The problem asks for the number of even integers in the range from 10 to 100 (inclusive) that are not divisible by 3.
We can solve this using the principle of inclusion-exclusion (or simply subtraction) by following these steps:
Step 1: Find the total number of even numbers in the range [10,100].
The even numbers are 10,12,14,...,100. This is an arithmetic progression. The number of terms can be found with the formula:
Count = {(Last Number−First Number)/Step} + 1
Total Even Numbers = {(100 - 10)/2} + 1 = 46
There are 46 even numbers in the range.
Step 2: Find the number of even numbers that are divisible by 3 in the range [10,100].
A number that is both even (divisible by 2) and divisible by 3 must be divisible by the least common multiple of 2 and 3, which is 6. We need to find the number of multiples of 6 in the range [10,100].
Let 6k be a multiple of 6. We are looking for integers k such that:
10 ≤ 6k ≤ 100
1.66 ≤ k ≤ 16.66
Since k must be an integer, the possible values for k are {2,3,4,...,16}.
The number of such multiples is:
Count=16−2+1=15
There are 15 even numbers (multiples of 6) that are divisible by 3 in the range.
Step 3: Find the number of even numbers that are not divisible by 3.
Subtract the number of even numbers that are divisible by 3 (from Step 2) from the total number of even numbers (from Step 1).
Even Numbers Not Divisible by 3 = Total Even Numbers − Even Numbers Divisible by 3
Even Numbers Not Divisible by 3 = 46 − 15 = 31
The final count is 31, which is option C.