SOLUTION: When a band marches in rows of 2,3,4,5 or 6 there is always one member left over. What is the least number of members in the band if there are no members left over when marching

Let the number be n.
Then:

n must be 1 more than a multiple of 2
n must be 1 more than a multiple of 3
n must be 1 more than a multiple of 4
n must be 1 more than a multiple of 5
n must be 1 more than a multiple of 6
n must be a multiple of 7

So there is some positive integer p such that n = 7p

The least common multiple of 2,3,4,5,6 is 60

So n must be 1 more than a multiple of 60.

So there is some positive integer q such that n = 60q+1

So we have the Diophantine equation:

7p = 60q + 1

Write 60 in terms of its nearest multiple of 7, which is 63

7p = (63-3)q + 1

7p = 63q - 3q + 1

Divide through by 7

p = 9q -3q/7 + 1/7

3q/7 - 1/7 = 9q - p

The right side is an integer so the left side must equal
that same integer.  Suppose that integer is A:

3q/7 - 1/7 = A;  9q - p = A

3q - 1 = 7A

Write 7 in terms of its nearest multiple of 3 which is 6.

3q - 1 = (6+1)A

3q - 1 = 6A + A

Divide through by 3

q - 1/3 = 2A + A/3

q - 2A = A/3 + 1/3

The left side is an integer so the right side must equal
that same integer.  Suppose that integer is B:

A/3 + 1/3 = B;    q - 2A = B

A + 1 = 3B

A = 3B - 1

q - 2A = B

q - 2(3B - 1) = B

q - 6B + 2 = B

q = 7B - 2

9q - p = A

9(7B - 2) - p = 3B - 1 

63B - 18 - p = 3B - 1
60B - 17 = p

Since n = 7p
      n = 7(60B-17)
      n = 420B-119

So the smallest positive value of n is when B=1

      n = 420(1)-119
      n = 301  

Edwin