mathematics - How many diamonds?

A group of Seven men robbed a diamond shop at night. They ran to a nearby forest and all slept there for the night.

1. One of them woke up and tried to run away with the bag full of diamonds, but another one woke up at the same time and caught him.


2. They both decided to divide the diamonds in half and run away, but when they divided them, they got one diamond as a remainder.


3. So they woke up the third man. They again divided the diamonds and got one as a remainder.



4. They woke up the fourth, fifth, and sixth men one by one, each time dividing the diamonds equally and getting exactly one diamond as a remainder.


5. But when they woke up the last man, all the diamonds were divided equally and completely.

Since there are multiple solution for this, how many lowest possible number [greater than 0] of diamonds were there?

Answer

Let's say $N$ is the number of diamonds:
After the first 6 woke up we can deduce:

$N \mod 2 = N \mod 3 = ... = N \mod 6 = 1$

This means that

$N - 1$ is a multiple of the smallest common multiple of $2,3,4,5,6$.
So $N - 1$ is a multiple of 60.

After the 7th wakes up we deduce...

$N$ is a multiple of 7.

So

we need the smallest number $N$ a multiple of 7 and $N-1$ a multiple of 60.

Starting at 1...wrong because it's not a multiple of 7.
61... wrong
121...Neh
181...Nope again
241...keep looking
301...Winner.