Question

The highest power of 2 that divides the sum of the numbers 4 + 44 + 444 + ..+ 444 ...100 fours
*please answer step by step*

Answer 1

Its easy to see that this is a diverging sequence and if you keep following it up to infinity as you show by the dots following the 4th term, the answer will approach infinity. However if you want to sum it up to n terms, there is an easy way provided you know the sum of a Geometric Progression.

Rewrite 4 + 44 + 444 + 4444 +…..up to n terms as

4/9 ( 9 + 99 + 999 + 9999 +….up to n terms)

4/9 ( 10 - 1 + 100 - 1 + 1000 - 1 + 10000 - 1 +…..up to n terms)

4/9 ( 10 + 100 + 1000 + 10000 +…up to n terms - n)

The next bit is easy, just apply the formula for calculating sum of a G.P. which is a(r^n - 1)/(r - 1) where a is the first term, r the common ratio and n the number of terms in the sequence.

So you finally get 4/9[ 10(10^n - 1)/9] or 40/81[ 10^n - 1 ]

HAPPY LEARNING!

Answer 2

Answer:

the answer is 3

Step-by-step explanation:

Ans is 8

Take 4 as common

It will be something like this

4(1+11+111........111.......11(100 ones)

When you will see the unit digit when adding, it will be 0 and the hundred place digit as 9

90 is only divided by 2^1

(in terms of power of 2)

4(1+11+111+.........+111(100 1s)/2

4*2= 1+11+111+...........111(100 1s)

2^3=8