0
$\begingroup$

I've had some problems with some math games. Let's say I have this 84 x 90, and the second number always ends in zero. How would I be able to calculate this faster than doing (84 x 9) x 10.

$\endgroup$
  • $\begingroup$ That's pretty fast: think of $84\cdot9$ as $80\cdot9+4\cdot9$. $\endgroup$ – symplectomorphic Mar 13 '18 at 16:30
1
$\begingroup$

$$84\times90=(80+4)\times9\times10=(720+36)\times10=756\times10=7560$$

gets the job done without any carrying or borrowing, which is what slows me down the most when trying to do arithmetic in my head.

Alternatively, if you've memorized all your two-digit squares (which I have not!), then

$$84\times90=(87-3)\times(87+3)=87^2-3^2=7569-9=7560$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I like to use $(a+b)^2=a^2+2ab+b^2$ to square in my head (for large squares I haven't memorized), but of course you're right it's much slower if you don't have them memorized... $\endgroup$ – symplectomorphic Mar 13 '18 at 16:53
0
$\begingroup$

$$84\cdot 90=84\cdot (100-10)=8400-840=7560$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$84\times 90 = 84\times(100-10)=8400-840 = 7600-40 = 7560$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$84 \times 9 = 84 \times 10 - 84 = 840-84 = 800-44=756$$

And now add a $0$ at the end.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.