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Start with the numbers 1,2,3,4,5,6,7,8,9,10 in that order and use the four common operations (+ − × ÷) and number concatenation (ie, you may write 123 concatenating 1, 2, 3) to obtain 2020.

Some clarifications:

  • All the ten numbers must be used, in that order, without repetitions.
  • Unary - is allowed; inserting decimal point is also allowed.
  • Number 10 cannot be split as 1 0; it can of course be concatenated with 9 to obtain 910.
  • Concatenation may only be applied to literal digits.
  • Exponentiation is not allowed, even though it is written without any explicit operators.
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  • 4
    $\begingroup$ There seem to be many many solutions to this question. It might be worth considering some tighter restrictions. $\endgroup$ – hexomino Dec 19 '19 at 9:50
  • $\begingroup$ In particular you should disallow concatenation $\endgroup$ – Adam Dec 19 '19 at 10:58
  • $\begingroup$ I am not sure that there is a solution without concatenation. $\endgroup$ – mau Dec 19 '19 at 14:42
  • $\begingroup$ Can we use parentheses ? $\endgroup$ – Adam Dec 19 '19 at 22:03
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Here's one way to do it if we disallow concatenation, division, unary and decimals

$(((1-(2\times 3)+4+(5\times 6)) \times 7) +8 - 9) \times 10 = 2020$

although there will be many others.

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2
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My two cents

(12*34)*5+6*7-8*9+10 = 2020

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  • $\begingroup$ Those brackets are not needed. :) $\endgroup$ – sudhackar Dec 19 '19 at 10:15
  • $\begingroup$ @sudhackar indeed, but it just helps me visually $\endgroup$ – rhsquared Dec 19 '19 at 10:17
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Here's some

12*34*5-6-7-8-9+10
123+45*6*7+8+9-10
12+34*5*6+78+910
1*2*34*5*6+7-8-9-10
1*23*45+67+8+910

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$((1+2) \div 3+45+67+89) \times 10 = 2020$

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