You may also like

problem icon

Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

problem icon

Ladder and Cube

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

problem icon

Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

Weekly Problem 12 - 2011

Stage: 3 and 4 Short Challenge Level: Challenge Level:1

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.

 
 

This problem is taken from the UKMT Mathematical Challenges.

View the previous week's solution
View the current weekly problem