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Use the four basic operators ×, ÷, +, − and if you want brackets to make:

8 _ 7 _ 6 _ 9 _ 2 _ 5 _ 4 _ 3 _ 1 = 2016.

You can use each operator as many times as needed. Concatenation is not allowed.

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  • 2
    $\begingroup$ Is each operator to be used exactly once? $\endgroup$
    – hexomino
    Sep 10 '20 at 8:54
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    $\begingroup$ Is concatenation allowed? $\endgroup$
    – Deusovi
    Sep 10 '20 at 8:57
  • $\begingroup$ no, you can use operators as much as u want $\endgroup$
    – oklahoma95
    Sep 10 '20 at 9:34
  • $\begingroup$ concatenation is not allowed $\endgroup$
    – oklahoma95
    Sep 10 '20 at 9:34
  • 1
    $\begingroup$ Also, puzzles where there are so many answers where one is not better/worse than any other is generally frowned upon in Puzzling.SE $\endgroup$
    – Ankit
    Sep 10 '20 at 15:03
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I tried a computer program to solve this problem.

I got 425 different expressions giving 2016.

So I added restrictions. I used only addition and multiplication. I stil get 26 expressions. You'll find them below.

$8*7*(6+9+2*(5+4)+3*1) = 2016$
$8*7*(6+9+2*(5+4)+3)*1 = 2016$
$8*7*(6+9*2+5+(4+3)*1) = 2016$
$8*7*(6+9*2+5+4+3*1) = 2016$
$8*7*(6+9*2+5+4+3)*1 = 2016$
$8*7*(6+9*2+(5+4+3)*1) = 2016$
$8*7*(6+9+(2*(5+4)+3)*1) = 2016$
$8*7*(6+(9+2*(5+4)+3)*1) = 2016$
$8*7*(6+(9*2+5+4+3)*1) = 2016$
$8*7+(6*9+2)*5*(4+3*1) = 2016$
$8*7+(6*9+2)*5*(4+3)*1 = 2016$
$8*(7+6+9*2+5)*(4+3*1) = 2016$
$8*(7+6+9*2+5)*(4+3)*1 = 2016$
$8*(7+((6+9)*2+5)*(4+3*1)) = 2016$
$8*(7+((6+9)*2+5)*(4+3)*1) = 2016$
$8*(7+((6+9)*2+5)*(4+3))*1 = 2016$
$8*(7*((6+9)*2+5)+4+3*1) = 2016$
$8*(7*((6+9)*2+5)+(4+3)*1) = 2016$
$8*(7*((6+9)*2+5)+4+3)*1 = 2016$
$8*((7+6)*9*2+5+4*3+1) = 2016$
$(8+7+6)*((9*2+5)*4+3+1) = 2016$
$(8+7*(6*9+2))*5+4*(3+1) = 2016$
$(8+(7*(6+9+2)+5)*4)*(3+1) = 2016$
$(8+((7+6)*9+2+5)*4)*(3+1) = 2016$
$(8*7+6*9+2)*(5+4*3+1) = 2016$
$(8*7+(6*9+2)*5*(4+3))*1 = 2016$

Note that over half of the solutions are just variations of the placement of the '*1'.

To answer the question about "no brackets" and "not ending in a product":

I found no solution without brackets.

But the solution
$((8+7)*6+9+2)*5*4-3-1 = 2016$
doesn't need any bracket if we ignore operator precedence. It can be computed on an old calculator just typing the operations from left to right.

It is also one of 12 expressions I found that doesn't end in a product or quotient.

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  • $\begingroup$ Without brackets? $\endgroup$
    – WhatsUp
    Sep 14 '20 at 1:10
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    $\begingroup$ I found no solution without brackets. $\endgroup$
    – Florian F
    Sep 14 '20 at 10:48
  • $\begingroup$ Did you find anything that isn't a product, i.e. last operation to be executed is multiplication? (I didn't, but I'm not a computer.) $\endgroup$ Sep 15 '20 at 8:46
  • $\begingroup$ @PaulPanzer I believe you can turn the last one here into a quotient (also the second one and the fifth one, if you wish). $\endgroup$
    – WhatsUp
    Sep 15 '20 at 11:09
  • $\begingroup$ @WhatsUp Quotient is technically a product ;-) $\endgroup$ Sep 15 '20 at 11:23
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This works:

8 x 7 x 6 x 9 / ( - 2 + 5 ) x (4 - 3 + 1)

or

8 x 7 x 6 x 9 / ((2 x 5 ) - 4) x (3 + 1)

or

8 x (7 - 6) x 9 x ( - 2 + 5 + 4) x (3 + 1)

or

(8 - 7 + 6) x 9 x (((2 + 5 + 4) x 3) - 1)

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Here are a few:

$8+7+6+9+2-5*4-3-1 = 2+0+1*6$ $8+7+6+9+2-5*4-3-1 = 2+0*1+6$ $8+7+6+9+2-5*4-3-1 = 2+0/1+6$ $8+7+6+9+2-5*4-3-1 = 2-0+1*6$ $8+7+6+9+2-5*4-3-1 = 2-0*1+6$ $8+7+6+9+2-5*4-3-1 = 2-0/1+6$ $8+7+6+9+2-5*4-3*1 = 2+0+1+6$ $8+7+6+9+2-5*4-3*1 = 2-0+1+6$ $8+7+6+9+2-5*4-3/1 = 2+0+1+6$ $8+7+6+9+2-5*4-3/1 = 2-0+1+6$ $8+7+6+9-2-5*4-3+1 = 2*0+1*6$ $8+7+6+9-2-5*4-3+1 = 2*0*1+6$ $8+7+6+9-2-5*4-3+1 = 2*0/1+6$ $8+7+6+9-2-5*4-3*1 = 2*0-1+6$ $8+7+6+9-2-5*4-3/1 = 2*0-1+6$ $8+7+6+9-2*5-4*3+1 = 2+0+1+6$ $8+7+6+9-2*5-4*3+1 = 2-0+1+6$ $8+7+6+9-2*5-4*3-1 = 2+0-1+6$ $8+7+6+9-2*5-4*3-1 = 2-0-1+6$ $8+7+6+9-2*5-4*3-1 = 2*0+1+6$

And very many more.

There are $2,762$ solutions in this form (without using parentheses).

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    $\begingroup$ Taking the no concatenation a step further, but technically correct, I guess :) $\endgroup$
    – klabuster_
    Sep 10 '20 at 10:22
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    $\begingroup$ I think 2016 was supposed to remain a single number. $\endgroup$
    – Florian F
    Sep 13 '20 at 8:35
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    $\begingroup$ @FlorianF At the time the question was posted there was no clarification as to where the operators were meant to be placed. The underscores were not there originally. $\endgroup$ Sep 15 '20 at 11:07
  • $\begingroup$ I see. I thought it was a curious choice to split the 2016 but given the original question that makes perfectly sense. $\endgroup$
    – Florian F
    Sep 15 '20 at 15:13
  • $\begingroup$ @FlorianF Original wording is "Add the four basic operators ×÷+− and optionally brackets to: 876925431 To get the total 2016." which is pretty unambiguous IMO. $\endgroup$ Sep 15 '20 at 16:01
1
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A possible solution:

8*7*9*4+3*1-2-6+5 Note that I didn't keep the original order of the digits.

This is how I arrived at it:

2016 is a lot bigger than single-digit numbers, so we need multiplication to get big
8*7*6*9*2*5*4*3*1
362880
This is approximately 150 times too big. We will want to divide by 150. 150 = 2*3*5*5. Let us divide by 5*5
8*7*6*9*2/5*4*3*1
14515.199999999999
Then divide by 3
8*7*6*9*2/5*4+3+1
4842.4
Now divide by 2
8*7*6*9/5*4+3+1+2
2425.2
This too big by 400, let's try replacing *6/5 with *5/6
8*7/6*9*5*4+3+1+2
1686.0
This is too small by 400. The average of 6/5 and 5/6 is approximately 1, let's remove 5 and 6 from the product.
8*7*9*4+3+1+2+6+5
2033
This is close! We need to remove 17. 17 = 6+6 + 2+2 + 1
8*7*9*4+3*1-2-6+5
2016

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    $\begingroup$ You shouldnot change the order of digits $\endgroup$ Sep 10 '20 at 9:52

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