this question is related to the maths subject
i have tried " [9-7-8+6]+[5-3-4+1][-1]" but this not the correct way.
2007-06-20 01:49:53 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The trick is you have to make at least one number with two or more digits. Since the sum of 1..9 is 45, which is odd, you can't get to zero with addition and subtraction of single digits only. (Proof: You could separate out all the digits added, from subtracted. Let's call the added total n, then the subtracted total would also have to be n, in order for the difference to be zero. However, the total of all digits used in the equation would then be 2n, which is necessarily even. Since the sum of 1...9 is odd, it's not possible to have a single-digit-only solution.)
Anyway, I came up with:
(31 + 72 + 5) - (96 + 8 + 4) = 0
Both the added and subtraced sums are 108.
A fancier one:
(598 + 3 + 4 + 7) - 612 = 0
2007-06-20 01:53:58 · answer #1 · answered by McFate 7 · 2⤊ 0⤋
(9*1+2+8+3) - (4+7+6+5)
=22-22
=0
2007-06-21 15:27:13 · answer #2 · answered by Happy 3 · 0⤊ 1⤋
may be is ï¼
36-ï¼1+2+4+5+7+8+9ï¼=0
2007-06-20 08:56:16 · answer #3 · answered by Xu D 2 · 0⤊ 0⤋
91-64-27+8-5-3=0
It contains all the digits from 1 - 9 & 0 is got by using addition & subtraction only.Digits also didn't repeated. If you think it's correct you can select my ans. as the best one
2007-06-20 11:13:21 · answer #4 · answered by niti 2 · 0⤊ 1⤋
Using single digit numbers there can not be any correct answer to this problem
2007-06-26 23:44:47 · answer #5 · answered by reshuorange 1 · 0⤊ 0⤋
(819) - (273 + 546) = 0
2007-06-22 03:32:56 · answer #6 · answered by Madhukar 7 · 0⤊ 0⤋