Question

A boy was asked to find the L.C.M. of 3, 5, 12, and another number. But while calculating, he wrote 21 instead of 12 and yet came with the correct answer. What could be the fourth number?

Answer 1

Hint: Here, we need to find what the fourth number can be. We will assume the fourth number to be \[x\]. First, we will write down the given numbers as a product of their prime factors. Then, using the product of prime, we will observe the greatest power of the prime factors of 3, 5, 12 and the greatest power of the prime factors of 3, 5, 21. Since the L.C.M. obtained in both cases is the same, the prime factors with greatest powers must be the same in both the cases. We will compare the prime factors with greatest powers in 3, 5, and 12 with the prime factors with greatest powers in 3, 5, and 21. We will multiply the factors not common in both cases to find the fourth number.

Complete step-by-step answer:
Let the fourth number be \[x\].
The fundamental theorem of arithmetic states that every composite number can be written as a product of its prime factors in a unique way.
A prime factor is a factor of a number which is divisible by 1 and itself only.
First, we will write the given numbers as a product of their prime factors.
We know that 3 and 5 are prime numbers.
Therefore, the prime factor of 3 is 3 and the prime factor of 5 is 5.
Now, 12 is the product of 4 and 3.
Therefore, we can write 12 as
\[ \Rightarrow 12 = 4 \times 3\]
4 is the square of the prime number 2. Thus, we get
\[ \Rightarrow 12 = {2^2} \times 3\]
The number 21 is the product of the prime numbers 3 and 7.
Therefore, we can write 21 as
\[ \Rightarrow 21 = 3 \times 7\]
Now, we will observe the greatest power of the prime factors in the list of prime factors of 3, 5, and 12. The greatest power of 2 is 2, the greatest power of 3 is 1, and the greatest power of 5 is 1.
The lowest common multiple of the numbers 3, 5, 12, and \[x\] is the product of the prime factors with the greatest powers, which include \[{2^2}\], 3, and 5.
Also, in the list of prime factors of 3, 5, and 21, the greatest power of 3 is 1, the greatest power of 5 is 1, and the greatest power of 7 is 1.
The lowest common multiple of the numbers 3, 5, 21, and \[x\] is the product of the prime factors with the greatest powers, which include 3, 5, and 7.
Since the L.C.M. obtained in both cases is the same, the prime factors with greatest powers must be the same in both the cases.
Now, we will compare the prime factors with greatest powers in 3, 5, and 12 with the prime factors with greatest powers in 3, 5, and 21.
We can observe that the prime factor \[{2^2}\] is not present in the product of primes of 3, 5, and 21.
Also, we can observe that the prime factor 7 is not present in the product of primes of 3, 5, and 12.
This means that the fourth number will include the prime factors \[{2^2}\] and 7 when expressed as a product of primes.
Therefore, we can multiply these factors to get the smallest possible value of the fourth number.
Thus, we get
\[x = {2^2} \times 7\]
Multiplying the terms, we get
\[ \Rightarrow x = 4 \times 7 = 28\]
Therefore, the fourth number can be 28.

Note: Here we were provided with the LCM of the numbers. The L.C.M. is the product of the prime factors with the greatest powers. We have found 28 as the smallest possible value of the fourth number. Since the fourth number must contain \[{2^2}\] and 7 when expressed as a product of primes, it must have 28 as a factor. Therefore, any multiple of 28 can be the fourth number. The L.C.M. of 3, 5, 12, and any multiple of 28 will always be equal to the L.C.M. of 3, 5, 21, and that multiple of 28.