Question

How many two digit numbers are divisible by 5 such that remainder is 1.

Answer 1

Hint: To solve the above question, we use the Arithmetic progression series formula but to apply it we require first, last terms and the difference between two consecutive terms, so we will fill out all these requirements first.
Formula used:
\[{A_n} = a + (n - 1)d\]

Complete step-by-step answer:
As per the question, we want to know how many two-digit numbers are divisible by 5 and leave 1 as the remainder. Which we can find out with the A.P. formula, but to apply the A.P. formula we need to find out the first, last two-digit numbers of the series, and the difference between two consecutive terms of the series.
We know that the first multiple which is divisible by 5 is 5, But it’s not a 2-digit number then We will take the second multiple of 5 is 10. It is a two-digit number and when we add 1 into it. It will become 11.
Similarly, the third multiple of 5 is 15, when we add 1 into it becomes 16.
Here, we have found the two two-digit numbers which are divisible by 5 and leave 1 as the remainder. Those are 11 and 16.
To find out the last two-digit number of the series, we will consider the last two-digit multiple of 5 that is 5, and add 1 into it, it becomes 96.
Below is the formula for A.P.
\[{A_n} = a + (n - 1)d\]
Here $ {A_n} $ is the $ {n^{th}} $ term of the series =96
The first term of the series: $ a $ =11
 The total number of terms: $ n $
The difference between two consecutive terms: $ d $ =5
We put all these values into the formula
 $ 96 = 11 + (n - 1)5 $
Simplify above equation
 $ 5(n - 1) = 85 $
 $ n - 1 = 17 \Rightarrow n = 18 $
So, the correct answer is “18”.

Note: An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence, here one may tend to take 15,20,... as sequence but it will be wrong , the reminder has to be 1 and hence should be taken accordingly.