Hard maths question :( urgent 10 points

A man died and has 7 kids olive trees so the kids wanted to share the trees
5 traveled so the two others shared the trees and every kid had the same amount , but they left one tree without sharing it
the 3rd kid came back and they did it again everyone had the same amount but 1 extra tree was left
same thing when the 4th came back
the same for the 5th
the same for 6th
but for the 7th they all had the same amount with no extra trees!
what's the number of trees plz , it's so urgent , if u have time for a quick explanation post it , if not i need the solution fast.

the number must be odd (not divisible by 2) it is also not divisible by 5 but only one remains, so it must end in 1, it cannot end in 6
let the number = x
x is divisible by 7
x-1 must end in 0, be divisible by 3, 4 and 6

301

x = 1 mod 2
x = 1 mod 3
x = 1 mod 4
x = 1 mod 5
x = 1 mod 6
x = 0 mod 7

We can pare this list down a bit

x = 1 mod 4
x = 1 mod 5
x = 1 mod 6
x = 0 mod 7

and a bit more

x = 1 mod 5
x = 1 mod 12
x = 0 mod 7

lcm of 5, 12, and 7 = 420

420/5 = 84, 420/12 = 35, 420/7 = 60

34(84) - 85(35) + 2(60) = 1

mod(34*84, 420) = 336
.....mod(336,5) = 1
.....mod(336,12) = 0
.....mod(336,7) = 0

mod(-85*35, 420) = 385
.....mod(385, 5) = 0
.....mod(385, 12) = 1
.....mod(385, 7) = 0

mod(2(60), 420) = 120
.....mod(120, 5) = 0
.....mod(120, 12) = 0
.....mod(120, 7) = 1

1*(336) + 1*(385) + 0*(120) = 721

mod(721, 420) = 301

mod(301, 2) = 1
mod(301, 3) = 1
mod(301, 4) = 1
mod(301, 5) = 1
mod(301, 6) = 1
mod(301, 7) = 0

301 trees