Evens vs. Odds

$(1 + 3) \times 5 = 2 \times (4 + 6)$

$1+5+3!=2+4+6$

1-3+5=2/4*6

$\large \frac{d}{dx} [1+3+5]= \frac{d}{dx} [2+4+6]$

$1!×3!×5! = (2-\sqrt{4})!×6!$

|1+3-5|=(2+4)/6

(1+5) 3!=(2+4) 6

13+5=18 and 24-6=18 so 18=18

1 x 3 + 5 = 2 - 4 +6

(-1 + 3!) / 5 = (2+4) / 6

(-1+3)*5=(2^4)-6

1 - 3! + 5 = 2 + 4 - 6
( 1 + 3 ) × 5 = 2 × ( 4 + 6 )
1 + 3! + 5 = 2 + 4 + 6

$log(1) + 3 + 5 = 2 (4! / 6) = 8$

1st solution 13+5=24-6,,, 2 nd solution ln(1)×5×3=2+4-6,,,,3rd 1×3!×5=24+6

ln(1) * 3 * 5 = 2 + 4 - 6 = 0

$(1 + 3) \mod 5 = (2^4) \mod 6$

$-1+3\cdot 5=2\cdot 4+6$ or $13+5=24-6$

$\boxed{(1+3)!\times5=\dfrac{(2+4)!}{6}}$

-1+3x5 = 2x4+6

1-3+5=2/4x6

1+3!+5 = 2+4+6

1+6+5 = 2+4+6

12 = 12