/sci/ - Science & Math

[Anonymous No.2226354]

[Anonymous No.2226359]

>implying number = digit
Reword the question to something coherent.

[Anonymous No.2226357]

0

[Anonymous No.2226364]

Within about 0.5 seconds it becomes obvious that the last digit is 0.

[Anonymous No.2226363]

lol y so fail?

[:D !!s2uh28knASO No.2226378]

>>2226364
why?

[Anonymous No.2226390]

>>2226378

>I'm a retard

[Anonymous No.2226394]

This number is too large to have a last digit.

[Anonymous No.2226399]

If it takes you anywhere near 30 seconds to answer that and you go to MIT you are probably a Communications Major.

[Anonymous No.2226401]

>>2226378

Because.

[:D !!s2uh28knASO No.2226410]

>>2226364


awesome way to give an actual answer.. but nm i already figure it out

[Anonymous No.2226408]

>What is the sum of all the natural numbers from 1 to 100?

>A third grader answered this in under 30 seconds

[Anonymous No.2226405]

too big lol no answer

[Josef !!nUf2NflSAyw No.2226412]

>>2226394 This number is too large to have a last digit.
I beg to differ.
http://pastebay.com/112328

[Anonymous No.2226452]

>>2226408

That takes about 3 seconds if you know the formula for the sum of an arithmetic series, regardless of age.

[Anonymous No.2226490]

>>2226412
what language?

[Anonymous No.2226491]

>>2226452
>implying it should take you longer than 3 seconds to figure out the formula

[Anonymous No.2226499]

>>2226354
I CAN INTO WOLFRAM APHA IN 30 sek.

http://www.wolframalpha.com/input/?i=last+number%282010^2010%29

U JELLY?

[Anonymous No.2226520]

you can easily predict the last 2011 digits...

[Anonymous No.2226527]

>>2226378
2010^ 2010 = 2010 * some number = 10*some number

[Josef !!nUf2NflSAyw No.2226536]

>>2226490
Mathematica, Ctrl+A, Ctrl+C, pastebay, Ctrl+V. :)

Apart from that, writing a long int class is a good beginner programming exercise.

[Anonymous No.2226548]

>>2226536
wow you need mathematica for this?

what a fucking engineer.

That number is obviously divisible by 10. More generally:

<span class="math"> a^b \mod n = (a \mod n)^(b \mod \phi(n)) \mod n[/spoiler]
where <span class="math">\phi[/spoiler] is the euler totient function

[Anonymous No.2226556]

Fucking retards look.

2010^2010 = 0^0 = 0

[Anonymous No.2226574]

n = str(2010**2010)
print n[-1]
0

[Anonymous No.2226581]

>>2226408
but that was Gauss and he was a fricken genius!

[Anonymous No.2226585]

>>> 2010**2010
...000000000000000000000000000000

[Anonymous No.2226592]

You guys are all morans first of all the 2010 cross out leaving the triangle thing(Lol we dont know what the hell it does so we throw it away) leaving 0.

[Anonymous No.2226593]

>>2226556
not shure if serious...

if y: fullretad.jpeg

if n: 5/10

[Anonymous No.2226594]

>>2226585
umad?

[Anonymous No.2226603]

9

[Anonymous No.2226673]

So, how do you know that the last number is 0? What procedure did you follow?

[Anonymous No.2226684]

>>2226673

2010^2010
=2010*(2010^2009)
=201*10*(2010^2009)

So it is divisible by 10 and therefore ust end in zero.

Yay :)

[Anonymous No.2226690]

You are all retarded if you don't realize within 20 milliseconds that the last 2010 digits are 0.

[Anonymous No.2226691]

>>2226673
first number is divisible by 10.

any number divisible by 10, multiplied by any whole number, is also divisible by 10.

numbers divisible by 10 end in zero.

you could prove a whole lot more, like the last 2010 digits are zero

[Anonymous No.2226717]

>>2226673

In a more general case e.g: 2008^2008 I might find 2008^1, then 2008^2 etc and look for a pattern.

[Anonymous No.2226727]

2010^2010 is just 2010 multiplied by itself 2010 times.
2010x2010x2010x2010x2010....

The list digit must therefore be zero.

[Anonymous No.2226729]

>>2226364
for MIT student it takes about 29 second to translate question to wapanese.

[Anonymous No.2226731]

>>2226684
>>2226691
Thanks, /sci/.

[Anonymous No.2226738]

>>2226717
see
>>2226548


<span class="math"> 2008^2008 mod 10 = 8^(2008 mod 4) mod 10 = 8^(0) mod 10 = 1[/spoiler]