A relevant post about a question

[Michael T]

Numsen tests:

(I don't know Latex, so excuse me)

(x-y)/(x) + (x/y) = _______ (Mixed number)

e.g.: (1/6) + (6/7) but larger numbers.

[stupidityismygam]

[Michael T]

It's 11:00, stephen, gimme a break here. I'm incrementing my post count, it's tiring work.

But okay. That works.


And how would you (estimate) the sum of the first 20 triangular numbers?

[stupidityismygam]

well....

the sum of 2 consecutive triangular numbers is a perfect square

so the first 20 is just [unparseable or potentially dangerous latex formula]

then [unparseable or potentially dangerous latex formula]

so since the sum of consecutive squares is [unparseable or potentially dangerous latex formula] we get [unparseable or potentially dangerous latex formula]

[Michael T]

Neat.

Oh, and, isn't there some 'universal' equations for the nth x-gon number?

[stupidityismygam]

there is...but...i never remember it...

basically what i do...is like

for triangular the coefficient is 1/2, square 1, pentagon 3/2, hexagon, 4....

so then with that info...you know the first of all number is 1, and the 2nd is the number of sides in the number

so then you can just derive the formula and it take like 5 seconds when you get good...

[PatriotAlanis]

Here's the formula:

[(n^2)(x-2) - (x-4)(n)]/2[/tex]

[stupidityismygam]

ok i think i got it....

ok for the kth n-gonal number the formula is



[unparseable or potentially dangerous latex formula]...you might wnat to check this...but i hope it works

[Kurt]

"Universal formula" for the nth k-gonal number is:

[unparseable or potentially dangerous latex formula].

It's better to just remember how to increment though. I've just memorized ones for n=3 and n=5, and then remember to add one to the n coefficient and subtract one after it for n=6, etc. Doesn't take too long.

(Lol, 3 answers at the same time...and the last two are the same but switched. :P)

[stupidityismygam]

wow...everybody knew that....i almost felt good for deriving that too...o well

yeah michael that is the formula just fyi

[Michael T]

13 x 539

besides foiling....?

[stupidityismygam]

[PatriotAlanis]

13 * 159

triple the last number--> 7 carry the 2
triple the 5, add it to 9 and the remainder --->6 carry the 2
triple the 1, add it to 5 and the remainder --->0 carry the 1
get the first number, add the remainder --->2

2067

If you are familiar with the 12's trick, it's the same but instead of doubling, its tripling... works for any "teen" number

[PatriotAlanis]

[Kurt]

Ooh, that trick has a name. I've always called it the "thirteen's trick" (or whichever "teen" I was using it for.)

It's hand to remember factors of 1001, 10101, etc. They appear on the test quite a lot, actually.

[zefuri]

uhhh what purpose do the ears serve...

[MistaPotta]

They're zeroes. They are place-holders so the pattern continues infinitely.

He could have said [unparseable or potentially dangerous latex formula], but then his answer would have too many leading and trailing zeroes.

~MP

[zefuri]

worlow, did jmal come up with that name?

[stupidityismygam]

nope, coach thornton called it the mickey mouse trick :) :) :)

[Englandude]

13*539

Divide 539 by 11. You get 49.

Now you have 11*13*49

11*13= 143

Now we have the 143 trick. divide 49 by 7 and multiply by 1001.

Answer=7007

Is that how yall would do it?

[Kurt]

Yeah, it's probably helpful to memorize your multiples of 143...they show up on the test so often.

That being said, I haven't done that. :P

On 13*539, I'd get to the '07' part using the 13's trick probably, and then realize I should have seen the factor of 143 and just write the '70' part then, lol.

[Englandude]

=D LoL

[zefuri]

didnt thorton quit/give up because of doerres lack of...iono how to say it

[stupidityismygam]

he is retiring this year...and yes doerre has become moronic now...

our infinite supply of smart white people has died (strange how infinite always seems to end)...

[zefuri]

lol you need william and diana to make children to get more asians lol