Fixed. I realised why that shifting is expensive.

... My mistake. If *p is false the next if wouldn't run at all

I saw this pattern under expected

} while (*p);
if (*p) do {

e.g ][
but OP told us to remove it yeah?

So why is that expected?

Fair enough. XD

Seems like it. I didn't learn that in my lesson with 8086.
bx work that way too back in 8086?

Well I updated the test condition, from 500 sentences 200 assertions to 20 sentences 1000 assertions.
And updated the desciption.
Some interesting find was whatever framework codewars is using, bitwise shift right is badly supported.
Instantly timeout during loops. Whereas division is working as per expected.

Yeah i called random() 100000 times I wonder if I should just bitwise shift and reuse it since one call of random() can provide for 10 loops

I wasn't sure if i need not provide that since it is quite obvious. But done anyways.

first kata I done I forgot its linked. Anyways, fixed

Any way to solve it witout understanding Euler's theorem?
I'm stuck because but I know to obtain the period of a number, but I don't know how to get the period of a large power.

Next, I am not sure if there is a pattern using the last digit of every term and period from array[size-1] to array[0] even if I can get the period of the last pair of pow(2nd last term, last term);

resource i'm referencing to: https://brilliant.org/wiki/finding-the-last-digit-of-a-power/