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