Match the post#

\(\displaystyle 0 + 12 + 3 \cdot 4 \cdot5 \cdot 6 + 789 = 1161 \)

\(\displaystyle (9 \cdot 8+7-6+543 \cdot 2) + 1+0!=1161\)
 
0! + 12 + 3*4*5*6 + 789 = 1162

√(9)*8 + 7*6*(5 + 4)*3 + 2 + 1 + 0! = 1162

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ksdhart2, your second equation would be correct without the parentheses.

(Everyone here who uses the "(number expression)*number" format or "number*(number expression)" format
is being redundant (as I used in my second equation), too, because the parentheses don't require the asterisk.
I suspect it is to make the prior and/or following number stand out by being a little distance away from the
parentheses.
 
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\(\displaystyle 0! + 1234 + 5 + 6 + 7 - 89 = 1164\)

\(\displaystyle 9 \cdot 87 + 6 + 54 + 321 + 0 = 1164\)
 
\(\displaystyle 0! + 1234 + 5 - 6 - 78 + 9 = 1165\)
\(\displaystyle 9*8 +(7 - 6) * 5+ 4^3 + 2^10 = 1165\)
 
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My last one should read:
\(\displaystyle 9*8 +(7-6)*5 +4^3 +2^{10} = 1165\)

\(\displaystyle 0+12^3 +4 -567-8+9 = 1166\)
\(\displaystyle 98+7*6-5+4+3+2^{10}=1166\)
 
\(\displaystyle 0!+12^3 +4 -567-8+9 = 1167\)
\(\displaystyle 98+7*6+(5+4)\div3+2^{10}=1167\)
 
\(\displaystyle 0 + 1 \cdot 23 \cdot 4 \cdot 5 + 6 + 78 \cdot 9 = 1168\)

\(\displaystyle 9 \cdot 8 \cdot 7 + 654 + 3^2 + 1 + 0 = 1168\)
 
\(\displaystyle 0! + 1 \cdot 23 \cdot 4 \cdot 5 + 6 + 78 \cdot 9 = 1169\)

\(\displaystyle 9 \cdot 8 \cdot 7 + 654 + 3^2 + 1 + 0! = 1169\)
 
-0! - 12 + 3 + 4^5 +67 + 89 = 1170

98 + (7 - 6)*5 + 43 + 2^10 = 1170
 
\(\displaystyle 0 - 12 + 3 + 4^5 +67 + 89 = 1171\)

\(\displaystyle 987 + 6 \cdot 5 \cdot 4 + 3 \cdot 21 + 0! = 1171\)
 
(0! + 1) * (2 * 3 - 4 + 567 + 8 + 9) = 1172

(-9 + 8 + 7*6 + 543 + 2) * (1 + 0!) = 1172
 
0 + 1 + 234*5 - 6 + 7 - 8 + 9 = 1173

-√(9) + (8 + 7)(6 + 5)(4 + 3) + 21 + 0 = 1173
 
0! + 1 + 234*5 - 6 + 7 - 8 + 9 = 1174

-√(9) + (8 + 7)(6 + 5)(4 + 3) + 21 + 0! = 1174
 
(-0! - 12 + 3*4*5) * (6*7 - 8 - 9) = 1175

(-9 - 8 + 7*6) * (5 + 43 - 2 + 1 + 0) = 1175
 
0! + (1 - 2 + 3 + 45) * (6*7 - 8 - 9) = 1176

(-9-8 + 7*6) * (5 + 43 -2 + 1) + 0! = 1176
 
\(\displaystyle 0! \cdot 1^2-3+4^5+67+89=1178\)

\(\displaystyle 9 + 87 + 6 \cdot 5 \cdot 4 \cdot 3^2 + 1 + 0! = 1178\)
 
-0! - 1 - 2^3! + 456 + 789 = 1179

987 - 6 - 5 - 4 - 3 + 210 = 1179
 
\(\displaystyle 0-1-2^{3!} +456 +789 =1180\)

\(\displaystyle (9+8+7+6-.5)*4*(3-2)*10=1180\)
 
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