@another approach..
The remainders when the following nos. are divided by 5 are :
Numbers Remainder
2^0 1
2^1 2
2^2 4 (-1)
2^3 3 (-2)
Now, we know that given a product of nos. a1, a2, ... aN when divided
by a no. K, the remainder is:
the product of (a1 mod K).... (aN mod K) i.e
[ Lets call this as NumTheoProp1]
a1* a2* a3*...*aN = x*K + (a1 mod K)*(a2 mod K)*....*(aN mod K)..
[ the above can be proved by representing ai = xi * K + (ai mod K),
and then multiplying all the ai's..]
We can then recursively apply the same approach to the product of
remainders until and unless the remainder is smaller than K.
But a complete reduction into a value < K is not the point here. We
are going to use the above fact for the following:
Say, the given no. is 2^R ...
Now, the above can be interpreted as,
2^R = 2^(4q) * 2^r, where R = 4q + r and 0<=r < 4
Now we know that, 2^4 when divided by 5 gives a remainder 1..
Hence applying the [NumTheoProp1] :
2^(4q) = (2^4)*(2^4)*... (q times)..
when 2^(4q) = x*5 + (2^4 mod 5)^q..
= x* 5 + 1..
Hence,
2^R = 2^(4q) * 2^r
= x*5 + (2^4q mod 5)*(2^r mod 5)
= x*5 + (2^r mod 5)..
Now, as 0<=r < 4, based on the remainder jolted at the starting..
2^r mod 5 = one of (1, 2, 4, 3)..
But if you closely observe there is a pattern..
For any no. 2^R where R>= 4 we can reduce it to 2^r where r < 4 by
using [NumTheoProp1]
Hence,
No.( 2^r) Remainder
0 1
1 2
2 4
3 3
4 1
5 2
6 4
7 3
.... and the cycle continues..
Now. say a no. N is given, then it can be represented as,
N = b0* (2^0) + b1* (2^1) + .... + bN* (2^logN)..
where bi is either 0 or 1 [ basically it mimics the bit pattern)
N mod 5 = ( [sum over all i ( (bi* (2^i)) mod 5 )] mod 5)
Now, the inner mod 5 can be directly calculated based on the index of
i (as there is a pattern). Btw it also depends on the value of bi. If
bi =0, then the remainder will be 0 otherwise we will get the
remainder based on the pattern..
The outer mod 5, can be handled by following an incremental process,
i.e whenever the current remainder becomes >=5 we will subtract 5 from
it.
Code:
int N; // given number..
int r[4] = {1, 2, 4, 3};
int currRem = 0;
int i = 0;
while (N)
{
if (N&1)
{
currRem += r[i%4];
if (currRem >=5)
currRem -=5;
}
++i;
N >>= 1;
}
if ( currRem == 0)
{
printf("N is divisible by 5");
}
On Jan 22, 2:42 am, Dave <[email protected]> wrote:
> @Arun: Proof that n = 4*a + b is a multiple of 5 if and only if a - b
> is a multiple of 5:
>
> Given n, write it as n = 4*a + b where 0 <= b <= 3.
> Note that 5*a is a multiple of 5, and 5*a - n = 5*a - (4*a + b) = a -
> b.
> If n is a multiple of 5, then 5*a - n is a multiple of 5, so a - b is
> a multiple of 5.
> Similarly if n is not a multiple of 5.
>
> QED
>
> Dave
>
> On Jan 21, 11:57 am, Arun Vishwanathan <[email protected]> wrote:
>
>
>
>
>
>
>
> > @dave: thanks for that..but i just wanted to know as to how u thot of
> > converting n to a-b in the iteration. when u say 4a +b is a multiple of 5
> > iff a-b is a muliple of 5 i was able to get that only when i tried an
> > example...if they ask divisbility by 3 or 6 or 7 how wud the logic change??
>
> > On Sat, Jan 21, 2012 at 9:34 AM, karthikeyan muthu <
>
> > [email protected]> wrote:
> > > check the last char ... it should be 0 or 5 , int to string without mod
>
> > > On Sat, Jan 21, 2012 at 10:05 PM, Dave <[email protected]> wrote:
>
> > >> @Karthikeyan: Is this supposed to relate to the question of
> > >> determining divisibility by 5?
>
> > >> Dave
>
> > >> On Jan 21, 9:25 am, karthikeyan muthu <[email protected]>
> > >> wrote:
> > >> > @dave
>
> > >> > int no=10;
> > >> > char ans[100];
> > >> > sprintf(ans,"%d",no);
> > >> > cout<<ans;
>
> > >> > On Fri, Jan 20, 2012 at 10:29 PM, Arun Vishwanathan
> > >> > <[email protected]>wrote:
>
> > >> > > @dave or anyone: can u pls explain the logic of n&3 in dave's
> > >> solution?
> > >> > > why is it subtracted from n(which is divided by 4 using >>2) and what
> > >> does
> > >> > > n& 3 indicate?
>
> > >> > > On Sat, May 7, 2011 at 9:38 AM, Dave <[email protected]> wrote:
>
> > >> > >> @Umer: Do you suppose that you can convert an int into a string
> > >> > >> without using division or mod, either directly or indirectly?
>
> > >> > >> Dave
>
> > >> > >> On May 4, 1:12 am, Umer Farooq <[email protected]> wrote:
> > >> > >> > I'm surprised to see that why are you guys making this problem so
> > >> > >> complex.
> > >> > >> > This problem can be solved in two steps only.
>
> > >> > >> > 1- Convert the given int into string
> > >> > >> > 2- Check if the last character is 0 or 5. // it yes, then return
> > >> true
> > >> > >> else
> > >> > >> > return false
>
> > >> > >> > for e.g.
>
> > >> > >> > 125 (last character is 5 ... therefore it is divisible by 5)
> > >> > >> > 120 (last character is 0 ... therefore it is divisible by 5)
> > >> > >> > 111 (last character is 1 ... therefore it is not divisible by 5)
>
> > >> > >> > The pseudo-code has been written in my above email.
>
> > >> > >> > On Wed, May 4, 2011 at 1:49 AM, Dave <[email protected]>
> > >> wrote:
> > >> > >> > > @anshu: Spoiler alert... I was thinking of something more along
> > >> the
> > >> > >> > > line
>
> > >> > >> > > int DivisibleBy5 (int n)
> > >> > >> > > {
> > >> > >> > > n = n > 0 ? n : -n;
> > >> > >> > > while( n > 0 )
> > >> > >> > > n = (n >> 2) - (n & 3);
> > >> > >> > > return (n == 0);
> > >> > >> > > }
>
> > >> > >> > > To see that it works, write n as n = 4*a + b, where 0 <= b <= 3.
> > >> Then
> > >> > >> > > the iteration replaces n by a - b. Consider (4*a + b) + (a - b),
> > >> the
> > >> > >> > > sum of two consecutive values of n. This simplifies to 5*a,
> > >> which is a
> > >> > >> > > multiple of 5. Thus, n is a multiple of 5 before an iteration if
> > >> and
> > >> > >> > > only if it also is a multiple of 5 afterwards,
>
> > >> > >> > > It is clearly log n because n is replaced by a number no greater
> > >> than
> > >> > >> > > n/4 on each iteration.
>
> > >> > >> > > Examples:
> > >> > >> > > n = 125. The sequence of iterates is 30, 5, 0. Ergo, 125 is a
> > >> multiple
> > >> > >> > > of 5.
> > >> > >> > > n = 84. The sequence of iterates is 21, 4, -1. Ergo, 84 is not a
> > >> > >> > > multiple of 5.
>
> > >> > >> > > Dave
>
> > >> > >> > > On May 3, 3:13 am, anshu <[email protected]> wrote:
> > >> > >> > > > algorithm:
>
> > >> > >> > > > if any number(a) is divisible by 5 it can be wriiten as 4*b +
> > >> b -->
> > >> > >> > > > this cleary shows the last two bit of a & b will be same.
>
> > >> > >> > > > lets understand by an example (35)10 = (100011)2
>
> > >> > >> > > > xx1100
> > >> > >> > > > + xx11
> > >> > >> > > > ---------
> > >> > >> > > > 100011
>
> > >> > >> > > > now this clearly shows we can calculate the unknowns(x) by
> > >> > >> traversing
> > >> > >> > > > right to left
>
> > >> > >> > > > code:
>
> > >> > >> > > > int main()
> > >> > >> > > > {
> > >> > >> > > > int n, m;
> > >> > >> > > > cin >> n;
> > >> > >> > > > m = n;
>
> > >> > >> > > > int a, b;
> > >> > >> > > > int i=2;
>
> > >> > >> > > > a = (m&3)<<2;
> > >> > >> > > > b = (m&3);
> > >> > >> > > > m >>= 2;
>
> > >> > >> > > > bool rem = 0,s,r;
>
> > >> > >> > > > while (m>3)
> > >> > >> > > > {
> > >> > >> > > > r = a&(1<<i);
> > >> > >> > > > s = r^(m&1)^rem;
> > >> > >> > > > b = b|(s<<i);
> > >> > >> > > > a = a|(s<<(i+2));
> > >> > >> > > > rem = (r&s)|(s&rem)|(r&rem) ;
> > >> > >> > > > i++;
> > >> > >> > > > m >>= 1;
> > >> > >> > > > }
>
> > >> > >> > > > if (a+b == n) cout << "yes\n";
> > >> > >> > > > else cout << "no\n";
>
> > >> > >> > > > return 0;
>
> > >> > >> > > > }- Hide quoted text -
>
> > >> > >> > > > - Show quoted text -
>
> > >> > >> > > --
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>
> > >> > >> > --
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>
> > >> > >> > - Show quoted text -
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