I had the exact same route to 3 guesses on today's puzzle. Very nice.
Square Numbers
Okay, strap in. I don't know the exact number I have memorized at this point, but it'd be probably in somewhere in the 80s before I'd have a do a quick sanity check to make sure. However, I can square rather large numbers in my head, though 5- and 6-digit ones are usually slow. 3-digit squares shouldn't take more than a few seconds.
I'll cut-and-paste and adapt stuff I've written elsewhere on how to do it. I'm confident that you could at least mentally calculate at least three-digit squares fairly quickly without too much practice. My method is one I developed while bored in high school English class years ago. Others have certainly independently come up with it.
First, you need to memorize squares up to 25, but that's it.
For squares larger than 25, I rewrite them in the format:
(50n ± y)², where y is between 0 and 25.
(50n ± y)² =
2500n² ± 2 * 50 * y + y² =
2500n² ± 100y + y²
This means the last two-digits are only affected by the y² term. The other two terms are hundreds and above. Moreover, the y² term only contributes a number between 0 & 6 to the hundreds digit, which is easy to handle.
For example,
64² =
(50 + 14)² =
50² + 2 * 50 * 14 + 14² =
2500 + 14 * 100 + 14² =
3900 + 14² =
3900 + 196 =
4096
283² =
(300 - 17)² =
300² - 2 * 300 * 17 + 17² =
90,000 - (6 * 17) * 100 + 17² =
100 * (900 - 102) + 17² =
100 * (798) + 17² =
79,800 + 289 =
80,089
Now, this does involve knowing what all the squares ending in -50 (i.e., 100n + 50)...350, 450 , 550, etc.
I've got a trick for that, too. To square a number ending in 5 (10n + 5), all you need to do is:
1) Remove the ending 5.
2) Take the remaining number and multiply it by one plus that number (n(n+1)).
3) Take that result and postpend 25 on the end. That's the solution.
So;
35² = 1225
45² = 2025
55² = 3025
65² = 4225
95² = 9025
2835² = 8,037,225
For that last one, I used the result above for 283² (80,089), added 283 (80,372), and then put a 25 on the end.
Now, if you want to calculate 763²:
(750 + 13)² =
(75²)(10²) + 2 * 750 * 13 + 13² =
5625 * 100 + 15 * 13 * 100 + 13² =
(5625 + 195) * 100 + 13² =
582,000 + 169 =
582,169
For 4-, 5-, and 6-digit numbers, instead of 50n, I'd use 500n, 5000n, 50000n, respectively. That means the range of y goes up by a factor of 10 each time, between 0 & 250, 0 & 2500, etc. .However, that means I have to repeat the process multiple times:
Leveraging the answer for 283² = 80,089, let's calculate 4283²:
(4000 + 283)² =
4000² + 2 * 4000 * 283 + 283² =
16,000 * 1000 + 8 * 283 1000 * + 283² =
(16,000 + 2264) * 1000 + 283² =
18,264 * 1000 + 283² =
18,264,000 + 80,089 =
18,344,089
Please note in all these calculations how easy all the addition steps are. There is usually only one or two non-zero digits that overlap. The 769² one was the hardest because I had 5625 + 195, which still isn't that bad.
I wouldn't worry about doing 4-digit numbers, anyway. I'd first start learning up to about 125², which isn't hard and is impressive and useful enough. Your next goal might be up to around 625². That way, the middle term in the expansion only involves multiplying a number from 1 thru 25 by a number 1 thru 12:
(550 + 3 )² =
302,500 + 2 * 550 * 3 + 3² =
3025 * 100 + (11 * 3) * 100 + 3² =
(3025 + 33) * 100 + 3² =
305,800 + 9 =
305,809
This can expand as far as you need. The problem is that once you get to 5- and 6-digits, you need to repeat the process multiple times and memory becomes a problem:
123,456² =
(100,000 + 23,456)²
10 billion + 2 * 100,000 * 23,456 + 23,456²
Which means I now have to do (20,000 + 3456)², which involves 456², which is (500 - 44)². There's no reason to need to do that for numbers that large unless you get bored easily. Calculating it in this iterative manner, starting with 456², minimizes how much short-term memory I need.
Finally, being able to square numbers quickly is useful because then you can often use the difference of two squares to multiply two different numbers very fast.
24 * 36 =
(30 - 6)(30 + 6) =
30² - 6² =
900 - 36 =
864
Leveraging the calculation of 283² from above:
274 * 292 =
(283 - 9)(283 + 9) =
283² - 9² =
80,089 - 81 =
80,008
Because I'm so adept with calculating square numbers, I use difference of two squares a lot. This quite a bit to digest. I'm obviously open to clarify any of that.
xdle BONUS: As part of that y² term being the only part that affects the last two digits, that means all 22 possible two-digit endings are found in 1² through 24². 24² and 26² share the same two-digit ending (576 & 676). Same with 23² and 27² (529 & 729). Same with 1², 49², 51², 99², 101², etc.
Moreover, the numbers 2, 3, 7, and 8 will never be in the units digit. That can help you quickly judge some of those "(x + 792) is a perfect square" type clues.