Question

How many 2 digits are divisible by 6?

Answer 1

Hint: In this question use the concept that 2 digits are formed from 10 till 99 only, and from this the minimum 2 digits number is 10, but 10 is not divisible by 6 so find the minimum number of two digit which is divisible by 6, if we keep on adding 6 to this then a arithmetic progression series if formed. The last number of this series is known that is 99, so directly use the formula for the nth term of an A.P to get the value of n.

Complete step by step answer:
As we know that the two digit number is from (10 to 99).
So the first two digit number which is divisible by 6 is 12, second two digit number which is divisible by 6 is (12 + 6 = 18) and the last two digit number which is divisible by 6 is 96.
So the sequence of two digit numbers which is divisible by 6 is
12, 18, 24, .................. 96.
Now as we see this follows the trend of an arithmetic progression (A.P) whose first term (a1) = 12, common difference (d) = (18 – 12) = (24 – 18) = 6 and the last term (an) = 96.
Now as we know that the last term or nth term of an A.P is calculated as
$\Rightarrow a_n=a_1+\left(n-1\right)d$ where symbols have their usual meanings and n is the number of terms.
So substitute all the values in this equation we have,
$\Rightarrow 96=12+\left(n-1\right)6$
Now simplify the above equation we have,
$\Rightarrow 96-12=\left(n-1\right)6$
$\Rightarrow 6\left(n-1\right)=84$
Now divide by 6 we have
$\Rightarrow n={\displaystyle \frac{84}{6}}+1$
$\Rightarrow n=14+1=15$
So the number of two digit numbers which are divisible by 6 are 15.
So this is the required answer.

Note – An arithmetic series is one in which the common difference that is the difference between the two consecutive terms remains constant throughout. It is advised to grasp all the series related formula either it be of sum of n terms or the nth term. There is another series which is mostly formed that is geometric progression, in this series the common ratio remains constant throughout the series.