This one is quick and interesting.
Natural number ends with 2.If we remove this 2 and place it at the beginning, the number gets doubled.Find the smallest possible original number.
(e.g. number is say 142. New number will be 214. 214 is not twice 142 so 142 is not such number).

Paul Clapham wrote:No need to wait... the title of this forum is Programming Diversions. While the question is certainly diverting, I haven't seen any programming yet.

Maybe you could interpret the title as 'diversions AWAY from programming', thus no programming required.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

So, we just look for a value of n that makes x an integer ending with 2. The n vs (n+1) doesn't matter, but the -4 vs -2 will produce different answers.

So the answer is 105263157894736842. The check for the last digit being 2 turned out to be unnecessary, but I think that's a coincidence.

ETA: In reading through Mike's solution, I see the -4 makes sense if you declare x to be the number without the final 2. Sorry about that!

I wonder if we can prove we don't have to check the final digit, as both Mike and I discovered?

Mike Simmons
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Eugene is quoting the same formula I used above. And we got the same answer - except that as I noted above, my 'x' represented all the digits other than the 2. So I got

I tried some thing very simple and may sound very stupid....

BigInteger i,a,b;
for(i=0;i<100000000000000;i++)
{
String x,y;
String num = Integer.toString(i);
x = num+"2";
y = "2"+num;
int a = Integer.parseInt(x);
int b = Integer.parseInt(y);

if(a==(b/2))
System.out.println(a+" "+b);
}

did not get any no...though....

SCJP

Mike Simmons
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Posts: 3020

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Rohini Sahuji wrote:did not get any no...though....

It's possible this is caused by the fact that your code does not compile. That's just a guess though.