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Question
Find, how many two digit natural numbers are divisible by 7.
Solution
The first two-digit number that is divisible by 7 is 14
So, the sequence starts from 14
And the highest two-digit number that is divisible by 7 is 98
So, the sequence becomes:
14, 21, ..., 98.
We, need to find the numbers in the given sequence
Using an = a + (n - 1)d
a is the first term, d is the common difference, n is the number of terms and is the nth term and an is the nth term
a = 14, d = 7, an= 98 substituting the values we get
98 = 14 + (n - 1)7
98 = 14 + 7n - 7
91 = 7n
n = `91/7`
n = 13
Therefore, there are total 13 two digit numbers that are divisible by 7.
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