Answers to “Can you solve these puzzles?”

This page contains the answers to these puzzles.

Make the Sums

make-the-sums-ans

Unit Octagon

Name the regions as follows:
octagon-a1
$E$ is a 1×1 square. Placed together, $A$, $C$, $G$ and $I$ also make a 1×1 square. $B$ is equal to $H$ and $D$ is equal to $F$.
Therefore $B+E+F=A+C+D+G+H+I$. Therefore the hatched region is $C$ larger than the shaded region. The area of $C$ (and therefore the difference) is $\frac{1}{4}$.

Reverse Bases

If $ab$ in base 10 is equal to $ba$ in base 4, then $10a+b=4b+a$. So, $9a=3b$. $a$ and $b$ must both be less than 4, as they are digits used in base 4, so $a=1$ and $b=3$. So 13 in base 10 is equal to 31 in base 4.

By the same method, we find that:

  • 23 in base 10 is equal to 32 in base 7.
  • 46 in base 10 is equal to 64 in base 7.
  • 12 in base 9 is equal to 21 in base 5.
  • 24 in base 9 is equal to 42 in base 5.

This page contains the answers to these puzzles.

Make the Sums

make-the-sums-ans

Unit Octagon

Name the regions as follows:
octagon-a1
$E$ is a 1×1 square. Placed together, $A$, $C$, $G$ and $I$ also make a 1×1 square. $B$ is equal to $H$ and $D$ is equal to $F$.
Therefore $B+E+F=A+C+D+G+H+I$. Therefore the hatched region is $C$ larger than the shaded region. The area of $C$ (and therefore the difference) is $\frac{1}{4}$.

Reverse Bases

If $ab$ in base 10 is equal to $ba$ in base 4, then $10a+b=4b+a$. So, $9a=3b$. $a$ and $b$ must both be less than 4, as they are digits used in base 4, so $a=1$ and $b=3$. So 13 in base 10 is equal to 31 in base 4.

By the same method, we find that:

  • 23 in base 10 is equal to 32 in base 7.
  • 46 in base 10 is equal to 64 in base 7.
  • 12 in base 9 is equal to 21 in base 5.
  • 24 in base 9 is equal to 42 in base 5.

Matthew is a postdoctoral researcher at University College London. He hasn’t had time to play Klax since the noughties, but he’s pretty sure that Coke is it!

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