Given that B=6 and E = 8 ‌‌‌‌‌A‌‌6‌‌‌C ×‌‌‌‌‌‌D‌‌‌8 –––––––––– A‌‌C‌‌‌F‌‌6 8‌‌A‌‌‌G‌‌‌0 –––––––––– F‌‌H‌‌‌F‌‌‌6 –––––––––– As at the Thousand’s place A+8=F, there is no carry over, thus the value of A can either be 0 or 1, but 0 will make it a 3-digit number. Hence, the value of A=1 and thus that of F = 9. Also, in the first multiplication, E×C=B, i.e., 8×C=6, So ‘C’ can either be 2 or 7. By hit and trial, we find that C=2. Therefore, ‌‌‌‌‌‌‌‌1‌‌6‌‌‌2 ‌‌×‌‌‌‌‌‌D‌‌‌8 –––––––––– ‌‌1‌‌‌2‌‌‌‌9‌‌‌6 +8‌‌1‌‌‌G‌‌‌0 –––––––––– ‌‌‌9‌‌H‌‌‌9‌‌‌6 –––––––––– As 9+G=9, therefore, G should be ‘0’ which can be obtained when D=5, thus the difference between D & F isF–D=9–5=4