Math word problem show work?

Question

If the eggs in a basket are removed two at a time, one egg will remain. If the eggs are removed three at a time, two eggs will remain. If the eggs are removed four, five, or six at a time, then three, four, and five eggs will remain, respectively. If they are taken out seven at a time, however, no eggs will be left over. Find the smallest number of eggs that could be in the basket.

Answer 1

let say the number = x

x/2 = n1 + 1/2 ---(1)
x/3 = n2 + 2/3 ---(2)
x/4 = n3 + 3/4 ---(3)
x/5 = n4 + 4/5 ---(4)
x/6 = n5 + 5/6 ---(5)
x/7 = n6 --(7)
where n1, n2, n3, n4, n5, n6 is integer.

from (3)
x/4 = n3 + 3/4
x/2 = 2 x n3 + 3/2
x/2 = (2 x n3 + 1) + 1/2 = n1 + 1/2 (same as (1))
thus, if number x divided by 4 remaining 3,
the number x is also divided by 2 remaining 1.

from (5)
x/6 = n5 + 5/6
x/3 = 2 x n5 + 5/3
x/3 = (2 x n5 + 1) + 2/3 = n2 + 2/3 (same as (2))
thus, if number x divided by 6 remaining 5,
the number x is also divided by 3 remaining 2.

so, smallest x is common factor of 4,5,6 less 1

smallest number x = 4 x 5 x 6 - 1 = 119

from (7),
119/7 = 17 (integer)

So, smallest number x = 119

Answer 2

We're told at the end that seven divides the number evenly. We're also told that the number can't be multiple of 2, 3, 4, 5 or 6. It can't be 7, because if we took 6 out of the basket we'd have one egg left over, not five. So try with the next multiple of 7 that isn't a multiple of the other mentioned numbers: 7*7 = 49 (we can't use any of the prime numbers less than 7 because we know they won't work). But 49 doesn't work with the division by 3.

Since the number divided by 5 gives us a remainder of 4, then the number must end in 4 or 9. So 7*11 and 7*13 won't work. Try 7*17 = 119. This seems to work with the other conditions ( [119 -1]/2 = an integer, as is [119-2]/3, [119-3]/4, [119-4]/5, and [119-5]/6).

Answer 3

In other words, if you add 1 egg to the basket, you would have a number which is divisible by 2,3,4,5,6

Right now the number is divisible by 7.

N + 1 = multiple of 6*5*2 = 60
N could be 59, 119 and so on.
119 is the smallest value which is divisible by 7.

Answer = 119

Answer 4

Since the remainders are all 1 less than the divisors, we need a number that is 1 less than the common multiple of 2,3,4,5, and 6, which is 60, and the number must also be a multiple of 7.

So try 59, which doesn't divide by 7, then 119, which does, and we have the answer on the second try: 119

Answer 5

119

119/2 remainder 1
119/3 remainder 2
119/4 remainder 3
119/5 remainder 4
119/6 remainder 5
119/7 remainder 0