Pearls

The faster method: let number of pearls be x, then,

x (mod 8)=5 ; then the values satisfying above restrictions were:13,21,29...... but as we know that ,

x(mod 5)=1 also,

then values satisfying above restrictions were:21,61,101.......

then again if u quickly notice 101 is a prime and 100 is a perfect square, then problem is solved ,but this was a trial and error method ,

So,I made a blue j program for it:

import java.io.*;

class xyz

{

public static void main() throws IOException

{

    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

    int a,b,c,d;

    int i,j,k,l;

    int count[] =new int[100000];

    for(i=1;i<=10000;i++)

    {   

        k=0;

        for(j=2;j<i;j++)

        {

            if(i%j==0)
                k++;
        }
        if(k==0)
            count[i]=i;
    }
    for(a=1;a<=1000;a++)
    {
        for(b=1;b<=1000;b++)
        {

            if((count[a]-1==b*b)&&((count[a]-2)%9==0)&&(count[a]%8==5)&&count[a]%5==1)                {
                System.out.println(count[a]);   

            }
        }
    }
}

}

First the array stores all primes numbers between 1 and 10,000, then the array of prime numbers checks for other restrictions. and Boom,it gives only 1 output which is 101.!!!

I wrote a python line to fulfill all conditions

Number = min ( ( (i * i) +1 for I in xrange(1000) if ((i * i)+1)%5 == 1 and ((i * i)+1)%8 == 5 and ((i * i)-1)%9== 0 ))

I first sorted out among the conditions the most useful one. In the ques the number let it be X , (X-1) is a perfect square and X mod 5 is 1. So that suggests (X-1) is a multiple of 5 and is a square number.
Now the problem is easy enough. square number divisible by 5 are 25,100,225...... 25 does not fulfill other conditions. Then moving on to 100 and yes it meets all conditions mentioned. So number is 100+1=101

We need to know the number of pearls which satisfy all four conditions.If we see condition number 3,when divided in 5 knites,1 pearl is left that means last digit is 1,this satisfies condition number 1 as well.Now look for condition number 2 that is if 2 number are talen out ,remaining is multiple of 9 so we reach to number 99 +2( this satisfies earlier two conditions as well. last one divided in 8 knites ,left over pearls are 5.So total number of pearls are 101 in the chest.Ans.

K.K.GARG,India

Let the number be X. We are told that k^2 + 1 = X, and that X is prime. We are also given that k^2 is a multiple of 5. Since 5 is prime, k^2 must be divisible by 25. We are also told that k^2 must be divisible by 4, as (k^2 - 4 ) divided by 8 = an integer. Therefore k^2 is a multiple of 100.

But k^2 - 1 is divisible by 9, so the digit sum of k^2 - 1 must equal a multiple of 9. This doesn't work for any multiple of a 100 below 1,000, except 1000 (which gives 999) or 100 (which gives 99). Since 1000 isn't a square number, the only remaining solution less than 1000 is 100.

If we try out 100, we find that it works, so the number of pearls is 101. I thought this was a rather nice solution, other than the guesswork at the end.

The faster method: let number of pearls be x, then, x (mod 8)=5 ; then the values satisfying above restrictions were:13,21,29...... but as we know that , x(mod 5)=1 also, then values satisfying above restrictions were:21,61,101....... then again if u quickly notice 101 is a prime and 100 is a perfect square, then problem is solved ,but this was a trial and error method , So,I made a blue j program for it: import java.io.*; class xyz {

Condition 3: Number ends in 1 or 6.

Condition 5: Number cannot end in 6, so it has to end in 1.

So, when I remove one, it should end in 0. Which takes us to Condition 1: If this number that ends in 0 is a perfect square, it has to end with two zeroes. Smallest possible value is 100, which means our answer could be 101.

Now verify Conditions 2 and 4. They hold. Hence, 101 is the answer.

x-1 is divisible by 5 with mod 1.x is a prime we can conclude last digit is 1. 101-1is a perfect square mod(101,8)=5. so it's correct

let the soln be of the form m^2+1 also this m^2+1=5k+1 so m^2=5k so k= 5a^2 and m= 5a now m^2+1=8l+5 therefore m^2= 4(2l+1) or 25a^2=4(2l+1) clearly 2l+1 is odd so a^2 is 4 or a is 2 so mis 5a i.e. 10 and no. is 10^2 + 1 i.e. the final ans is 101

Let N be the number of pearls. Since N is prime it has to be odd (as 2 can be ruled out easily). Using condition 3 (that is, N-1 is divisible by 5), N=5p+1 for some integer p. If p is odd then N will be even leading to a contradiction. Hence, p is even thus N should end in 1, that is, N=10q+1 for some integer q. Using condition it can be further concluded that N=100n^2+1. The smallest of these is 101 which satisfies all the conditions.

For sake of completeness: we can show that condition 4 is satisfied for all odd n. Hence, N=100(2m+1)^1+1=400m(m+1)+101. This satisfies all the conditions if m or m+1 is divisible by 9. The smallest of course is 101 (with m=0). The next two are 28901 and 36101 and so on.

Consider x=no. of pearls. 1)Then x-1 is perfect square 2)x-2 is divisible by 9 3)x-1 is divisible by 5 4)x-5 is divisible by 8. Now from condition 1 & 2 we have to find perfect square which is divisible by 5.. its simple like 5^2, 10^2... Now check other conditions...

take first 10 perfect squares 1 4 9 16 25 36 49 64 81 100

than add 1 to each value to get supposed number of pearls 2 5 10 17 26 37 50 65 82 101

now take prime numbers from these numbers 2 5 17 37 101

by subtracting 2 from each number

0 3 15 35 99

as from the above values only 99 is the multiple of 9

also if we divides 101 by 5 we shall get 1 left

and if we divide 101 by 8 , 5 will be left

hence 101 is the exact value that satisfies each point