How many positive four-digit integers N have the property that the three-digit number obtained by removing the leftmost digit is equal to N24?
Let N = 1000 a + 100 b + 10 c + d. Then, we want to have the relationship: (1000 a + 100 b + 10 c + d)/24 = 100 b + 10 c + d.
Let M = 100 b + 10 c + d. It follows that N = 1000 a + M. We also know M = N/24. Thus, we have M = (1000 a + M)/24. Solving for M, we have M = 1000a/23. Now, no matter what integer from 1-9 a takes on, M will never be an integer. Therefore there are 0 positive integers N that has the property stated.