Hi fellow Brilliantians out there! As any kid studying for a competitive exam, an arsenal of vast number of tricks, which make calculations simple, is a must. Why not try & write some of these here in this feed?
Here are some of mine -
Do it this way!
\[11 \times 23 = 2 (2+3) 3 = 253\]
Hey presto! You got the right answer!
Take another one- \[11 \times 49 = 4 (4+9) 9 = 4 (13) 9 = (4+1) 3 9 = 539\]
Just Brilliant! (Pun intended).
Take the case of \(145^2\).
1. Take the number other than 5 & consider it as a whole (in this case 14).
2. Multiply it by its consecutive integer (here \(14 \times 15 = 210\)).
3. Join 25 at the back of the number you just got (here 21025).
And that, my dear friend, is the answer! (Check if you don't believe me!).
Remember: This is only valid for the squares of integers ending with 5.
I hope to add more!