(25 + 1) * (10 + 7) + 5 = 447
is the best I can do
448 = (25 + 7) * (10 + 5 - 1)
Next up: 2, 4, 7, 9, 10, 25. Target is 881
9 x 4 x 25 - (10 + 7 + 2)
1, 1, 3, 3, 7, 7. Target 500.
Maybe I’m confused, but isn’t this the best possible?
(7 × 7 × 3 × 3) + 1 + 1 = 443
Nope. It’s solvable.
No. Max is (9 + 1 + 1) x 7 x 3 x 3 = 77 x 9 = 693
This is easier than some of the earlier ones.
((3 x 3) + 1) x ((7 x 7) + 1) = 500
Next up!
The target is 951 and the numbers are 2 2 5 6 25 75
Technically, I think the max is 7 x 7 x (3+1) x (3+1) = 784. But anyway.
((2 x 6) x (75 + 2)) + 25 = 949. Alternatively, ((5 + 6 + 2) x 75) - 25 = 950. But I can’t get 951.
Does anyone know an algorithm for finding the maximum total of 6 numbers using just multiplication and addition?
I should note that I do not know if this one is solvable. I gave it a quick try to ensure that the solution wasn’t trivial.
I can’t get 951 either. (75 x 25 + 5)/2 + (6 x 2) = 952, for the sake of completeness.
Pretty simple (assuming the set we have to work with) - first add any 1s to the lowest non-1 numbers, then multiply all the numbers by each other.
Come to think, actually (1+1) x 3 x 3 x 7 x 7 = 882, so in fact if there are 2 1’s you should add them to make a 2 then multiply out; if there’s only one 1 you should add it to the otherwise-lowest number and then multiply.
((3*2) + 2 + 1 +1)^3 = 1000.
Wait, that was cheating.
If the rules allow three or more 1’s, then I think the rule would be to add them in groups of three until you have two, three or four left. Then:
With two left, group as (1+1+1)…(1+1+1)(1+1)…
With three left, group as (1+1+1)…(1+1+1)(1+1+1)…
With four left, group as (1+1+1)…(1+1+1)(1+n)…
(where n is the smallest number apart from 1)
:smack::smack: I’m not sure where I got the 9 from.
Well, if we allow exponents I can do you one better:
((2^3)+2)^3 -1 = 999, our target
edit:
or
((3^2)+1)^3 -1 = 999
It appears to me that Dead Cat has come in first with a close answer to the last puzzle, and it also appears to me that I am arithmetically inept (see posts 44, 46, 47).
If we allowed exponents, who knows what heights we could reach!
True, but I am wondering, if you use the rules of the OP, how many of the possible puzzles disallow an “exact” answer. Quartz’s last puzzle
The target is 951 and the numbers are 2 2 5 6 25 75
might be one of them.
951 = ((75 + 6 - 5)*25 +2) /2
Target: 901
Numbers: 2,2,4,25,50,100
4*50 22 + 100 = 900. Don’t see a way to get that sodding 1.
If I win anyway :
Target : 557
Numbers : 1,3,6,9,10,100
I think you probably will, based on previous posts it’s probably solvable, but I can’t do it.
(100 x (6 - 1)) + ((10 + 9) x 3) = 557.
New target: 484
Numbers: 1, 2, 5, 7, 9, 9.
Don’t know if this is solvable (or trivial).
Not trivial anyway 9 x 9 x (5 + 1) - 2 = 484
5, 5, 10, 10, 25, 50
target 373