Question 524261
1- One row consists of three consecutive digits in ascending order reading from left to right.
2-F is not a prime number.
3-H is larger than M & both H & M are divisible by I, which is not 1.
4-N is divisible by both J & L, neither of which is 1
<pre>
F=? G=? H=?    
I=? J=? K=?    
L=? M=? N=?    

We first look at clue 4.
</pre>
4-N is divisible by both J & L, neither of which is 1
<pre>
We observe that:
6 is divisible by both 2 and 3
8 is divisible by both 2 and 4

6 and 8 are the only digits which are divisible by two digits other than 1,
(or themselves), thus:  

(i)  6 and 8 are the only candidates for N.

Since in either case, 2 is one of the divisors, we know that 

(ii)  either J or L must be 2. 

Next we look at clue 3:
</pre>
3-H is larger than M & both H & M are divisible by I, which is not 1.
<pre>
I can't be 2 because of (ii).  So we look for a digit for I other
than 2 (or 1) which two other digits are divisible by.  There is only
one possibility, 6 and 9 are divisible by 3, so 

(iii) I must be 3.   

That makes H & M be 6 & 9 and since H is larger than M, 

(iv)  H=9 and M=6   

Now since M=6, (i) tells us that 

(v)   N=8

F=? G=? H=9    
I=3 J=? K=?    
L=? M=6 N=8

Now we look at clue 1
</pre>
1- One row consists of three consecutive digits in ascending order reading from left to right.
<pre>
That can't be the bottom row because 6 and 8 are not consecutive.
It can't be the top row because G can't be 8 since N is 8. Therefore
the row that contain the consecutive digits in ascending order is the
middle row, so we have J=4 and K=5.  Now we have:

F=? G=? H=9    
I=3 J=4 K=5    
L=? M=6 N=8

By (ii) above L=2 since J=4, so we have:

F=? G=? H=9    
I=3 J=4 K=5    
L=2 M=6 N=8

We have now used every digit but 1 and 7.

So we look at clue 2: 
</pre>
2-F is not a prime number.
<pre>  
7 is prime, and 1 is not, so F=1 and G=7.

Final solution:

F=1 G=7 H=9    
I=3 J=4 K=5    
L=2 M=6 N=8

Edwin</pre>