- Thread starter Denis
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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.

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

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

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.

Last edited:

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\(\displaystyle 0! + 1234 + 5 - 6 - 78 + 9 = 1165\)

\(\displaystyle 9*8 +(7 - 6) * 5+ 4^3 + 2^10 = 1165\)

\(\displaystyle 9*8 +(7 - 6) * 5+ 4^3 + 2^10 = 1165\)

Last edited:

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\(\displaystyle 9*8 +(7-6)*5 +4^3 +2^{10} = 1165\)

\(\displaystyle 0+12^3 +4 -567-8+9 = 1166\)

\(\displaystyle 98+7*6-5+4+3+2^{10}=1166\)

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\(\displaystyle 0!+12^3 +4 -567-8+9 = 1167\)

\(\displaystyle 98+7*6+(5+4)\div3+2^{10}=1167\)

\(\displaystyle 98+7*6+(5+4)\div3+2^{10}=1167\)

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\(\displaystyle 0*1^2-3+4^5+67+89=1177\)

\(\displaystyle 9+8+7+65+4^3+2^{10}=1177\)

\(\displaystyle 9+8+7+65+4^3+2^{10}=1177\)

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\(\displaystyle 0-1-2^{3!} +456 +789 =1180\)

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

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