Christmas = Halloween

$\LARGE \text{Oct} \ 31 = \text{Dec} \ 25$

How to make the equation above "appear true"?

#Logic

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

31 in base 8 (oct) = 25 in base 10 (dec)

Richard Grayson - 5 years, 11 months ago

Yes , you are right. Can you find another logic? (its a bit out of box though).

Nihar Mahajan - 5 years, 11 months ago

They are both "holidays" :D

Seth-Riley Adams - 4 years, 12 months ago

first is 31/10 and 31=1 mod10

second is 25/12 and 25=1 mod12

Dev Sharma - 5 years, 2 months ago

Oct 31 is a Monday, Dec 25 a Sunday. So both cannot be equal! Hence the given equation cannot be true.. ;-)) Sad but it is false..;)

Sunny Sharma - 5 years, 1 month ago

Or you can just say they are in the same year and call it a day. (i.e. 2018=2018)

Trevor Greenlee - 2 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$