Puzzle: Unique equal distribution among 7 friends?

Question

There are seven friends, They steal diamonds from a diamond merchant and run away in forest. While running, night sets in and they decide to rest in the jungle

When everybody’s sleeping, two of the best friends get up and decide to distribute the diamonds among themselves and run away. So they start distributing but find that one diamond was extra. So they decide to wake up 3rd one and divide the diamonds again …..only to their surprise they still find one diamond extra. So they decide to wake up fourth one. Again one diamond is spare. 5th woken up……still one extra. 6th still one extra. Now they wake up 7th and diamonds are distributed equally.

Now the question is how many diamond they stolen?

Answer 1

Lets say there are x diamonds, now these diamonds are exactly divisible by 7.
and
x = 1 + N1*2;
x = 1 + N2*3;
x = 1 + N3*4;
x = 1 + N4*5;
x = 1 + N5*6;
x = N6*7;
where N1, N2, N3, N4 and N5 are integers.
From above we can also say
N1*2 = N2*3 = N3*4= N4*5 = N5*6 = y
Now y should be divisible by 2, 3, 4, 5 and 6, its nothing but common multiple of all.

LCM of 2, 3, 2*2, 5, 2*3 => 2*3*2*5 => 60

at the same time common multiple + 1 should be divisible by 7 as well.

60 + 1 is not divisible by 7
120(60*2) + 1 is not divisible by 7
180(60*3) + 1 is not divisible by 7
240(60*4) + 1 is not divisible by 7
300(60*5) + 1 is divisible by 7

Thus they must have stolen minimum 301 diamonds.

Answer 2

Answer should be 301

Answer 3

2 x 150 +1
3 x 100 + 1
4 x 75 +1
5 x 60 +1
6 x 50 + 1
7 x 43 exactly.