Puzzle asked in Oracle : What is the number?

[sk_kireeti]

There are 5 digits (XXXXX) in a number and each digit of the number has a specific meaning.

1 st digit -> represents number of 0's in the number
2 nd digit -> represents number of 1's in the number
3 rd digit -> represents number of 2's in the number
4 th digit -> represents number of 3's in the number
5 th digit -> represents number of 4's in the number

What is the number?

Also explain your approach to solution if there is any logical approach.

Thanks,
Krishna

[romeovictor04]

21200 Am I Right

[sk_kireeti]

You are correct romeovictor. It is the only possible number.

Is there any logical approach for this? If you can explain how you scoped the number of possibilities to 1 would be of very useful to us.

Thanks,
Krishna

[sk_kireeti]

Here is the java program which does the same....


/* **************************************************** */
/* here is the puzzle for which the code finds solution.*/
/* */
/* There are 5 digits (XXXXX) in a number and each digit*/
/* of the number has a specific meaning. */
/* */
/* 1 st digit -> represents number of 0's in the number */
/* 2 nd digit -> represents number of 1's in the number */
/* 3 rd digit -> represents number of 2's in the number */
/* 4 th digit -> represents number of 3's in the number */
/* 5 th digit -> represents number of 4's in the number */
/* */
/* What is the number? */
/* **************************************************** */

public class puzzle
{

static int a[]=new int[5];

static int getNumberCount(int searchnumber)
{
int count=0;
for(int i=0;i<5;i++)
{
if(a[i]==searchnumber)
count++;
}
return count;
}

public static void main(String args[])
{
for(a[0]=1;a[0]<=4;a[0]++)
for(a[1]=0;a[1]<=4;a[1]++)
for(a[2]=0;a[2]<=4;a[2]++)
for(a[3]=0;a[3]<=4;a[3]++)
for(a[4]=0;a[4]<=4;a[4]++)
{
if(a[0]==getNumberCount(0) && a[1]==getNumberCount(1) &&a[2]==getNumberCount(2) &&a[3]==getNumberCount(3) &&a[4]==getNumberCount(4))
System.out.println("Number : "+a[0]+a[1]+a[2]+a[3]+a[4]);
}
}
}

[romeovictor04]

the only possible logical solution is elimination.
i started with the first number and eliminated all other numbers except 1 & 2
if i give 1 then the second number will be 2 or 3
and the third number will be either 2 or 3
then it wont be possible to solve the equation

thei i put 2 in the first place and in the third place
i had only one place left and that could be only the first place and the number could be only 1.

thats all

RV

[deesam]

HEY I TOO GOT IT..IT IS 21200..THANQ 4 NICE PUZZLE

[Nips]

21200
31000
40000

are the possible solution (there might be few more ).

Logic is just arrangement and 0 plays great deal of role in it .

Originally Posted by Nips

21200
31000
40000

are the possible solution (there might be few more ).

Logic is just arrangement and 0 plays great deal of role in it .

21200 is correct, but ...
31000 says there are no 3's.
40000 says there are no 4's.

[aegis007]

Originally Posted by Nips

21200
31000
40000

are the possible solution (there might be few more ).

Logic is just arrangement and 0 plays great deal of role in it .

"21200" is alone correct.

In "31000" u used number 3, then u should specify the number of 3's in the 4th digit.
Similarly in "40000" u used 4 in the first digit but should specify number of 4's in the 5th digit.

[Annapurna_Kulkarni]

Hi Kireeti.

Could you please explain me clearly i am confused !!!!!!!!!!

please explain me the method how you selected those numbers only adn how you concluded ..... Please do reply