We are required to find the least possibe square number that is exactly divisible by the numbers 10, 12 and 16. In this question, I found out the required least possible square number is 3600. I did that by hit and trial method by taking out number randomly and then squaring and checking to see if the square number is exactly divided by a the three numbers. I just want to know the definite way to find the answer if there is any.

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    $\begingroup$ $900$ isn't divisible by $16$. $\endgroup$ – Daniel Fischer Jun 2 '14 at 14:01
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    $\begingroup$ Look at the prime factorizations of $10$, $12$, $16$. The answer is in there. $\endgroup$ – Jyrki Lahtonen Jun 2 '14 at 14:02

$10=2\times 5$

$12=2^2\times 3$


From here, we obtain $\text{lcm} (10,12,16)=2^4\times 3\times 5$, hence the least square number required is $2^4\times 3^2\times 5^2=3600$.

  • $\begingroup$ Nice, the smallest number divisible by any quantity of numbers is the lcm of those numbers, and then you just made all the exponents even so it was a square right? $\endgroup$ – Jorge Fernández Hidalgo Jun 2 '14 at 14:04
  • $\begingroup$ taking number for exampe 10,12 and 18, the lcm of those numbers is 180 and squaring 180 makes it 32400. However, 900 is also a square number and is divisible by a three numbers. $\endgroup$ – Ufomammut Jun 3 '14 at 0:14
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    $\begingroup$ Newer question asked by OP to clarify here. $\endgroup$ – 6005 Jun 4 '14 at 3:00

The key observation is that if a prime $p$ divides $n^2$ then $p$ divides $n$.

Now consider the given numbers:

$10 = 2 \cdot 5$

$12 = 2^2 \cdot 3$

$16 = 2^4$

This means that you need $n$ to be a multiple of $2^2 \cdot 3 \cdot 5$.

The smallest such $n$ is $60$ and square you seek is $3600$.


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