This is the third part of the Solving Consecutive Number Problems Series. In this post, we solve more problems about consecutive numbers. We have already discussed four problems in the first part and second part of this series, so we start with the fifth example.
There are 3 consecutive odd numbers. Twice the smallest number is one more than the largest. What are the numbers?
In the first post in this series, we have learned that odd numbers increase by 3 (e.g. 7, 9, and 11). So, let
= the smallest odd number
= the second odd number
= the largest odd number.
Now that we have represented the numbers, we now go to the second sentence. The second sentence says that twice the smallest number is one more than the largest. The smallest number is , so twice the smallest number is . Now, is one more than the largest number . This means that if we add 1 to the largest number, then they will be equal. That is,
Solving, we have
So the consecutive numbers are 5, 7, and 9.
Check: Twice the smallest is 2(5) = 10 is one more than the largest which is 9. Therefore, our answers are correct.
There are three consecutive even integers. The sum of the first two integers is 16 more than the largest. What are the numbers?
As we have discussed in the previous posts, the representations of consecutive odd numbers and consecutive even numbers are the same. Consecutive even numbers such as 18, 20, 22 increase by 2 inch time.
= the smallest even number
= the second even number
= the largest even number
The second sentence states that the sum of the first two integers is 16 more than the largest. The sum of the first two integers is and the largest integer is . Since the sum of the first two integers is 16 more than the largest integer, if we add 16 to the largest integer, then they will be equal. That is,
Therefore, the consecutive integers are 18, 20, 22$.
Check: The sum of the first two integers 18 + 20 = 38 is 16 more than 22. Therefore, we are correct.
The video of this post can be watched on Youtube as shown below.
To know more about number problems, read the Solving Number Problems Series.