Question

In the sequence 20, 202, 2020, 20202, 202020, ... each subsequent number is obtained by adding the digit 2 or 0 to the previous number, alternately. Calculate the sum of the digits of the first 100 numbers of that sequence which are divisible by 202.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The sum of the digits of the first 100 numbers of the sequence which are divisible by 202 is 10200.

Given:

A sequence = 20, 202, 2020, 20202, ...

To Find:

The sum of the digits of the first 100 numbers of the sequence that are divisible by 202.

Solution:

We can solve the problem using the following mathematical process.

Every number of interest consists of an even number of the digit 2 and the number of two’s are of the list 2, 2, 4, 4, 6, 6, 8, 8…, 100, 100.

The sum of those numbers is

2 (2 + 4 + 6 + … + 100)

4(1 + 2 + 3 + … + 50)

Using the formula for the sum of first n natural numbers i.e.

$\frac{n(n+1)}{2}$

4 ( $\frac{(50)(51)}{2}$)

5100

Now we double that and the answer is

5100 × 2

10200

Hence, the sum of the digits of the first 100 numbers of the sequence which are divisible by 202 is 10200.

#SPJ3