How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?
Correct Answer: 70
Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.
c-a = 2.
Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.
Thus, there can be 7 cases.
B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be $$7\times\ 10=70$$.
Create a FREE account and get: