Weekly Problem 12 - 2011
The following seven pairs add to $16$ so at least one of each pair
must be removed:
$(1,15)$, $(2,14)$, $(3,13)$, $(4,12)$, $(5,11)$, $(6,10)$,
$(7,9)$.
If removing these seven is sufficient, then we would be left with
$8$, $16$ and seven others.
But
$16+9 = 25$
So we must remove $9$ and keep its partner
$7$
$7+2 = 9$
So we must remove $2$ and
keep $14$
$14+11 = 25$ So we
must remove $11$ and keep $5$
$5+4 = 9$
So we must remove $4$ and keep
$12$
$12+13 = 25$ So we
must remove $13$ and keep $3$
$3+1 = 4$
So we must remove $1$ and keep
$15$
$15+10 =
25$ So we must remove $10$ and
keep $6$
But we have kept $3$ and
$6$ which add to $9$.
Hence it is not
sufficient to remove only seven. If we remove the number $6$, we
obtain a set which satisfies the condition:
$\{8,16,7,14,5,12,3,15\}$ or in ascending order
$\{3,5,7,8,12,14,15,16\}$.
Hence eight is the
smallest number of numbers that may be removed.