There are 10 integers between 28 and 98 that are exactly divisible by 7.
Understanding Divisibility
Divisibility refers to whether a number can be divided by another number without leaving a remainder. In this case, we're looking for numbers divisible by 7, meaning they leave no remainder when divided by 7.
Finding the Solution
The provided references state that the numbers between 1 and 100 divisible by 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
To answer the question, we need to consider only those numbers between 28 and 98. Let's examine the list:
- 28 is divisible by 7, but it's the lower limit, so we exclude it from our count.
- 98 is divisible by 7, but it's the upper limit, so we include it.
The numbers divisible by 7 between 28 and 98 are: 35, 42, 49, 56, 63, 70, 77, 84, 91, 98. Therefore, there are 10 such numbers.
Alternative Approach: Using Arithmetic Sequences
We can also solve this using the concept of arithmetic sequences. The sequence of numbers divisible by 7 starts at 35 and ends at 98. The common difference is 7.
The formula for the nth term of an arithmetic sequence is: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the number of terms, and d is the common difference.
In our case:
- an = 98
- a1 = 35
- d = 7
Solving for n:
98 = 35 + (n-1)7
63 = (n-1)7
9 = n-1
n = 10
Therefore, there are 10 numbers between 28 and 98 divisible by 7.
