Question

i) How many two digit numbers are divisible by 13?
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Answer 1

Step-by-step explanation:

The first two digit number divisible by 13 is 13, the next two digit number divisible by 13 is 26 and next is 39 and so on. The last two digit number divisible by 13 is 91. Therefore, the required sequence is:

13,26,39,.......91

From the above sequence, we get that the first term is a

1

=13,a

2

=26 and the n

th

term is T

n

=91

Now, we find the common difference as follows:

d=a

2

−a

1

=26−13=13

We know that the general term of an arithmetic progression with first term a and common difference d is T

n

=a+(n−1)d

To find the number of terms of the A.P, substitute a=13, T

n

=91 and d=6 in T

n

=a+(n−1)d as follows:

T

n

=a+(n−1)d

⇒91=13+(n−1)13

⇒91=13+13n−13

⇒13n=91

⇒n=

13

91

⇒n=7

Hence, there are 7 two digit numbers which are divisible by 13.

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