Continue the search inside a DO loop - if-statement

Question

I wrote a simple FORTRAN code in order to do the following: assume we have to integer numbers n1 and n2 which have common divisors. For example 3 and 6 both are divided by 3. Here is the code
PROGRAM test
INTEGER i,n1,n2
WRITE(*,*)' Please enter two numbers: '
READ(*,*)n1,n2
DO i=2,10,1
IF(MOD(n1,i).EQ.0.AND.MOD(n2,i).EQ.0)THEN
n1=n1/i
n2=n2/i
ENDIF
n1=n1
n2=n2
ENDDO
WRITE(*,*)n1,n2
PAUSE
END
This works fine for the example (3,6). However, there are cases like (4,8) in which the numbers have more than one common divisor, in this case 2 and 4. Another example (16,24). I want to compute the maximum common divisor of the two numbers and then reduce them (i.e. 3,6 to 1 and 2), but the code returns the first one (4,8 returns to 2, 4 instead of 1,2). How should it be modified in order to calculate the maximum divisor?
Many thanks in advance!

Answer 1

You could stay with an i, till your if-statement is false.
In other words:
If a number can be divided by i, then don't immediately go to i+1, but try to divide by i again.
EDIT: I think the easiest way is to use a DO WHILE-loop. To calculate the divisor, you have to multiply all your i.
gcd = 1
DO i=2,10,1
DO WHILE (MOD(n1,i).EQ.0.AND.MOD(n2,i).EQ.0)
n1=n1/i
n2=n2/i
gcd = gcd * i
ENDDO
ENDDO
WRITE(*,*) gcd

Answer 2

What you are looking for is the greatest common divisor. You may do this:
function gcd(a, b)
implicit none
integer a, b, aa, bb, cc, gcd
aa = abs(a)
bb = abs(b)
do while (bb .ne. 0)
cc = mod(aa, bb)
aa = bb
bb = cc
end do
gcd = aa
end
Note: it is written in Fortran 77 + MIL-STD-1753 (for the DO WHILE construct and IMPLICIT NONE).