Question

@ikram002p

Find all positive integers \(n\) such that
\(1! + 2! + … + n! | (n+1)! \)

Answer 1

False

Answer 2

nice

Answer 3

this are all consecutive integers from
\[(n+1)!+1\text{ to }(n+1)!+n+1\]

Answer 4

this are all consecutive intergers from\[(n+1)!+1\]
\[(n+1)!+2\\(n+1)!+3\\.\\.\\.\\(n+1)!+n+1\]

Answer 5

its funny cos i am working on the same problem a few seconds ago....

Answer 6

oh ive tried this few days back and gave up... looks ikram also worked on it today :)

Answer 7

my version says prove for any integer n there exists a set of consecutive intergers such that none are prime
so we se \[(n+1)!+2\] is divisible by 2
\[(n+1)!+3\] is divisible by 3

.......
\[(n+1)!+n+1\] is divisible by n+1
@ganeshie8 ealier i startded with 1 instead of 2...typo,here \[n\ge 1\]

Answer 8

sry \[n>1\]

Answer 9

try to make conjecture from it by doing ittaration
n sum n! (n+1)!
1 1 2
2 3 6
3 9 24
4 33 120
so 4th ittaration faild n<5
@Enas_sh see this

Answer 10

Nice :) so how to conclude it olly works for n = 1, 2 ?

Answer 11

by induction

Answer 12

interesting... ive seen proofs to prove something using induction,
never worked a proof to disprove @myininaya

Answer 13

so keep continuing ittaration from n =4, coz by induction its clozed way u cant do that

Answer 14

http://openstudy.com/users/myininaya#/updates/5276cbf2e4b0275c53855b4a

Answer 15

n sum n! (n+1)!
4 33 120 33 is not multiple from 120
5 33+120 120(6)
6 153+720 720(7)
7 873+5040 5040(8)
.
mmmm still work on on it ... i can see that we need congrunce....

Answer 16

thanks @ikram002p @myininaya @Jonask ...
@myininaya 's proof by induction in that link makes sense to me ! sorry i took bit long to digest... it was really mouthful to me :o

Answer 17

please help me in maths quesiton