Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

Question

Homi · Accepted Answer

Understand the Question

To find the smallest number divisible by $4$ , $9$ , and $10$ , we first compute their Least Common Multiple (LCM).
For a number to be a perfect square, every prime factor in its prime factorisation must have an even exponent (occur in pairs).
If any prime factor in the LCM has an odd exponent, we multiply the LCM by that factor to make all powers even, giving the smallest required square number.

Step 1 · Find the LCM of 4, 9, and 10

Prime factorise each number

\begin{aligned} 4 &= 2 \times 2 = 2^2 \\ 9 &= 3 \times 3 = 3^2 \\ 10 &= 2 \times 5 = 2^1 \times 5^1 \end{aligned}

Taking the highest power of each prime factor

\begin{aligned} \text{LCM}(4, 9, 10) &= 2^2 \times 3^2 \times 5^1 \\ &= 4 \times 9 \times 5 \\ &= 36 \times 5 \\ &= 180 \end{aligned}

Step 2 · Make the LCM a Perfect Square

The prime factorisation of the LCM is $180 = 2^2 \times 3^2 \times 5^1$

The prime factors $2$ and $3$ are paired (even power of $2$ ), but the prime factor $5$ is unpaired (power of $1$ ).

To make $180$ a perfect square, multiply it by $5$

\begin{aligned} \text{Smallest square number} &= 180 \times 5 \\ &= 900 \end{aligned}

Checking the prime factorisation of $900$ $900 = 2^2 \times 3^2 \times 5^2$

Since all prime factors have even powers, $900$ is the smallest perfect square divisible by $4$ , $9$ , and $10$ .

Answer

$900$

Common Mistakes

Stopping at LCM: Assuming the LCM ( $180$ ) is the final answer without checking whether it is a perfect square.
Multiplying Unnecessary Factors: Multiplying the LCM by all prime factors instead of only the unpaired factor ( $5$ ).
Odd Powers in Perfect Squares: Overlooking that every prime factor in a perfect square must have an even exponent.

Question 5