Seven Robbers loot a bank and hide the coins at a deserted factory. They decide to divide the money equally next morning

Question

Seven Robbers loot the bank and hide the coins at a deserted factory. They decide to divide the money equally next morning. Two greedy robbers decided to cheat the others and reach the factory at night. They equally divided the coins between them, one coin left. So they called another robber and then they decided to divide equally among three. Sadly again one coin left. Same thing happened for 4th 5th and the 6th robber. However when 7th robber reached in the morning, they are able to divide the coins equally.

How many coins were there in total?

Answer 1

Let the number of coins be - N

Then, N - 1 is a multiple of 1, 2, 3, 4, 5, 6 but not 7. The LCM of 1, 2, 3, 4, 5, 6 is 60
Again N is a multiple of 7.
So, N should be any number which is a multiple of 7 and N - 1 is a multiple of 60. So upon finding the multiples of 60 and adding 1 to them we find 60 * 5 = 300 + 1 = 301 is such a number.

Therefore N = 301.

Answer 2

The solution has to be a multiple of 7 because the seven robbers have to divide the coins equally.
I have tried multiples of 7 until I reached 301 which complies allow the robbers to divide the coins equally.
The answer is 301 coins