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Given two natural numbers N and K where N >= 2, you have a list of available numbers as all numbers in inclusive range K to K+N-1. You can sum any two natural numbers in available list to create a new natural number, which is then added to list of available numbers. How many natural number exist which cannot be created by this method?

I'm tempted to say none.

Pick a number. Let's say 18743286432987543895.

I can make that easily. My starting values are K=2 and N=7584750849753846743087565784367894.

That means my list is every number from 2 to 7584750849753846743087565784367894 + 2 - 1, which contains that first number.

Basically, there is no number you can name that I cannot get onto my list, because I just pick my starting values to be 2 and some number bigger than your number.

Or are you saying that you get a given a set, fixed N and K, and you have to identify all the numbers that cannot be added to the list for that particular N and K?
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there is set of all the natural numbers upto infinity and the N and K are fixed
closed account (STD8C542)
@Repeater in your case you cannot get 1 in your list
can anybody tell the answer for n=2 and k=6.

plz explain.
With n=2 and k=6, here are the numbers you can get:

2,3,4,5,6,7 to begin with.

Now you can get all remaining numbers to infinity, by adding 2 to the second highest number in your list, forever.

2 +6 = 8
2 +7 = 9
2 + 8 = 10
2 + 9 = 11
2 + 10 = 12

and so on.

So with the numbers n=2 and k=6, you can get every positive integer from 2 to infinity.

this is wrong i think..
We have a range k to k+n-1 both inclusive,i.e,6 and 7.
So why we are starting from 2?

here is the link to the question..

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Oh, right. I assumed that it didn't matter which was n and which was k, so I just swapped them round. I thought this was a number theory question, but it seems it's another codechef.

can u plz tell the approach
I think thats a problem from an ongoing contest.....use a pen and a paper and write down answers for 2 or 3 examples you will see a pattern....give it time if u want to learn!!
can u plz explain the test case
n=2 k=6?
6 7 12 13 14 18 19 20 21 24 25 26 27 28 30 31 32 33 34 35....(rest all)

So number of missing numbers are 5+4+3+2+1

I guess that explains a lot!!
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So number of missing numbers are in these....

So what's the answer? (And can't you make 12 and 14, for example?)
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yes 12 and 14 can be made.
closed account (STD8C542)
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@iotaa yes sure....i missed that....i mean to say just keep adding numbers already in the array...and after a while all the numbers will start appearing!!
@iotaa fixed....just realised its a super easy problem!!
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closed account (STD8C542)
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still getting wa
the pattern is quite simple if you write it down with examples, still I think there is a boundary value of something that is giving me WA in 1 task.

If any of you guys have completed the problem, can you provide some inputs and there right outputs, like large values and such?
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