3 Consecutive integers will always be a multiple of 3...

pkw209 wrote:does the same hold true for every number?

In other words, will 5 consecutive integers always be a multiple of 5, will 100 consecutive integers always be a multiple of 100?

Thanks!

Its simple:-
let the no.s be x,x+1,x+2,x+3,x+4
so,sum = 5x + 4*5/2 = 5x + 2*5 which is divisible by 5.

Just tweeking harsh,champ's example.
If the no of integers is odd then , yes
If there are odd number of interger take the middle value as x
so in case of 5 integers the series will be
x-2,x-1,x,x+1,x+2 it is obvious that the sum is 5x which is clearly divisible by 5
in case of even no of elements say 4 the series is
x-1,x,x-1,x+2 sum is 4x+2/4 which will leave a remainder 2

n case of even no of elements say 4 the series is
x-1,x,x-1,x+2 sum is 4x+2/4 which will always leave a remainder 2

4x+ "2/4"..? plz explain


Thanks!

x-1,x,x-1,x+2 sum is (4x+2) /4 or the dividend (4x+2) when divided by 4 will have a quotient x and remaninder 2

Take the no 4,5,6,7 = (5*4) + 2 or 22/4 = quotient 5 and remainder 2

if we take 6 nos x-2,x-1,x,x+1,x+2,x+3 sum is (6x+3)/6 remainder is 3
8 nos (8x+4)/8 remainder is 4

I have edited the word always leave a remainder in the previous post which is an error.

the key is POSITIVE consecutive integers. 4 consec. positive integers will always have a mult of 4.

However, x consec. integers will not necessarily have a mult of x. Ex. 0,1,2,3 (4 integers) none of which are multiples of 4.