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.
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.
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.
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)
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).
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 by5*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 of6/5
and5/6
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