 # Week 10

Example 1. A boy was asked to find the LCM 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?

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Let the fourth number be x. First, we will write the given numbers as a product of their prime factors.

The fundamental theorem of arithmetic states that every composite number can be written as a product of its prime factors in a unique way.

Using prime factorization, we can write 12 = 22 × 3 and 21 = 3 × 7.

Thus, the first set of numbers are 3, 5 and 12 = 3, 5 and 22 × 3.

The second set of numbers are 3, 5 and 21 = 3, 5 and 3 × 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 L.C.M of the numbers 3, 5, 12 and x is the product of the prime factors with the greatest powers, which includes 22, 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 L.C.M 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 22 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 22 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 = 22 × 7 = 4 × 7 = 28.

Therefore, the fourth number is 28.

Note. Here we were provided with the L.C.M 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 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.

Clearly, xa and xb. Otherwise, the given equation becomes undefined. Transposing the right side elements to the left side, we have

Now, we are done with the various possible factorisation. Our task is to find the value of the variable x.

Therefore, either x = 0 or a + bx = 0, whence x = a + b.

Alternatively, we can rewrite the given equation as

Notice carefully that, both L.H.S and R.H.S are the sum of a number and its reciprocal with the equality in between them.

This is possible only when

As before, either x = 0 or x = a + b.