the modulus operation in some situations. In the percentage form, the o's in the percentage represent the 0's from the 100, so dood might have a point there.
But if we are going to be this picky anyway, the negative sign cannot be used because it creates a new number. -(2) is actually -1 x 2 so we are introducing a new number -1. -2 is not the same as -(2) if we are using specific digits.
Originally posted by Gastelbut to be that picky would also eliminate both subtraction and division since a - b is actually shorthand for a + (-b) whilst a/b is shorthand for a*(1/b), it would also rule out exponentiation for a^b introduces b instances of a whilst factorial for any a > 0 introduces a '1' and we don't have any of themππ
But if we are going to be this picky anyway, the negative sign cannot be used because it creates a new number. -(2) is actually -1 x 2 so we are introducing a new number -1. -2 is not the same as -(2) if we are using specific digits.[/b]