good...

What is the largest positive integer n so that n! can be expressed as the product of n - 3 consecutive positive integers?

18 Answers

62
Lokesh Verma ·

6! = 720 = 9x8x10

7! = !! (Dont know how to write it as product of 3 integers..(

I have no proof.. but I think 6!

24
eureka123 ·

i dont have the answers dude...

9
Celestine preetham ·

eureka
chk ur posts and finish of ur other open recent threads to a close pls

9
Celestine preetham ·

no byah i din do by trial and error :)

62
Lokesh Verma ·

oh sorry celestine

I started to take n! = product of 3 consecutive :(

62
Lokesh Verma ·

wow celestine.. 23.. how did you reach that far.. my mind stopped working beyond 6 ;)

9
Celestine preetham ·

argument is based on logic not very convincing

u need to replace 4 terms by 1 term in n!

so replacing lower order to n+1 will give max n

ie 1.2.3.4 =24

9
Celestine preetham ·

see 23 ! bhaiyya

23! = 5.6.......23.24 20 terms

11
Mani Pal Singh ·

as the product of n - 3 consecutive positive integers?
sir aapki counting mein 3x4x5 nahi aana chayiye kya??
please clear it!!!

11
Mani Pal Singh ·

FIVE
PAANCH
PANJ
5

11
Mani Pal Singh ·

[11][11][11][11][11]

provide ur arguemeny
i am in sink with #2
give ur justification[1]

9
Celestine preetham ·

23 ????? its easy to verify but argument is long

11
Mani Pal Singh ·

tell us wats the answer dude!!!!!!

1
Thedarkknight ·

Wats the answer?

1
Thedarkknight ·

is it 6

11
Mani Pal Singh ·

the 2 options r
4 and 5
when we have 4 we have 4X3X2X1 =24
and incase of 5
we have 5!=5X4X3X2X1=5X4X3X2

24
eureka123 ·

ya how??

1
Philip Calvert ·

HOW [11]

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