Comic # 400 ?? How the... I guess if you only consider the "good" comics then I'm actually on Comic # 7.

I don't think I did a shout out to The Nerdvana yet (it's listed in the math comic list over at the sidebar though). Anyways, here are a few mathy ones:

- Cross Product (my favorite)
- This mom needs to learn integration!
- Decreasing curves
- How do you find the volume of a cricket ball?

Link to the 'Saint Math Tricks Day' trick from 2010.

Awesome!!!

And hence would only work for 47 through 99

CHEERS!! Here's to another 400 more!!

Well, technically it will work for 1-99 being cubed... but you may have 000 as the first group of 3 digits.

Actually, if you have 000 as the first 3 digits, the number will only be the last 3 digits. 000,658 = 658

Btw could you make the iphone app compatible with iOS 3.1.2? I'm stuck with that

The trick isn't as impressive if you use an app.

I knew this trick!!! xD

Guuuyz... you should listen to Van der Graaf Generator's last album, A Grounding In Numbers (and, in fact, other Peter Hammill's works). Or at least the song from that album, "Mathematics".

it doesn't work for 56 cubed :-( or 58 cubed :-(

It should. 56 * 56 * 56 = 175,616. The 175 lies between 125 and 216 (and take the lower which corresponds to FIVE). The 616 ends in 6, and 216 ends in 6 which corresponds to SIX. So, 56.

A good exercise is to verify mathematically why it always works. For example, if I gave you the number (10x+y)^3 then your goal is to determine x and y. First ask, what is the last digit of (10x+y)^3? Well it's easy to see that it is y^3 mod 10, but does this uniquely give you y? For x, how do the first 3 digits of (10x+y)^3 correspond to the table?

I like this.

Reminds me of Richard Feynman's race vs. an abacus. He was a quick natural mental calculator, so when in a shop he saw the owner using a Chinese Abacus, he challenged him to a race. He tried addition, multiplication, and even a square root--but the abacus was faster.

However, when he suggested a cube root, the owner chose 1729, which was 12^3 + 1, which allowed Feynman to do successive approximations and beat out the abacus.

Hi there. This is awesome.

So i have to work on my thesis. Consequently I don't do that and, instead, I try to prove that the method above always works.

Whoever wants to figure this out for himself: stop reading this post.

(10x+y)^3=z

z mod 10 = y^3 mod 10 is obvious and unique ({0^3;..;9^3} mod 10 consists of numbers 0 to 9).

Now to the first 3 digits:

The expansion of (10x+y)^3 is easy enough, you can also easily identify how the first 3 digits are formed.

They are:

x^3 + floor(3x^2*(y/10)+3x*(y/10)^2+(y/10)^3) [floor() .. rounding down]

So, for this method to work, the following must hold:

(x+1)^3 >= x^3 + floor(3x^2*(y/10)+3x*(y/10)^2+(y/10)^3)

or

(x+1)^3 - x^3 >= floor(3x^2*(y/10)+3x*(y/10)^2+(y/10)^3)

where

(x+1)^3 - x^3 = 3x^2+3x+1

since y € {0;1;..;9} it follows that (y/10) is always smaller than 1

thus

3x^2+3x+1 >= floor(3x^2*(y/10)+3x*(y/10)^2+(y/10)^3)

So the method introduced in this strip will work for every positive integer smaller than 100.

Fun stuff.

Nice. Now go work on your thesis!!

I should mention that this trick is well known and has appeared in numerous books. It is also one that Scott Flansburg (http://scottflansburg.com/thc/media/video/morethanhuman/) has performed numerous times.

Well, I will. Eventually.

Didn't know this was a well known trick, first time I came across it.

Any1 here has some recommendations regarding websites where I can find some more of those neat tricks?

Met my highschool math teacher the other day and I enjoy tormenting him by now being a better calculator than he is (I used to be really bad at quickly doing actual calculations).

Given \emph{any} 6-digit number, the probability that your answer is correct is kinda nowhere dense, because most of the numbers are not perfect cubes at all.

Yup, but when you perform the trick you tell someone to take a 2 digit number and cube it on the calculator... they then read of the resulting 6 digit number and in less than a second you can name the 2 digit number they started with.

That is a nice trick, but only works up to 90^3 (729,000).

For any number bigger than 90^3 it fails.

91^3 = 753,571. By the table the number should be 98.

92^3 = 778,688. By the table the number should be 98.

You can try. I checked up to 99^3 and all failed.

Regards... Walter from Brazil.

Ehh no?

1^3 ends in 1, so does 91^3

2^3 ends in 8, so does 92^3

3^3 ends in 7, so does 93^3

...

and since 91^3 > 729000 and 99^3

Walter from Brazil, you are wrong.

far easier way to do this - the last digit corresponds to the last digit of the cubed number - x4 cubed always gives a number ending in 4 - however, for 2,3,7, and 8, it is slightly differnt - 8's are 2's, 2's are 8's, 7's are 3's, 3's are 7's. the first digit of the number is found as you suggested. to use walters example, 92^3 = 778688, 8 = 2, so ends in 2. 778 > 729, therefore 92

Please look at the item 4 in the figure. The method is different from the item 3.

That walter guy is wrong, it still applies after that.

awesome trick dude.... hope it works for other numbers other than that

nice trick.... awesome

WHOOAAA this is sooo cool!! Awesome lol.

this "trick" doesnt help at all unless you memorize the 6 digit perfect cubes

Wow! That comes in handy to know day to day.

I use it in work like...........never.

6+4=THIIS IS FASYY

Umm... I've tried it several times so far with different 6 digit numbers and all of them have failed to be correct.

According to your table:

The cube root of 189677 should be:

189 - Between 5 and 6, take 5

677 - Ends in 7, therefore take 3

That makes 53, but the correct answer is actually 57.

Another random attempt:

282410

282 - Between 6 and 7, take 6

410 - Ends in 0, therefore take 0

That gives us 60, however the correct answer is 65.6

Not really impressive enough if it even fails to come close to the target number. It sure my be useful to work out estimates, but as for suprising your friends, it doesn't really work at all!

Umm... I've tried it several times so far with different 6 digit numbers and all of them have failed to be correct.

According to your table:

The cube root of 189677 should be:

189 - Between 5 and 6, take 5

677 - Ends in 7, therefore take 3

That makes 53, but the correct answer is actually 57.

Another random attempt:

282410

282 - Between 6 and 7, take 6

410 - Ends in 0, therefore take 0

That gives us 60, however the correct answer is 65.6

Not really impressive enough if it even fails to come close to the target number. It sure my be useful to work out estimates, but as for suprising your friends, it doesn't really work at all!

The comic says 'This trick works for six-digit perfect cubes.' Your two numbers are not perfect cubes.

Please, tell me you are not serious

Cool, didn't know that trick :)

It seems to work because n^3 covers every residue mod 10. That's how we know the last digit.

So, cuberoot1953125=125, 9th root1953125=cuberootcuberoot1953125=5.