Dark room of treasure or doom

Logic Level 5

A dark room full of valuable orbs with superpowers : 1 orb inscribed 1, 2 orbs inscribed 2, ... , 50 orbs inscribed 50 on them [orbs inscribed 1 to 50 on them, with orb numbers respectively 1 to 50].

As soon as Jason Langdon entered the room, the lights turned out and the door got close. A genie asked Jason Langdon to pick at least 10 orbs with the same integer inscribed on them.

If he did so, he would get all orbs of that room. If he failed, he'd be stuck in this room forever.

Additionaly, the genie told him that for each orb he picked, he will remain asleep for a day at the middle of nowhere. JL didn't want to lose an extra day of his life, but he didn't wish to get stuck either.

In this dark room where a man can never determine which orb is being picked, what's the least number of orbs JL should pick, to remain safe and to win all those orbs?

The answer is 415.

2 solutions

Sheikh Asif Imran Shouborno
May 26, 2014

This was a problem from BdMO 2008 , where I've imposed a new story.

As we can see, JL cannot pick 10 orbs of 1 to 9 . Let him count his worst luck, say he has picked all orbs from 1 to 9 and 9 orbs of each value from 10 to 50 . JL is now certain that if he picks another orb, the campaign will come successful.

So, the answer is: $\frac{9(9+1)}{2}+9\times41+1=415$ . :D

2 2

3 3 3

4 4 4 4

.........................

......................................

50 50 50 50 ................... 50 50 50 50

Like pointed out by Sheikh, let us prepare for the worst. Consider the columns arranged as above - With the first column starting at 1 and ending at 50, second starting with 2 and ending with 50 and so on. Hence, the last column has only one item - 50.

Let JL have a thoroughly unlucky day so that he starts choosing along the columns beginning from top left each time. So, he begins to choose from 1 and goes to choose on till 50 to finish the first column. Moving onto the second column he begins to choose from 2 and ends with 50; and hence, successively for each column.

After he is done with the column whose top-most element is 9 he should know that as soon as he moves on to the next column (the one whose top-most element is 10) he would have at least 10 orbs with the same integer (i.e. 10 orbs with the number 10).

Therefore, we will count the number of orbs he picks after he is done with the column whose top-most element is 9.

Hence, first column = 50 orbs ; 2nd column = 49 orbs and so on till the 9th column = 42 orbs.

Our answer lies in the sum of 50, 49, 48, 47, 46....42 and 1. (Goes on till only 42 because the column whose top-most element is 9 has 42 orbs; Add one to accomodate for the last choice) which is 415!

Saurabh Dhole - 6 years, 5 months ago

oh! did it the same way but added 41 instead of 1 :p

Ashwin Upadhyay - 6 years, 4 months ago

Wonderful answer!! Thank you so much......☺

Phoenix Amyra - 5 years, 5 months ago

Exactly my thought! :)

Felipe Perestrelo - 5 years, 9 months ago

Noel Lo
Jul 18, 2015

Hello, this is my solution:

We first need to identify the worst case scenario. That Jason picked a large number of orbs and yet still did not manage to get at least 10 orbs with the same number. Specifically, we need to ask ourselves, what is the largest possible number of orbs picked such that Jason has at most 9 orbs with the same number?

To answer this question, we imagine a scenario where he picks orb 1, orb 2, orb 3 all the way until orb 50. We consider this round 1. For round 2, Jason would sart with orb 2 and finish at orb 50. For round 3, he will start at orb 3 and end at orb 50. and so on and so forth. We realise that after round 9, Jason will have the largest possible number of orbs where he has at most 9 orbs with the same number. This maximum possible number is 50 + 49 + 48 + 47+46+45+44+43+42 =414.

We see that if Jason picks JUST ONE more orb, the number on it will DEFINITELY be the same as that on 9 other orbs. This additional number can only be 10 to 50 and he already has 9 of each (from 10 to 50). So the minimum number required is 414+1 = 415.

Dark room of treasure or doom

The answer is 415.

2 solutions

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