Memorizing the first 100 perfect squares (2022)

https://dzhu.page/mathematics/memorizing-100-perfect-squares/

by aragonite • 2 years ago

88 55 2 years ago

thechao2• 2 years ago

(10a + b)^2 = 100a^2 + b^2 + 2ab

So... 73^2 is 4900 + 9 + 420 = 5329. The really nice part is getting estimates for square roots of numbers.

So, sqrt(3895)? 60^2 + 120n = 3600 + 120n => n=2; that's 3844 (from above); the difference is 51; the residual estimate is then: 62 51/(62*2).

metacritic12• 2 years ago

20ab?

thechao2• 2 years ago

Yes

quantum_state• 2 years ago

73-50=23 => 23x200=4600, 50-23=27, 27^2=729 => 73^2=4600+729=5329, all can be done in one’s head.

For 27^2, one can just memorize, or: 27-25=2 => 2x100=200, 25-2=23, 23^2=529. 27^2=200+529=729.

As long as one knows square of numbers up to 25, it is done for all up to 100 and more … :-).

rustybolt• 2 years ago

One nice thing from knowing squares is that you can calculate for example 23 * 47 as (35 - 12) * (35 + 12) = 35^2 - 12^2.

But this doesn't always work and you still need to be good at adding/subtracting.

JadeNB2• 2 years ago

But this doesn't always work and you still need to be good at adding/subtracting.

It does always work. If you mean that not every integer product can be written this way, you're right; 23 × 46 can't, unless you're willing to memorise squares of half-integers. But, if avoiding non-squaring multiplication is really key, then you can still just write 23 × 46 = 23 × 47 - 23, and then compute 23 × 47 = 35^2 - 15^2 as you suggest.

shiandow2• 2 years ago

Or you could do ((46+23)^2 -(46-23)^2) / 4

Or one of several similar formulae, but each has its own pros and cons.

amelius2• 2 years ago

What is wrong with computing it just as 7*23 + 40*23 ?

bibanez• 2 years ago

Nothing, but if you know perfect squares you will do 3 additions/subtractions instead of 2 products and 1 addition

kstrauser5• 2 years ago

I'm good at arithmetic, to the point that my buddy called from across the country a few days ago to have me perform "hard" math questions for his kids. It's always a relief when they pick random numbers like "what's 83 times 97?" Whew. Well, 90**2 - 7**2 = 8051. As a bonus, when you explain that method, it sounds even more impressive to them: "you mean, you just know what 63 squared is?" Sure, but it didn't happen overnight.

bee_rider2• 2 years ago

I think your star star just rendered as a star.

teachrdan• 2 years ago

That close to 100, I'll just do 83*100 - 83*3

nine_k• 2 years ago

Hmm, this looks like a neat trick to make hardware multiplication faster. Say, for 8-bit numbers it would require 3 8-bit additions/subtractions and one 16-bit addition, one 9-bit shift, and two lookups in a 512-byte table of squares, and zero conditional processing, as opposed to 8 16-bit additione, 8 16-bit shifts, and 8 LSB tests that a naive iterative algorithm would do.

I wonder what real integer multiplication hardware uses.

amelius2• 2 years ago

It all depends on how you can parallellize it.

ysofunny3• 2 years ago

I memorized all primes up to 127. and with a bit of effort I may come up with all the primes up to 307 (then 311,313, iirc) but I need pen and paper to double check.

the problem is that I don't know why do this.

nonetheless I can report that I've memorized 3 instances of two consecutive twin primes

11,13,17,19, then 101,103,107,109 (which just raises questions that I can only aspire to ask, nevermined answering, about the what, why, and how of decimal system),

and then 191,193,197,199. the next prime is 211. but the cool thing is how 210 = 2*3*5*7, which are all primes before the first double twin prime

your_friend• 2 years ago

reading this just reminded me how i miss talking to my friends with scientific background

kstrauser5• 2 years ago

I know exactly why I do this. I use to deliver pizza. My store was on the edge of a medium-sized city, and a lot of our customers were quite the drive away. I'd get incredibly bored on these long drives. Somewhere along the way, I'd start calculating what percent of my shift I'd worked that minute, and I'd try to get it to 3 decimal places before the next minute came. Again, no reason whatsoever, just bored. And then at some point I thought it'd be more interesting to calculate (number of minutes worked) / (number of minutes remaining), which is fun because the ratio grows really slowly at first and then very quickly. I realized the first step in that was dividing out the common factors, so 248 minutes / 232 minutes was 31 / 29. (Yes, I'd been taught that in class, but it's one thing to have a teacher make you do something and another to realize why you'd ever want to do it voluntarily.)

Do that long enough and you find patterns, like 7 * 11 * 13 = 1001. If you ever end up calculating n / 11, it's approximately the same as n * 7 * 13 / 1000. E.g., 3 / 11 ~= .273. Or take 27 * 37 = 999. Now n / 37 ~= n * 27 and shuffle the decimals. 7 / 37 ~= 7 * 27 / 1000.

And that's how I ended up reasonably good at mental arithmetic, and memorizing a frankly unnecessary number of squares, and being able to factor lots of numbers at a glance (or recognize that they're prime). I was awfully bored for an awfully long time.

FredPret3• 2 years ago

Boredom can be great.

Imagine being an ancient Babylonian or Greek with nothing to do but wait for your grapes to grow. No wonder they came up with lots of great stuff.

lupire2• 2 years ago

191, 193, 197, 199 are also twin twin prime pairs

These primes have to be 11,13,17,19 + 30k, due to division by 3.

5 is half of 10, so we easily rule out exactly the number 10x+5.

49 = 7²,

77 is a multiple of 7, of course.

98 = 2 • 7²

In order to avoid a multiple of 11 interfering, we need the gap preceding 11•10

99 is (10+1)(10-1) = 10²-1 and 100 is 10²

, so that clears out space for primes after 100.

Also, You might enjoy "Paterson Primes" https://m.youtube.com/watch?v=jhObLT1Lrfo

ysofunny3• 2 years ago

I'm intersted in why (and how) you reason that thing about "multiple 11s interferring"

so I'll tell you how I memorized the primes between 2 and 127.

to begin: all primes less than ten: 2,3,5,7. I seem to have learned these by rote memorization. but then, the 'fun coincidences' begin.

because 5 is a multiple of "ten", all primes after 10 can only end in 1,3,7,9. this is a key somehow.

this gives the first fun coincidence: that 11 and 101 are both prime.

after 19 comes 23. similarly (and this is the part where every self-respecting numberphiliac waves their hands a little, in excitement I hope), 113 ("one one three") is the next prime after "one oh nine".

notice that 23,29,31,37 are as much prime as 30+{23,29,31,37} = 53,59,61,67

this covers almost all of them. but we're missing primes in the 40s, and the 70s. as well as primes in the 80s and 90s (only 3 primes: 83, 89, and 97. I have no tricks to remember these 3. only rote repetition)

the forties and seventies, are a very compact: both 1,3. but only 47, and only 79; because, well, what fun! seven squared and seven eleven are just so simple, like low-hanging fruit in the garden of prime-number coincidences

absolute7• 2 years ago

This is cool

freedomben• 2 years ago

Why is it cool? What is it about TFA that interests you? What does it do that other resources have not done? Does it help you think about the problem in a different way?

I've always been fascinated with the perfect squares, and various patterns that arise from observing them. I love how the article examines the patterns and then extrapolates the study of the patterns to a tool for memorization.

vinay_ys• 2 years ago

This is beautiful. Maybe useless, but still a lot of fun to learn.

jurassicfoxy• 2 years ago

I previously memorized pi to 100 decimals. Was fun, and now I'll never forget 3.1415926535897932384626433 ... how many is that? 25?

There are a several little triplet "patterns" in this first batch that make it easy to this point:

3.1415 926 535 8 979 323 84 626 433.

FredPret3• 2 years ago

I'm guessing the answer to this question is no, but... for an irrational number like pi, can we guarantee that a certain digit sequence will occur somewhere in its infinite reaches?

That would be a wildly impractical but very fun way to encode information - if everybody had a couple of petabytes of pi on their harddrive one day, you could just send the starting and ending digit to communicate an arbitrary amount of information this way.

Of course you'd first have to search through the whole universe of digits to find a sequence that's just right.

Arnavion• 2 years ago

Such numbers are called "normal numbers". Pi likely is one based on all the digits we've computed for it, but we don't have any way to prove a given number like pi is normal or not.

Relevant Numberphile video: https://www.youtube.com/watch?v=5TkIe60y2GI

thfuran• 2 years ago

1.01001000100001... is infinite and non-repeating, but doesn't contain all substrings. A number that does is called a "normal" number, and it's not known whether pi is. It seems to be pretty normal though.

kurisufag• 2 years ago

this in practice: https://github.com/ajeetdsouza/pifs

bitwize• 2 years ago

That reminds me of how on The IT Crowd, the new -- exceedingly long -- Emergency Services number replacing 999 was fairly easy for Moss to remember. Being divided into groups of at most 5 digits probably helped.

frankus2• 2 years ago

Our math prof (who moonlighted as a "mathemagician") taught us a trick for two-digit squares:

For e.g. 23 x 23, subtract three from the first number, add three to the second, and then add 3² to the product. So 20 x 26 + 9, the idea being that multiplying by a multiple of 10 is easier to do mentally.

JohnMakin2• 2 years ago

This would have made my college career significantly easier. We were limited on scratch paper (dumb policy) on many tests, and I have to write out arithmetic like this long-hand style, which took up a ton of space.

bee_rider2• 2 years ago

I think these sort of mental math tricks are mostly useful when you have a case where you’ve got a formula that you often have to run, and it is similar enough every time that you can pre-do the algebra.

The one everyone knows is calculating a tip (bump the decimal place left once and double the result, round depending on how nice you feel). Might be applied multiple times per day, depending on your dining habits.

I bet the formulae you had to use on your tests were worth doing by hand.

akovaski• 2 years ago

I recently leafed through "Mathematics for Engineers"[0], originally from 1926, which begins with many such methods for making arithmetic easier. Though maybe not always more space efficient.

Luckily I got to use a calculator for school.

[0]https://archive.org/details/dli.ernet.11725/page/n9/mode/1up

hinkley2• 2 years ago

If you look at the way they teach kids math on paper today, they are just teaching them how to do math in your head. The idea being that if your hands were free you’d use the calculator on your phone.

When I first encountered the outrage over New Math my first thought was that this is how I avoid embarrassing myself in checkout lines. Do I have enough cash to pay for this stuff in my hands?

hinkley2• 2 years ago

I do the less efficient quadratic form. It works but is slower.

20² + 20 x 3 x 2 + 3²

pranaysy2• 2 years ago

Neat! This is (a+b)(a-b) + b² = a²-b² + b² = a² and in your eg, a=23 and b=3

In addition to using this trick for getting to multiples of 10, I used it to compute the product of two numbers by leveraging their proximity to a number in between whose square I knew. For eg if I need to multiply 23 by 27, I instead see it as (25-2)(25+2), which is 25²-2² = 621.

(using that same trick from the article to calculate squares of numbers that end in 5)

pranaysy2• 2 years ago

For ref: https://en.m.wikipedia.org/wiki/Difference_of_two_squares#Me...

koolba• 2 years ago

That’s a neat one!

(a + b)(a - b) = a^2 - b^2

a^2 = (a + b)(a - b) + b^2

So pick a value of b that makes a - b end in a zero.

frankus2• 2 years ago

FWIW you can also pick a value of b to make a + b end in a zero, which keeps b at 5 or less.

lupire2• 2 years ago

No need to memorize

(10a+5)² = [a * (a+1)][25]

(a+1)² = a² + a + (a+1)

(a-1)² = a² - a - (a-1)

(a+2)² = a² + 4(a+1)

(a-2)² = a² - 4(a-1)

apalmer• 2 years ago

I am not knocking the article but seems like if you are going to dedicate the effort of learning quick mental math it's probably more efficient to 'just' know how to multiple 2 digits quickly than specifically focusing on 2 digit squares...

dxbydt2• 2 years ago

I give amc & mathcount mocks to kids every weekend & time them. There was one question everyone got under 5 seconds. I was like how did you guys do it so fast. Apparently aops had told them every year will have a special property that'll forsure be on the test. So that's why they all knew 45^2 was 2025.

madcaptenor• 2 years ago

When I did math competitions I remember getting the advice to know the prime factorization of the current year.

dxbydt2• 2 years ago

45^2 = 2025 => must be the sum of first 45 odds => 1 + 3 + ... + 87+ 89 = 2025

while its not pythagorean, 40^2 + 20^2 + 5^2 = 2025.

877 is another bullcrap that has shown up on a bunch of these tests. 877 is a prime, and four times 877 is 3508 = sum of all the divisors of 2025.

there's a bunch more i shared with them.

Memorizing the first 100 perfect squares (2022)

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