When genius failed
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I'm stuck with a problem. Either there's no solution or I haven't got it.
Problem
1 2 3 4 5 = 20
Insert + - x / between the numbers without changing the order of the numbers. Each sign must be used at least once and only once (no order assigned for use of the signs). Brackets can be used.
How can we get the equation to be = 20?
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I'm not very clever or intuitive so I'm going for a brute-force method of exhausting all possible arrangements.
First cut:
There can be 4x3x2x1 = 24 different permutations for the order of the 4 operators (+-x/).
A simple check would show that without using brackets, none of the 24 permutations will give a result equal to 20.
Hence, the key thing is the brackets, which, unfortunately, is so much trickier.
I don't know how to compute the total number of meaningful permutations for the order of the brackets. So I manually list them.
One bracket
1. (1 2) 3 4 5
2. (1 2 3) 4 5
3. (1 2 3 4) 5
4. 1 (2 3) 4 5
5. 1 (2 3 4) 5
6. 1 (2 3 4 5)
7. 1 2 (3 4) 5
8. 1 2 (3 4 5)
9. 1 2 3 (4 5)
Two brackets (exclusive)
10. (1 2)(3 4) 5
11. (1 2) 3 (4 5)
12. (1 2)(3 4 5)
13. (1 2 3)(4 5)
14. 1 (2 3)(4 5)
Two brackets (overlapping)
15. ((1 2) 3) 4 5
16. (1 (2 3)) 4 5)
17. ((1 2) 3 4) 5
18. ((1 2 3) 4) 5
19. (1 (2 3) 4) 5
20. (1 2 (3 4)) 5
21. (1 (2 3 4)) 5
22. 1 ((2 3) 4) 5
23. 1 (2 (3 4)) 5
24. 1 ((2 3) 4 5)
25. 1 ((2 3 4) 5)
26. 1 (2 (3 4) 5)
27. 1 (2 (3 4 5))
28. 1 (2 3 (4 5))
29. 1 2 ((3 4) 5)
30. 1 2 (3 (4 5))
Three brackets
31. (1 2)((3 4) 5)
32. (1 2)(3 (4 5))
33. ((1 2) 3)(4 5)
34. (1 (2 3))(4 5)
35. ((1 2)(3 4)) 5
36. 1 ((2 3)(4 5))
37. (((1 2) 3) 4) 5
38. 1 (((2 3) 4) 5)
39. 1 (2 (3 (4 5)))
40. (1 (2 (3 4))) 5
41. (1 ((2 3) 4)) 5
42. 1 ((2 (3 4)) 5)
43. 1 (2 ((3 4) 5))
44. ((1 (2 3)) 4) 5
That's all I can think of so far. If my universe of bracket permutations is complete, i.e. there can only be 24x45=1080 various ways we can arrange the 4 operators and brackets.
And so far, none of these 1080 permutations produced a result equal to 20.
There can only be two reasons:
1) There is no solution to this problem, i.e, can never achieve 20
2) My universe of bracket permutations is incomplete. Hence, here's where I need some help. Can anyone think of a meaningful bracket permutation for the 5 numbers with 4 operators which I haven't thought of? -
i see maths i..... *fainted*
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the one that says the signs - + x / must be used once and once only kills it..
are there any other operators we can use? sorta like INT(X) or ABS(X) or something like that? -
U said that bracket can be used.
and only once..wah.difficult sia -
uhhh.. this condition:
Each sign must be used at least once and only once (no order assigned for use of the signs). Brackets can be used.
so yours cannot stand...Originally posted by AMD2004:U said that bracket can be used.
so ( 1+2-3) + (4x5) = 20.
unless cannot use bracket. I kept it in order and i only use -,+ & x. -
Er... no... only the 4 common operators + - x / may be used, and only once.Originally posted by the Bear:the one that says the signs - + x / must be used once and once only kills it..
are there any other operators we can use? sorta like INT(X) or ABS(X) or something like that?
Everyone want to solve the main problem... how about my bracket permutation problem??? That's the key... -
i may start looking at a "cheat"..
sort of like (2)(3) = 2x3 = 6 -
That's not possible since you need an operator in between 2 numbers as each operator must be in use at least once.Originally posted by the Bear:i may start looking at a "cheat"..
sort of like (2)(3) = 2x3 = 6 -
i hate maths too.
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nah.. in math, the convention is that 2(3) = 6
besides, you forgot another "operator" -6 = 0 - 6 or (-1)(6) or (1)(-6)
math is fun
Originally posted by EricDraven:That's not possible since you need an operator in between 2 numbers as each operator must be in use at least once. -
Nope, it says insert the operators between the numbers so (4)(5) is out.Originally posted by the Bear:nah.. in math, the convention is that 2(3) = 6
besides, you forgot another "operator" -6 = 0 - 6 or (-1)(6) or (1)(-6)
math is fun
It didn't allow for 0 to be added in front or anywhere.
But I agree that (4)(-5) is probably acceptable. -
The more I think about it, the more I reject (4)(-5)...
I think it's too cheating... -
hey... we could be looking at something like Fermat's Last Theorem which just got disproved recently
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(1+2)/3) (4*5) = 20
err issit you can only use them once.. or you MUST use every sign once?
Ahh.. I read that MUST use every sign once.. hmm lemme try again.. *thinking* -
it's the first thing i thought of.. but you omitted the (-) or minus sign...
you have to use them all once...Originally posted by ShrodingersCat:(1+2)/3) (4*5) = 20
err issit you can only use them once.. or you MUST use every sign once? -
Use forum to help solve maths question. Not a bad idea...Originally posted by EricDraven:I'm stuck with a problem. Either there's no solution or I haven't got it.
Problem
1 2 3 4 5 = 20
...which I haven't thought of?
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put decimal point in between those numbers lor..
maybe u'll get 20
they say all must use once ?
or the x-+/ are optional and may be used once ? -
[{(1*-2)+3}/(4)(5)] to the power of -1
like tt can? -
This is the only thing I can think of.....and I don't know if its legal, but if you think of it, I haven't broken any rules.
[1/ (-2+3)] [4x5]
= [1/1][4 x 5]
= 20 -
Intriguing.Originally posted by gentlerock:This is the only thing I can think of.....and I don't know if its legal, but if you think of it, I haven't broken any rules.
[1/ (-2+3)] [4x5]
= [1/1][4 x 5]
= 20 -
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the question did say you gotta insert the operators in between the numbers so i think the "cheat blank operator" of multiplication between the 2 parentheses between 3 and 4 makes the answer wrong
Originally posted by gentlerock:This is the only thing I can think of.....and I don't know if its legal, but if you think of it, I haven't broken any rules.
[1/ (-2+3)] [4x5]
= [1/1][4 x 5]
= 20 -
(1^(-2/3) x 4) (+5) = 20
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the raising of the power as an operative won't be allowed either
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hmm try this one
log(1^(1-2/3)) + 4*5 = 20
otherwise
()/(1-23) + 4*5 = 20
only thing is i'm not sure if () = zero since matlab gives an error for that but the calculator gives it as 0.