Question

The number of factors (excluding 1 and the expression itself) of the product of ${{a}^{7}}{{b}^{4}}{{c}^{3}}def$ where a, b, c, d, e, f are prime numbers is
\[\begin{align}
  & a)634 \\
 & b)1872 \\
 & c)1278 \\
 & d)2078 \\
\end{align}\]

Answer 1

Hint: In the given question we have a number which is formed by multiplying prime numbers. Now we know that if we have \[N={{p}_{1}}^{{{a}_{1}}}{{p}_{2}}^{{{a}_{2}}}.....{{p}_{n}}^{{{a}_{n}}}\] such that all of the number ${{p}_{1}},{{p}_{2}},...........,{{p}_{n}}$ are prime numbers then Total number of factors is given by $\left( {{a}_{1}}+1 \right)\left( {{a}_{2}}+1 \right).....\left( {{a}_{n}}+1 \right)$. Hence we can calculate the total number of factors of the given expression by the above formula.

Complete step by step answer:
Now we are given with the number ${{a}^{7}}{{b}^{4}}{{c}^{3}}def$ where a, b, c, d, e, f are all prime numbers.
Now if we have a number N such that \[N={{p}_{1}}^{{{a}_{1}}}{{p}_{2}}^{{{a}_{2}}}.....{{p}_{n}}^{{{a}_{n}}}\] and all the numbers ${{p}_{1}},{{p}_{2}},...........,{{p}_{n}}$ are prime numbers then we know that the total number of factors of N is given by $\left( {{a}_{1}}+1 \right)\left( {{a}_{2}}+1 \right).....\left( {{a}_{n}}+1 \right)$ .
Let us first understand how the formula arrives. We will understand this with help of an example.
Now let us say we have a number N and it is formed by prime numbers p, q, r such that $N={{p}^{2}}{{q}^{3}}r$
Now to find the total number of factors we will find how many number of ways a factor can be possible.
Now we have 2 p’s, 3 q’s and 1 r.
Now to form a number we can either take 0 p’s, 1 p, or both p’s. hence we have 3 options.
Similarly we can take 0 q’s, 1 q, 2 q,’s or 3 q’s hence here we have 4 options.
And finally for r we have either we take 0’ r or 1 r. hence we have 2 options.
Hence we can see the pattern that for ${{p}^{n}}$ we will get n + 1 options.
Hence the total options for our number N will be 3 × 4 × 2 = 24.
Now let us apply the formula on our given number ${{a}^{7}}{{b}^{4}}{{c}^{3}}def$
Here the total number of factors will be $\left( 7+1 \right)\times \left( 4+1 \right)\times \left( 3+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right).$
Hence we have a total number of factors = 8 × 5 × 4 × 2 × 2 × 2 = 1280.
Now note that the total number of factors contains 1 and the number itself too.
Hence to calculate the number of factors (excluding 1 and the expression itself) we will subtract 2.
Hence we have required a number of factors as 1280 – 2 = 1278.

So, the correct answer is “Option C”.

Note:
Now note that the total number of factors contain the number itself as well as 1. Now since we are choosing primes to find a number of factors, the possibility where we chose none of the prime numbers makes us select 1 and the possibility that we chose all the prime numbers makes us select the number itself. For understanding consider our example. If we take 0 p’s, 0q’s and 0 r’s then we are taking the number 1 as ${{p}^{0}}{{q}^{0}}{{r}^{0}}=1$ similarly if we take 2 p’s, 3 q’s and 1 r we have formed the number ${{p}^{2}}{{q}^{3}}r$ which is the number itself.