What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

$$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr} $$
Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39