Answer
a) 7 and Integers are 56, 63, 70, 77, 84, 91 , and 98.
b) 5 and Integers are 55, 66, 77, 88, and 99.
c) 1 and only integer is 77
Work Step by Step
Neither 50 nor 100 is divisible by either 7 or 11.
a)There are $\frac{100}{7}$ (floor function) = 14 integers less than 100 that are divisible by 7, and $\frac{50}{7}$ (floor function)= 7 of them are less than 50 as well. This leaves 14 - 7 = 7 numbers between 50 and 100 that are divisible by 7. They are 56, 63, 70, 77, 84, 91 , and 98.
b)There are $\frac{100}{11}$ (floor function)= 9 integers less than 100 that are divisible by 11, and $\frac{50}{11}$ (floor function) = 4 of them are less than 50 as well. This leaves 9 - 4 = 5 numbers between 50 and 100 that are divisible by 11. They are 55, 66, 77, 88, and 99.
c) A number is divisible by both 7 and 11 if and only if it is divisible by their least common multiple, which is 77. Obviously there is only one such number between 50 and 100, namely 77.
Another way to do this is as we did in previous parts:
$\frac{100}{77}$ (floor function) - $\frac{50}{77}$ (floor function) = 1 - 0 = 1.
Precisely we are looking for the intersection of the sets found in parts a and b.
