Comments on ScribblinG: Puzzle: Help Jim

@Ramkumar Yup that was a bug, thanks for pointing ...

2011-02-18T12:44:04.975+05:30

@Ramkumar Yup that was a bug, thanks for pointing it.

Correct me if I am wrong, but Vijesh, should not t...

2011-02-18T12:29:34.664+05:30

Correct me if I am wrong, but Vijesh, should not the code have another line:
reminder = a%b, after you reinitialize a as a = reminder*10 ;

Otherwise no matter what the loop will run only once because reminder is already added to the Hash Set.

Spot on Vijesh - the same remainder is the key. :)...

2011-02-15T20:07:33.060+05:30

Spot on Vijesh - the same remainder is the key. :)

Finding Q would be straight forward. Now for the r...

2011-02-15T15:48:31.340+05:30

Finding Q would be straight forward. Now for the reminder part (D), keep iterating until the reminder number reccurs, that will help in identifying the recurring decimal number.
Something like 4/3 will give 1 as reminder first, then again it will give reminder 1. So how many times we repeatedly divide 1 it will give the same recurring pattern. For this one needs to remember the reminders that occurred previously.

HashSet hSet = new HashSet();
int reminder = a%b;
while(!hSet.contains(reminder)) {
hSet.add(reminder);
a = reminder * 10;
}

Let me know if there is a better solution.

If A / B = Q.D where a and b are 2 + integers Q...

2011-02-13T14:35:38.825+05:30

If

A / B = Q.D

where a and b are 2 + integers
Q is the integer part of the result and D the decimal part of the result.

Prob spec. Write a program that will accept 2 integers A and B as input and print Q.D and terminate with a newline. If D is a recurring decimal print the recurring part only once.

examples:
a=10 b=5 Q.D = 2.0
a=5 b=2 Q.D = 2.5
a=10 b=6 Q.D = 1.6
a=7 b=6 Q.D = 1.16

Pseudocode is fine, C or Java for implementation.

The Visitor, that's right.

2011-02-13T11:18:11.470+05:30

The Visitor, that's right.

Correction: ... in a sequence of 3 continuous n...

2011-02-12T21:42:12.650+05:30
Correction:
... in a sequence of 3 continuous numbers atleast one will be divisible by 3 ...

to be read as:

... in a sequence of 3 continuous numbers one will be divisible by 3 ...

That the number will be divisible by 2 is obvious,...

2011-02-12T21:35:37.751+05:30
That the number will be divisible by 2 is obvious, because every 2nd number will be divisible by 2. As the previous and next numbers are primes the middle number will be divisible by 2.

Similarly the number will be divisible by 3 because in a sequence of 3 continuous numbers atleast one will be divisible by 3. By the same argument as above the middle number will be divisible by 3.

Any number divisible by both 2 and 3 will be divisible by 6.