A poster measures 64 cm by 36 cm and is to be cut into identical squares with no material left over. What is the largest possible side length of the squares, and what is the smallest number of squares?

Step-by-step solution

1. 1

   Translate “nothing left over” into a divisibility condition. If the square has side $s$, then whole squares must fit along both edges, so $64$ and $36$ must each be a whole multiple of $s$:

   $s\mid 64\quad\text{and}\quad s\mid 36$

   In other words $s$ is a common factor of $64$ and $36$, and the largest permissible $s$ is their greatest common factor. This is the whole content of the problem — the geometry only supplies the divisibility.

2. 2

   Find the GCF by prime factorisation.

   $64=2^6,\qquad 36=2^2\times3^2$

   The GCF takes each shared prime to the smaller of its two exponents. The prime $3$ appears only in $36$, so it contributes nothing; the prime $2$ appears with exponent $6$ and $2$, so it contributes $2^2$:

   $\gcd(64,36)=2^2=4$

   So the largest square has side $4$ cm. (The Euclidean algorithm gives the same result faster: $64=1\cdot36+28$, $36=1\cdot28+8$, $28=3\cdot8+4$, $8=2\cdot4+0$, so the last non-zero remainder is $4$.)

3. 3

   Count the squares along each edge.

   $\frac{64}{4}=16\ \text{squares along the length},\qquad \frac{36}{4}=9\ \text{squares along the width}$

   Both divisions come out whole, confirming the tiling really works with no offcuts.

4. 4

   Multiply to get the number of squares.

   $16\times9=144\ \text{squares}$

   Check it against the areas: the poster is $64\times36=2304$ cm² and each square is $4\times4=16$ cm², and $2304\div16=144$. ✓ The two counts must agree because the pieces exactly cover the sheet.

5. 5

   Explain why the largest square gives the fewest squares. The number of pieces is $\dfrac{2304}{s^2}$, which decreases as $s$ grows. So maximising the side length and minimising the piece count are the same question asked twice — the two parts of the problem are answered by the single number $\gcd(64,36)=4$. Any smaller common factor works too but wastes cuts: $s=2$ gives $576$ squares and $s=1$ gives $2304$.