19!

A very elegant solution to this question would be similar to that posted by Michael Tang at the link here: Pin the tail on the factorial .

You can apply his solution to this question which is very similar to one that was posted just last week...

Here is the solution but applied to this question:

There are 3 factors of 5 and 16 factors of $2$ in $19!$ . $5^{3} * 2^{3} = 10^{3}$ , which accounts for the last 3 zeros behind the long string of digits. Hence, we get $2^{16-3} = 2^{13} \mid \overline{121645100408abc}$

We know that for a number to be divided by $2^{n}$ , its last $n$ digits must be a number that is divided by $2^{n}$ . In this case, the number formed by the last 13 digits, i.e. $\overline{1645100408abc}$ must be divisible by $2^{13} = 8192$ .

Note that $\overline{1645100408abc} =\overline{1645100408000} + \overline{abc}$

Therefore, $\overline{1645100408000} + \overline{abc} \equiv 7360 + \overline{abc} \equiv 0 \pmod {8192}$ . Hence, since $\overline{abc}$ is a $3$ digit number, we get $\overline{abc} = 8192 - 7360 = \boxed{832}$ .

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The mod function is used when we are interested in the remainder, instead of the quotient. For example $5 \equiv 2 \pmod{3}$ because when $5$ is divided by $3$ , it leaves a remainder of $2$ .

Similarly, $\overline {1645100408abc} \equiv 7360 \pmod{8192}$ . That is how I arrive at $7360$ :)

On a side note, I just want to clarify the motivations behind using the mod function in this case. This is because we know that a number is divisible by another if the remainder is $0$ . Since we are interested in the remainder only, it may be appropriate to use the mod function.

$\overline{121645100408abc000} = \overline{1216451004508abc} * 1000$

As Given $1216451004508abc*1000=19!$ This implies $\overline{121645100408abc}=2 * 3 * 6 * 7 * 8 * 9 * 11 * 12 * 13 * 14 * 3 * 16 * 17 * 18 * 19$ (as $4 * 5 * 10 * 5=1000$ ) is divided on both sides when we multiply the R.H.S only the last 3 digits contribute to $\overline{abc}$ .

$2 * 3 * 6 * 7 * 8 * 9=1144$ the last 3 digts(144) contribute to $\overline{abc}$ .
$144 * 11 * 13$ will have last 3 digits 592 now we will multiply further $592 * 12 * 14$ will have last 3 digits 456 similarly, $456 * 17 * 19$ will have last 3 digits 288, $288 * 16 * 18$ will have last 3 digits 944 and finally $944 * 3$ will have last 3 digits 832 now we have multiplied all the numbers and got the last 3 digits to be 832

[Latex edits]

To begin with, let us expand the factorial for illustrational purposes

$19!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18\cdot19$

The first thing we should take into account is that $a=b=c=0$ is not a solution to the problem, since then the integer would be evenly divisible by $1000000$ , which is a number that cannot be formed by multiplying solely the numbers included in $19!$ .

Additionally, we may divide the actual integer with $100$ by diving the expanded factorial as well, resulting in losing the integers $10,2$ and $5$ from the expansion (since $2\cdot5=10$ ). Consequently, the integer we are looking for has now acquired the following form

$\overline{121645100408abc0}$

We are now going to use the divisibility rules for $4$ and $8$ .

An integer is divisible by $4$ only if the integer formed of its last $2$ digits is divisible by $4$ .

On the other hand, an integer is divisible by $8$ only if the integer formed of its last $3$ digits is divisible by $8$ .

By using the divisibility rule for $4$ we can easily come to a conclusion that $c\in\{2,4,6,8\}$ , due to the fact that the only $2$ -digit integers that end in $0$ and are divisible by $4$ are $20,40,60$ and $80$ . Similarly, it is also correct that the only $3$ -digit integers that end in $0$ and are divisible by $8$ are $160,240,320,400,480,560,640,720,800,880$ and $960$ . However, taking into account that $c\neq0$ we may disregard the integers $400$ and $800$ .

Let us now turn our attention to the divisibility rule for $11$ , which can be found thoroughly explained here .

By using it, we come to the expression

$((1+1+4+1+0+0+a+c)-(2+6+5+0+4+8+b))\mid11$ $\Rightarrow$ $(a+c-b-18)\mid11$

Now, considering that $1\leq a,b\leq9$ and $c\in\{2,4,6,8\}$ , the minimum of the above expression is $-24$ ( $a=0,b=9$ and $c=2$ ), whereas the maximum is $-2$ ( $a=9,b=1$ and $c=8$ ).

Now, the only integers in the interval $(-24.-2)$ that are divisible by $11$ are $-22$ and $-11$ .

We are now going to implement the divisibility rules for $3$ and $7$ . The one for $3$ states that an integer is divisible by $3$ only if the sum of the digits of that integer is also evenly divisible by $3$ .

Therefore, if the sum of the digits of our integer is $a+b+c+32$ , it is evident that

$3\mid(a+b+c+32)$

The divisibility rule for $7$ can be found here .

We shall now discuss $2$ cases:

Case 1:

$a+c-b-18=-22 \Rightarrow a+c-b=-4$

From the possible numbers stated from the divisibility rule for $8$ (which again, are $160,240,320,480,560,640,720,880$ and $960$ ) the only pair $(b,c)$ that make the above equation at all possible is $(7,2)$ , since $1\leq a\leq9$ .

As a result of that, we can see that

$a-7+2=-4 \Rightarrow a=1$

The possible solutions for $a,b,c$ are $1,7,2$ accordingly. To check if this is indeed true we will use the equation from the divisibility rules for $3$ and $7$ .

Although $1+7+2+32=42$ which is divisible by $3$ , using the divisibility rule for $7$ we come to a conclusion that the integer $121645100408172$ is not divisible by 7, therefore

$a+c-b\neq-4$ , which proves this case wrong.

Case 2:

$a+c-b-18=-11 \Rightarrow a+c-b=7$

From the possible numbers stated from the divisibility rule for $8$ (which again, are $160,240,320,480,560,640,720,880$ and $960$ ) the pairs $(b,c)$ that make the above equation at all possible are $(1,6),(2,4),(3,2),(4,8)(5,6),(6,4),(8,8)$ for $a\in\{2,5,8,3,6,9,7\}$ accordingly, since $1\leq a\leq9$ .

Using the divisibility rule for $3$ , we may exclude all possible solutions for $a,b,c$ apart from $a=8,b=3,c=2$ and $a=9,b=6,c=4$ since those are the only ones that satisfy $3\mid(a+b+c+32)$ .

Having to finally choose from these two, we apply the divisibility rule for $7$ one last time to see that only $1216451004088320$ is divisible by $7$ , whereas $1216451004089640$ is not.

Therefore, $1216451004088320$ is the correct number which means that

$\overline{abc}=832$

which is in fact the correct solution of the problem.

Note: Another possible way is to simply divide the possible numbers with $3$ and/or $7$ and see the result. However, for purposes of proving this I added the divisibility rules of those two.

19! = 1 2 3 ... 18*19 Then, 19! ≡ 0 (mod 1000) and 19! ≡ 0 (mod 9) => 1+2+1+6+4+5+1+4+0+8+a+b+c+0+0+0 ≡ 0 (mod 9) 32+a+b+c ≡ 0 (mod 9) a+b+c ≡ 4 (mod 9)

If a ≡ n (mod x) and b ≡ m (mod x) Then a b ≡ m n (mod x)

19!/1000 ≡ 2 3 6 7 8 9 11 12 13 14 3 16 17 18 19 19!/1000 ≡ 2 3 6 7 8 9 1 2 3 4 3 6 7 8 9 ( mod 10) 19!/1000 ≡ (2^2) (3^3) 4 (6^2) (7^2) (8^2) (9^2) (mod 10) 19!/1000 ≡ 4 7 4 6 9 4 1 ≡ 8 4 6 ≡ 2 ≡ c (mod 10)

Then c=2 And: a+b+c ≡ a+ b +2 ≡ 4 (mod 9) a + b ≡ 2 ≡ 11 (mod 9)

Solutions (a,b) : (2,0),(0,2),(8,3),(3,8)

And (8,3) is the only solution

To solve this problem, you can do two (other way) ways below:

1. You can do manual the factorial The factorial is defined as, n the positive integer, n! = n (n-1) (n-2) ...... 3 2 1 .
2. Write the code using reccurence about factorial

Just multiply.

divide by 5 first (twice or three times if necessary), and then multiply by the quotient. Also, for every factor of 5 that you skip, be sure to also skip a factor of 2. You have to take the last 3 digits after every multiplication. So you would end up with something like this:

001 X 2 = 002 , 002 X 3 = 006 , 006 X 2 = 012 (drop a 2) , 012 X 1 = 012 (drop a 5) , 012 X 6 = 072 , 072 X 7 = 504 , 504 X 8 = 032 , 32 X 9 = 288 , 288 X 1 = 288 (drop a 2*5) , 288 X 11 = 168 , 168 X 12 = 016 , 016 X 13 = 208 , 208 X 7 = 456 (drop a 2) , 456 X 3 = 368 (drop a 5) , 368 X 16 = 888 , 888 X 17 = 096 , 096 X 18 = 728 , 728 X 19 = 832

For this problem, one can easily code this. I used python to code this problem. Using this code:

number=input("Enter a nonnegative integer to take the factorial of:")

product=1 for i in range(number): product = product*(i+1)

print(product)

I found out what "abc" was and you get that "abc" is 832.