Let's play with 2013 - a numbers game

fun2come's Avatar
2013 - a numbers game
May be some of you know this .... Well here is my version...
You can ONLY use each number 0, 1, 2, 3 once and can use any mathematical formula.
Let's see if we can get to 2013. We'll see if we get stuck somewhere...
I start with the easy and obvious ones to us all started:
0
1
2
3
1+3 = 4
2+3 = 5
1+2+3 = 6
10-3 = 7
10-2 = 8
12-3 = 9
10
13-2 = 11
12
13
(10-3)*2 = 14
12+3 = 15
20-3-1 = 16
20-3 = 17
21-3 = 18
20-1 = 19
20
21
23-1 = 22
23
23+1 = 24
square of (2+3) = 25
square of (2+3) + 1 = 26 (my birthday)
:
ok, who wants to continue to
:
2013

Put them in order please !!!

And Happy New YEAR Y'all !!!
Heh heh - Heh heh if I wanted to do math , I'd still be in school ! Heh heh Cool !
fun2come's Avatar
I thought you were getting schooled every day, lol

30-2-1 = 27

give it a try, each post needs to have a new number, you can do it !!!
The next one is an easy one, even for you....
9 x 4 - 8 = 28

Is this what you're looking for ?
fun2come's Avatar
No, and you know it... just stay out of it, will you, I clearly see your qualities lie somewhere else...

30-2 = 28
ozmosys's Avatar
30 - 1 = 29
3 * 10 = 30
30 + 1 = 31
30 + 2 = 32
30 + 2 + 1 = 33
30 + the square of 2 = 34
30 + the square of 2 + 1 = 35
square of 3*2 =36
square of 3*2 + 1 = 37
30 + 2 cubed = 38
30 + 2 cubed + 1 = 39
2 squared * 10 = 40
fun2come's Avatar
Great, now for those Damn 40 prime numbers.... took me a few minutes

summation (3 cubed) - the square of 2 =41
32+10 = 42 (I LUV that number)
summation (3 cubed) -2 =43
summation (3 cubed) -1 = 44
summation (3 cubed) = 45
summation (3 cubed) +1 = 46
summation (3 cubed) +2 = 47
square of (square of 2) * 3 = 48
SUM (3 cubed) + square of 2 = 49

on to the 50s any one ?
ozmosys's Avatar
I had this. Maybe you could clarify what you're doing with the summation. I'd expect that to be a sequence, but not sure where it begins or ends to give us 45.

2^5 (32) + 3^2 (9) = 41
where 2^5 means 2 to the fifth
and 3^2 means 3 squared
32 + 10 = 42
2^4 (16) + 3^3 (27) = 43
2^4 (16) + 3^3 (27) + 1 = 44
30 + 2^4 (16) - 1 = 45
30 + 2^4 (16) = 46
30 + 2^4 (16) + 1 = 47
2^4 (16) * 3 = 48
2^4 (16) * 3 + 1 = 49
(3+2) * 10 = 50
2^6 (64) -13 = 51
fun2come's Avatar
summation may not be the correct mathematical term, been too long ...
What I mean is (1+2+3+4+5+6+7+8+9) = 45
May be sigma(9)?

I would rule that 2^5 or similar is not allowed BUT 2 to the fifth is ok, LOL.
Just don't use different numbers (we only want to see 0, 1, 2, 3), so as we already had them correct up to 50, we only need to correct your last one:

2 to the sixth (64) - 13 = 51
13 * square of 2 = 52
13 * 2^2 + 1 = 53
3^3 * 2 = 54
13 * 2^2 + 3 = 55
2^3 * (2^2 + 3) = 56
3 * (2^2 + 2^2 + 3) = 57
ozmosys's Avatar
13 * 2^2 + 1 = 53
3^3 * 2 = 54
13 * 2^2 + 3 = 55
2^3 * (2^2 + 3) = 56
3 * (2^2 + 2^2 + 3) = 57 Originally Posted by Smokin Joe
Unless I've misunderstood the rules, you've already used the 1 in 13, so you can't use it again in + 1. Likewise, if you start with 13, you can't use the 3 again in + 3. Also some duplicate 2's and 3's in the last two equations.

But 3 cubed times 2 looks solid, and it's nice to have another participant!
I see. You're right. Back to the drawing board.

By the way, the summation notation is to add integers from 1 to 100 in this case.
ozmosys's Avatar
summation may not be the correct mathematical term, been too long ...
What I mean is (1+2+3+4+5+6+7+8+9) = 45
May be sigma(9)?

I would rule that 2^5 or similar is not allowed BUT 2 to the fifth is ok, LOL.
Just don't use different numbers (we only want to see 0, 1, 2, 3), so as we already had them correct up to 50, we only need to correct your last one:

2 to the sixth (64) - 13 = 51
13 * square of 2 = 52 Originally Posted by fun2come
Ok, here's where I'm confused. 3 cubed is 27, not 9, so did you mean summation of 3 squared? And I can live with spelling out powers to keep the equations cleaner looking.
ozmosys's Avatar
13 * 2^2 + 1 = 53
3^3 * 2 = 54
13 * 2^2 + 3 = 55
2^3 * (2^2 + 3) = 56
3 * (2^2 + 2^2 + 3) = 57 Originally Posted by Smokin Joe
How about this, using 54 as a baseline:
3 cubed * 2 - 1 = 53
3 cubed * 2 = 54
3 cubed * 2 +1 = 55

The next one is harder.
How about this?

summation (10) - 2 = 53
3^3 * 2 = 54
summation (10) = 55
summation (10) + 3 - 2 = 56
summation (10) + 2 = 57
summation (10) + 3 = 58