dagmat wrote:

My 2 cents:

Since 18 and 70 are included, the range of numbers allowed is (70-18) + 1. This becomes 53.

Next, divide 53/7, we get 7*7=49 as the closest. Hence only 7 numbers are present.

Similarly for 9, 53/9, we get 9*5=45 as the closest. Hence only 5 numbers are present.

63 is the first number divisible by both 9&7, but its outside the range. So total numbers = 7+5=12 (B). This method is similar to iPinnacle's method but I think its a bit faster.

Hope this helps...

this method would not work in all situations

eg : had the no. range been 19-69 inclusive then by this method the ans would still have been 12 whereas the correct ans is 10