X is the smallest positive integer greater than 100 such that, if its leftmost digit is deleted, the resulting integer is X19. How is it divisible by 2 prime numbers?

Hi Om, the resultant number is X/19.

So, the part that is subtracted ie., the leftmost digit * some power of 10 should be a multiple of 18, so that that part plus the remainder (X/19) as a whole is divisible by 19.

So, the leftmost number of X * power of 10 = multiple of 18.
The only possible value for the leftmost digit is 9. The number is 950.
So, X = $$2*5^2*19$$ => it is divisible by 3 prime numbers. So, the statement that X is divisible by 2 prime numbers is false.