ByErik7264
For every integer n ≥ 2, consider the last non-zero digit in the decimal representation of nl. The infinite sequence of these digits begins with 2, 6, 4, 2, 2, because 2! = 2, 3! = 6, 4! – 24, 5! = 120, and 6! = 720. Determine all digits that occur at least once in this sequence and show that each of these digits occurs infinitely often. Notes: nl is the product of all natural numbers from 1 to n.
I don't understand why 2, 4, 6 and 8 appear infinite
(2 votes)
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I do not know at what level this task has been made, i.e. which tools may be used.
A possible approach would be to consider a “period” of 10, i.e. the faculties at multiples of 10, i.e. 10!, 20!, ….
10! = 3628800 has the 8 as the last number different from zero. The next 10 multiplications change these 8, only the last digit of the multiplier being massive, 1, 2, …… (1)0. It therefore continues with 1*8 -> 8, 2*8 -> 6, 3*6 -> 8, …
In addition, a new zero is added to the 5 at which 10 likewise.
The effect on the 8 via these 10 multiplications is therefore equal to the multiplication with the last non-zero digit of 10!, this is the 8th.
In the series of faculties 10!, 20!, .. we see the end numbers 8, 4, 2, 6, 8, 4, 2, 6, …. which repeat periodically.
That’s because it goes into a circuit. Why don’t you just figure out a few followers? Sometime you see it.
I have to make a start-up I have seen that they come so often but why I don’t understand
because the last blurring of the associated factors are also repeated
I have been happy thanks beautiful
what do you mean now is a bit lost