Kiriti's powers of 2

Based from the given. We consider a=4k where k is also a variable positive number. So, we substitute

$(4k)^{100}$ + $(4k)^{101}$ + $(4k)^{102}$ + $(4k)^{103}$ +⋯+ $(4k)^{1000}$

thus, that the largest factor of $(4k)^{100}$ + $(4k)^{101}$ + $(4k)^{102}$ + $(4k)^{103}$ +⋯+ $(4k)^{1000}$ is $4^{100}$ which is equal to $2^{200}$

therefore,

$2^{200}$ = $2^{n}$ , n=200

rewrite a=2^2×(an odd number), a^100+a^101+a^102+.........+a^1000=a^100×(1+a+a^2+a^3+........a^10), (a+a^2+a^3......a^10) is even, and (1+a+a^2+a^3+........a^10) is odd a^100×(1+a+a^2+a^3+........a^10) =a^100×(an odd number) =[2^2×(an odd number)]^100×(an odd number) =2^200×(an odd number) ∴n=200.

let x=4k where k is an odd integer as it is not divisible by 8.Then a^100+a^101+a^102+....+a^1000=(4k)^100+(4k)^101+.....+(4k)^1000=(4k)^100(1+(4k)+(4k)^2+.....+(4k)^900)=4^100*k^100(1+2K) where 2K=(4k)+(4k)^2+.....+(4k)^900....k being odd the highest power of 2 dividing a^100+a^101+a^102+....+a^1000 is 4^100=2^200....thus n=200.

Only the odd multiples of 4 divide 4 but not 8. So, 4, 12, 20.... and so on. As such, $4$ should work as $a$ . First determine the highest $n$ such that $2^n$ divides $a^1 + a^2 + a^3...... a^{10}$ . Using $4$ as $a$ , this comes out to $1,398,100$ , which is only divisible when $n$ is equal to $2$ . Becuase your test exponents were $\frac {1}{100}$ of the questions, multiply your value of $n$ by $100$ , and this is the answer.
( $100*2 = 200$ )

since a is divisible by 4 but not by 8, it can be said that a = (2^2)k, where k is an integer. a^100+a^101+a^102+a^103+⋯+a^1000 is a geometric series that can be rewritten as a^100((1-a^901)/(1-a)). the problem then can be restated as [a^100((1-a^901)/(1-a))] / 2^n. we only need to look for n that will make a^100 divisible by 2^n.

using a = (2^2)k, we can rewrite a^100 as (2^200)k^100. to make 2^n be a divisor of (2^200)k^100, n has to be at most equal to 200. so n = 200.