17.4k views
3 votes
S = 2^0 + 2^1 + 2^2 + 2^3 + ...... 2^99a) Show that S can be divided by 15b) Show that S has at least 30 digits

1 Answer

5 votes

Answer:

Step-by-step explanation:

Here, we want to show that the sum is divided by 15

From what we have, the given sum is a geometric sequence

The first term is 1

Now, the pattern of ending afterwards will be 2, 4, 6 and 8

This ending keeps repeating itself

This 2,4,6,8 pattern could repeat itself 24 times

So we have a total of 1 + 24(4) = 97 terms

To make it 100, we have the last three terms as 2,4 and 8

So we have the ending number ONLY sum as follows:

1 + 24(2+4+6+8) + 2 + 4 + 8 = 1 + 480 + 14 = 495

We can divide this by 15 and that gives 495/15 = 33

That shows that the sum is divisible by 15

Secondly, we want to show that S has at least 30 digits

We can infer this from the last terms

We can write 2^99 as 2^33 * 2^33 * 2^33

A single 2^33 has a value of 8,589,934,592

That means 10 digits

The other two has 10 digits too

The sum of all possible digits in the largest term is 10 + 10 + 10 = 30

That makes a total of 30

The question states 30 or more

Hence, this is correct

User Dule Arnaux
by
6.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.