Adding digits and divisibilities

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

The number, $2018!$ , is definitely divisible by $9$ . The sum of its digits is as well, etc. So the last remaining digit too has to be divisible by $9$ , which can only happen if it is $\boxed9$

Multiples of 9 has digit sums divisible by 9 as well. 2018! contains multiple of 9 as 9 itself is contained in the factorial. i.e. 1x2x....9x...2018

2018! is divisible by 9 as 2018! = 1x2x3x...9x10x11....2018 and the divisibility rule of 9 says that the sum of the digits till the digit remains a single one must be 9. Therefore the answer is 9

Extra: Every factorial of a number above 5 has the same property

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import math

# Define our recursive function, digSum(), that takes an int
def digSum(num):
    # End condition: if there's only one digit, print it
    if num < 10:
        print num
    # Else, keep summing the digits
    else:
        digSum(sum(int(i) for i in str(num)))

# Call our function with 2018! as the int
digSum(math.factorial(2018))

In this solution, we begin with $2018!$ as our input integer and repeatedly sum the digits using Python's powerful string capabilities.

The heavy lifting is done here:

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else:
        digSum(sum(int(i) for i in str(num)))

Here, we recursively call our function, digSum(num) , with the result of its most recent input. Starting with $2018!$ , we sum() all of the integers, i , in the number, num . To accomplish this, we convert our number into a string with str(num) , allowing us to iterate over the number one digit at a time with a for statement, making summing the digits easy! So essentially, this line is saying:

For all of the characters in the number string, turn them into integers and sum them all together.

This is done multiple times until we check and find that there is a single digit remaining, bringing us to our answer, $\boxed{9}$ .

Any multiple of 9 has a digital root of 9. 2018! is a multiple of 9, so its digital root must be 9.

In the given set, every integer from 1 to 2018 is given. (This can also be expressed as 2018! )Given that every number is being multiplied together and one of those numbers is 9, we can conclude that the final number will be 9. How? Divisibility rules tell us that the digits of any multiple of 9 will add up to be a multiple of 9. From that logic, we can conclude that when this process is repeated, the end result will be 9.

If $x \in \mathbb{N}$ , then the digital root $d(x)$ of $x$ is the single-digit number obtained by iteratively summing up the digits of $x$ and its subsequent iterations. Moreover, $d(x) \equiv x \hspace{-.3cm} \mod{9}$ . To see this, write the decimal expansion of $x$ and note that the $n^{th}$ term in the expansion is congruent to the $n^{th}$ digit of $x$ modulo 9. Since $2018!$ is divisible by 9, so is the single-digit number $d(x)$ . Therefore, $d(x) = 9$

The number 2018! is congruent to 0 (mod 9) since 9 appears in there at least once. Let n = 2018!. The canonical representation of n in base ten is $n = n_0 10^0 + n_1 10^1... + n_r 10^r$ . Notice that since $10 \equiv 1 \pmod 9 , n \equiv n_1 + n_2 + \dots + n_r \pmod 9$ . This means that the sum of the digits of a number is congruent to the number itself mod 9. So when we get the number m from doing this process repeatedly we should note that $m \equiv n \pmod 9$ and since m is one digit and n is 0 (mod 9) [or 9 (mod 9)] we can be sure of our answer m =9

A simple python code with recursion does the trick:

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from math import factorial
def digit_sum(n):
    if n<10:
        return n
    return digit_sum(sum(int(i) for i in str(n)))

print digit_sum(factorial(2018))

And of course that the output is:

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9

The 'digit sum' of any number that's divisible by 9 is 9.

2018! is divisible by 9. So repeatedly obtaining the digit sum gives 9

In 2018! ,it is divisible by 9 then the sum so.So, out of 4 option 9 is divisible by 9.So it is correct

Here is how I did it: 2018! Is divisible by 9

I just used my calculator untill E(error) appeared. I typed in 1×2× ... ×21 then 5.10909422E+19 appeared and the last number was 9!

Watch VSauce's fixed-point video .

It seems that the result is always 9 from $6!$