There are 127 numbers divisible by 7 between 105 and 1000.
To determine the number of integers divisible by 7 between 105 and 1000, we need to find the first multiple of 7 greater than 105 and the last multiple of 7 less than 1000.
First, let's find the first multiple of 7 greater than 105. Dividing 105 by 7 gives us 15, so 105 = 7 15. We are looking for numbers between 105 and 1000, so 105 is not included. The next multiple of 7 is 7 16 = 112.
Next, let's find the last multiple of 7 less than 1000. Dividing 1000 by 7 gives us approximately 142.86. So, we take the integer part, 142, and multiply it by 7: 7 * 142 = 994.
Now, we have the first term (a = 112) and the last term (l = 994) of an arithmetic progression with a common difference (d = 7). We can use the formula for the nth term of an arithmetic progression:
l = a + (n - 1)d
where:
- l = last term (994)
- a = first term (112)
- n = number of terms (what we want to find)
- d = common difference (7)
Plugging in the values:
994 = 112 + (n - 1)7
994 - 112 = (n - 1)7
882 = (n - 1)7
882 / 7 = n - 1
126 = n - 1
n = 126 + 1
n = 127
Therefore, there are 127 numbers divisible by 7 between 105 and 1000.
