

Is it possible to get the rightmost NONZERO digit of a factorial without calculating for it?
Say 1000! = 2
Regards,
Ian
*NO, this is not a homework.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





I assume you mean without calculating the factorial but some simpler, faster calculation? You can rewrite a factorial as a polynomial in n but I'm not sure if that gives you an easy pattern for the last digit. And by definition, all factorials except 1! and 0! are even. Does this have a practical application or are you just curious?
If you don't have the data, you're just another a**hole with an opinion.





Nope. This is part of the question set my adviser gave me to prepare for a certain competition.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





Ian Uy wrote: It is said that the most complex structures built by mankind are software systems.
I've heard that bandied about but I think I'll disagree. They usually quote Vista at 50 million lines of code yet the latest generation of processor chips boast over 2 billion transistors. And you could argue that much of what gets counted in an operating system are simply utilities that are applications running on the base system so the core is smaller still.
If you don't have the data, you're just another a**hole with an opinion.





Hi,
I tend to disagree. Hardware, especially CPU design, mainly consists of very repetitive
structures, whereas software, all kinds of it, is very inhomogeneous. And a new CPU gets
designed in a matter of months, a new OS takes several years.
That helps explaining why you would find less than 10 bugs in your CPU having
2 billion transistors, and several thousands of bugs in just 50 million lines of Windows code.
So if you want to compare complexity, I suggest you look at design effort, and number
of bugs.





I still will disagree. In programming we keep using the same structures over and over again but insist on lovingly hand coding them each time. The chip makers have better tools and love to reuse their building blocks. A large part of their reuse is automated which explains the smaller number of bugs. They prove their components and then build up. Programmers make the same mistakes at ever increasing scales.
On another note, humans make machines with millions of parts. I will contend that one line of code is not equivalent to designing one mechanical part in complexity.
If you don't have the data, you're just another a**hole with an opinion.





Ian Uy wrote: Is it possible to get the rightmost NONZERO digit of a factorial without calculating for it?
Yes, for example the last nonzero digit of 1,000,000! is 4.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Got it!... after about 10 mins of tinkering...
static int fac_last_digit(int x)
{
int res = x;
int mod = 10;
while (x > 1)
{
if ((res * (x  1)) % mod == 0) mod *= 10;
res = (res * (x)) % mod;
}
return (res / (mod / 10));
}
The trick is to use the % operator with increasing powers of 10 depending on if the result is a multiple of the original modulus value...
(Needs tidying up a little)... why you need this is a mystery though.
Matthew Butler





Nice try, but you will quickly fall over without a large integer class. Even 1000! ends with 249 zeros, so mod will grow to about 10^249 or so.
If you get bored, try to write a routine to compute the last nonzero digit of googol! = (10^100)!, you should get 6.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Hi Peter,
IMO this will behave much better; it gives 4 for 10000000, does not overflow,
has no loops but uses a recursion (scaling down by a factor of 5).
private int rightmostNonzeroInFactorial(int n) {
int result = 1;
int count2 = (n+8) / 10;
int count3 = (n + 7) / 10;
int count4 = (n + 6) / 10;
int count5 = (n + 5) / 10;
int count6 = (n + 4) / 10;
int count7 = (n + 3) / 10;
int count8 = (n + 2) / 10;
int count9 = (n + 1) / 10;
int count10 = (n + 0) / 10;
count5 += count10;
if (count5 != 0) {
result = (result * rightmostNonzeroInFactorial(count5)) % 10;
count2 = count5;
}
count2 += 2 * count4; count2 += 3 * count8; count2 += 4 * count6;
count3 += 2 * count9; count3 += 3 * count7;
int[] fact3 = new int[] { 1, 3, 9, 7 }; result = (result * fact3[count3 % fact3.Length]) % 10;
int[] fact2 = new int[] { 1, 2, 4, 8 }; int factor2=fact2[count2 % fact2.Length];
if (factor2 == 1 && count2 != 0) factor2 = 6; result = (result * factor2) % 10;
return result;
}
Main principle is factors that are not multiples of 5 don't care about their
digits except the lowest one (and also: there will be more powers of 2 than
powers of 5).





Hi Luc,
I haven't worked through your code (are you sure the countN's do the right thing?).
There is a little known fact (don't ask me how I know it!) that
if n = 5q+r with 0 <= r < 5
then n! has the same rightmost nonzero digit as 2^n x q! x r!
Using this the problems reduce easily.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Hi Peter,
1.
countN tells me how many factors of n! end on digit N
2.
the recursion takes care of your q!
3.
I ran a check for all [1,500] and for 10000000
regards,





Luc Pattyn wrote: countN tells me how many factors of n! end on digit N
OK now I see what you are doing. I'm sure you have some very clever reason for this, but it is not clear to me why you do
count5 += count10;
if (count5 != 0)
{
result = (result * rightmostNonzeroInFactorial(count5)) % 10;
count2 = count5;
}
and not what I would expect:
if (count5 != 0)
{
result = (result * rightmostNonzeroInFactorial(count5)) % 10;
count2 = count5;
}
if (count10 != 0)
{
result = (result * rightmostNonzeroInFactorial(count10)) % 10;
}
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Hi Peter,
the way you propose the factors that end on 5 don't get represented by count5 factorial
(they only need odd multiples of 5), hence the recursion would not be correct.
I expect the theory that explains my code would be the same as the one that led to the
formula you've mentioned before.





Ah now I see (it's late here
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





cp9876 wrote: ...There is a little known fact (don't ask me how I know it!) that
if n = 5q+r with 0 <= r < 5
then n! has the same rightmost nonzero digit as 2^n x q! x r!
That should be 2^q rather than 2^n, and then it's valid for all n > 4. Removing recursion, here's one way to write it (in Python):
def f(n):
result = 1
while n > 1:
n, r = divmod(n, 5)
result = result * (6, 2, 4, 8)[n & 3] * (1, 1, 2, 6, 4)[r] % 10
return result
and that can be simplified a bit by using a larger table:
FACTAB = 6, 6, 2, 6, 4, 2, 2, 4, 2, 8, 4, 4, 8, 4, 6, 8, 8, 6, 8, 2
def f(n):
result = 1
while n > 1:
result = result * FACTAB[n % 20] % 10
n return result





I believe this to be an ACM question from either 96' or 97'
bool nonzerodigit(const unsigned int& n, unsigned int& result)
{
if (n > 100000) return false;
unsigned int result = 1;
for (unsigned int i = 1; i <= n; ++i)
{
unsigned int f = i;
while (0 == (f % 5))
{
f /= 5;
result /= 2;
}
result = (result % 100000) * f;
}
result %= 10;
return true;
}
100000 is an upper sanity bound and also a quantizer for the minimum required 6 "least significant" digits.





Thank you!
I am training for the next ACM ICPC Regionals.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





Hello,
I'm putting together a Silverlight 2 demo application. It is going to be a simple 2D tank game where you can use the arrow keys to rotate and move the tank. The tank can point in any direction (360 possible directions, right?). I've written the code to rotate the tank, but I'm not sure where to begin regarding moving the tank. How can I take the angle of the tank (say, 156 degrees for example) and use that to determine the next x,y position of the tank when it moves forward?
I'm sure there must be a simple equation for this, but I don't know what it would be called or even how to google this problem.
Any ideas?
Thanks,
Ian





Define the motion using a vector. The position coordinates are given by:
x = r*cos(theta)
y = r*sin(theta)
where r is the distance traveled and theta is the angle. You can get more description here[^]. You'll also have to assign a velocity.
I'm the ocean. I'm a giant undertow.





Beat me to it.
(Within 3 minutes, more than 3 hours from the original question, what are the chances of that?)
Matthew Butler





Pretty low I would imagine....wacky!
I'm the ocean. I'm a giant undertow.






Just use basic geometry/maths...
nextXpos = lastXpos + speed * sin(angle);
nextYpos = lastYpos  speed * cos(angle);
[angle is measured from the vertical, clockwise]
The only things you need to remember are:
> Math.Sin and Math.Cos take the angle argument in radians not degrees.
> Sin and Cos return double precision numbers: which may round to 0 or 1 if you are using integer positions and therefore the tank will not move as you want.
Hope this helps.
Matthew Butler




