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There are 13 people who hold a number, the first person holds an integer number, the second holds a number greater than the first and the next person hold a number greater than the previous one and so on. At the end:

  • The sum of the first 12 people's number is greater than 0.
  • The sum of 13 people's number is $13$.

What is the minimum value of the first person's number?

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  • $\begingroup$ @apm yes minimum. $\endgroup$ – Oray Jul 4 '18 at 7:27
  • $\begingroup$ Okay looks like a good puzzle. Let me try :) $\endgroup$ – apm Jul 4 '18 at 7:29
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(Looks like @apm got the same answer while I was writing, he was the first one to publish this particular answer.)

The first number can be as small as

$\bf{-142}$

Because of the restrictions, the 13th number cannot be more than $13 -\epsilon$.

Because of the "ordered numbers" restriction, all the earlier numbers must be smaller than that, for example $13-2\epsilon$, $13-3\epsilon$ and so on. (The exact choice of the numbers isn't important, as long as they are "very close" to 13)

The second smallest number must make for an integer sum with the larger numbers, so its non-integer part must sum up to an integer with the epsilons (all $-66$ of them) from earlier. Therefore the second number cannot be larger than $12+66\epsilon$, so it can contribute only $12$ to the total, while the numbers 3 to 13 will be able to contribute a whole $13$. Therefore the maximum sum from the the last 12 numbers is

$11 \times 13 + 12 = 155$

and therefore, the sum can add up to 13 when

the first number is $13 - 155 = -142$.

If we choose $\epsilon = 0.01$, the numbers become

$-142$
$12.66$
$12.89$
$12.90$
$12.91$
$12.92$
$12.93$
$12.94$
$12.95$
$12.96$
$12.97$
$12.98$
$12.99$

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If every number is integers, it will be

-65

So numbers will be.

-65,1,2,3,4,5,6,7,8,9,10,11,12

If we support non integers.

Smallest will be

-142

Numbers will be like(not exact numbers in between)

-142,12.99990<12.99991<12.99992<12.99993<12.99994<12.99995<12.99996<12.99997<12.99998<12.99999<12.999991<12.999999

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  • $\begingroup$ only the first number requires to be integer, the rest doesnt have to be. $\endgroup$ – Oray Jul 4 '18 at 7:43
  • $\begingroup$ Ok. Added that case also $\endgroup$ – apm Jul 4 '18 at 7:52

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