$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}} = {\mathtt{151}}$$
Now we have one less, one more, and one right on
I cannot say I do.....I don't......
It is NOOOO
NO!
(BTW...if you can't beat 'em, play with 'em)
NOOOOOOOOOOOOOOOOOO
Be back soon
The answer is the answer
Highly doubt it
NOOOOOOOOOOOOOOOOOOOOOOOOOOOO
) 