Question

Look at the following pattern:
(a) \[28={{2}^{2}}\times {{7}^{1}}\], Total number of factors (2 + 1) (1 + 1) = 3 \[\times \] 2 = 6. 28 is divisible by 6 factors i.e. 1, 2, 4, 7, 14, 28. Find the pattern.

Answer 1

Hint: Assume any composite number ‘N’ whose prime factors are p, q and r. Write N = \[{{p}^{a}}\times {{q}^{b}}\times {{r}^{c}}\], where a, b and c are the exponents to which p, q and r must be raised, respectively, so that their product must be equal to N. Now, observe the given pattern and find the total number of factors given as: - (a + 1) (b + 1) (c + 1).

Complete step-by-step solution
Here, we have been provided with a relation: - \[28={{2}^{2}}\times {{7}^{1}}\] whose total number of factors is given as: - (2 + 1) (1 + 1) = 3 \[\times \] 2 = 6. We have to find a pattern for any composite number.
Let us assume a composite number ‘N’ whose prime factors are assumed to be p, q, and r. These prime factors can be obtained by using the prime factorization method. We know that any composite number can be written as the product of their prime factors. These prime factors can appear more than one time so here we are assuming that ‘p’ appears ‘a’ times, ‘q’ appears ‘b’ times and ‘r’ appears ‘c’ times. Therefore, we can write,
\[\Rightarrow N={{p}^{a}}\times {{q}^{b}}\times {{r}^{c}}\]
Now, observing the pattern given in the question, \[28={{2}^{2}}\times {{7}^{1}}\], we can see that the total number of factors of 28 is given as (2 + 1) (1 + 1) = 3 \[\times \] 2 = 6. So, 1 is added to the exponents of each prime factor and then multiplied together to get the total number of prime factors. Therefore, for the assumed composite number ‘N’, we can write, using equation (1),
\[\Rightarrow \] Total number of factors = (a + 1) (b + 1) (c + 1)
Hence, ‘N’ is divisible by (a + 1) (b + 1) (c + 1) factors.

Note: One may note that we can check our answer by assigning any composite number to the value of ‘N’. You can see that here we have assumed only 3 prime factors of N. It may be possible that it has more than 3 factors, so in that case, we have to assume some more prime factors like s, t, u, etc and their exponents as d, e, f and so on. We have to find the total number of factors accordingly by adding 1 to each prime factor and then taking their product.