Question

2013 Related Questions

Answer 1

type any question related to the number 2013 then we will solve it together

Answer 2

\[\int\limits_{}^{}e^{x^{2013}} dx\]lol jk

Answer 3

\[\fbox{1} \]What is the remainder when \[1^2+2^2+...+2013^2\] is divided by 7

Answer 4

lol @Euler271 i meant solveable questions,lol that one will kill me cos its not elemtntary function

Answer 5

@Jonask xD

Answer 6

lol

Answer 7

for now dont post anything before the current question is solved,thank you

Answer 8

\[1^2+2^2+\cdots+2013^2\]\[=\frac16\times2013\times2014\times4027\]\[=671\times1007\times4027\]\[\equiv(-1)\times(-1)\times2\hspace{50pt}(\mbox{mod }7)\]\[=2\]

Answer 9

ok so we used\[\sum i^2=\frac{n(n+1)(2n+1)}{6}\]
awesome thanks @kc_kennylau

Answer 10

no problem :)

Answer 11

For positive integers n, let the numbers \[c(n)\] be determined by the rules \[c(1)=1,c(2n)=c(n)\] and \[c(2n+1)=(-1)^nc(n)\]. Find the value of
\[ \sum_{n=1}^{2013}c(n)c(n+2). \]

Answer 12

basically it means for even numbers we have a number then for odd we have its negetive,mening most stuff cancel?

Answer 13

\[c(1)=1\]\[c(2)=c(1)=1\]\[c(3)=-c(1)=-1\]\[c(4)=c(2)=1\]\[c(5)=c(2)=1
\]\[c(6)=c(3)=-1\]\[c(7)=-c(3)=1\]lemme think about this later

Answer 14

http://math.unt.edu/problem-of-the-month

scroll down to October 2013.

Answer 15

wow excellent solutions here
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3315676&sid=5ecb28f21a08d6171eae51920be551fd#p3315676

Answer 16

\[\fbox{3}\]A positive integer n is such that n(n+2013) is a perfect square.

Answer 17

a) Show that n cannot be prime.

Answer 18

lolz i don't know how to do i'm so dumb

Answer 19

n(n+2013) = k^2
n^2 + 2013n - k^2 = 0

by using quadratic formula, we get n :
n = (-2013 + sqrt(2013^2 - 4(1)(-k^2))) /2
n = (-2013 + sqrt(2013^2 + 4k^2)) /2

see n must be positive integer, therefore
4k^2 + 2013^2 = m^2
m^2 - (2k)^2 = 2013^2
(m+2k)(m-2k) = (3 * 11 * 61)^2

so far so good ? :)

Answer 20

this way i get from you, @mukushla :)