Question

find the sum of smallest and highest power of 13 which is completely divisible by 100?
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Answer 1

Answer:

The smallest and the largest numbers of three digits, which are divisible by 13 are 104 and 988 respectively.

So, the sequence of three digit numbers which are divisible by 13 are 104,117,130,...,988.

It is an AP with first term a=104 and common difference d=13.

Let there be n terms in this sequence. Then,

a

n

=988⇒a+(n−1)d=988⇒104+(n−1)×13=988⇒n=69

Now, Required sum =

2

n

[2a+(n−1)d]

=

2

69

[2×104+(69−1)×13]

=

2

69

[1092]

=69×546=37674