### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Calendar Capers

Choose any three by three square of dates on a calendar page...

### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

# No Square Sums

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

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.