We call a two-digit number powerless if none of its digits can be written as an integer to a power greater than one. For example, 53 is powerless, but 54 is not powerless since 4 equals 2 squared. Which of the following is a common divisor of the smallest and the largest powerless numbers? So let's first think about how we might build our powerless numbers. So we know that 0 is 0 squared, 1 is 1 squared, 4 is 2 squared, 8 is 2 cubed, and 9 equals 3 squared. Therefore, these are all powerful numbers in a sense. Powerful digits, I should say. So then when we want to build a powerless number, we can't use any of these five digits. Then our only digits that we can use are 2, 3, 5, 6, and 7. To make the smallest powerless number, it would be best if we took the smallest possible digits possible. And in this case, that would be 22. In the same way, we want to take the biggest digit possible, which is 7, and we make 77. Clearly, the only of our answer choices that is a common divisor of both of these numbers is 11. So our correct answer is 11d.

powerless numbers

common divisor

two-digit numbers

eligible digits

mathematics