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11. Find the smallest square number which is completely divisible by each of the numbers 10, 16 and 24.

Answer»

The PF of 10 is 2*5. 10 is not a perfect square, because 2 and 5 only appear once, and one is odd.

The PF of 16 is 2*2*2*2. 16 is a perfect square, because 2 appears four times, and four is even.

The PF of 24 is 2*2*2*3. 24 is not a perfect square, because 2 appears three times, and 3 appears once. Three and one are both odd numbers.

Any multiple of 10, 16, and 24mustcontain each of their prime factorsat leastas many times as they appear in the numbers being multiplied. So for example, the LCM (least common multiple) of 10, 16, and 25 must contain four factors of 2 (since 2 appears four times in the PF of 16), one factor of 3 (since 3 appears once in the PF of 24, and nowhere else), and one factor of 5 (since 5 appears once in the PF of 10, and nowhere else. Here is the PF for the LCM of 10, 16, and 24:

2*2*2*2*3*5 = 240

But 240 isn’t a perfect square, since the factors 3 and 5 appear only once.

If we multiply 240 by any other number, the result willstillbe divisible by 10, 16, and 24. And since the number we’re seeking has even numbers of prime factors, we can simply multiply 240 by 3*5 (or 15) to get the smallest perfect square that is divisible by 10, 16, and 24:

2*2*2*2*3*3*5*5 =3600.



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