Question

Let \[x \cdot y \cdot z = 105\] where \[x,y,z \in N\]. Then number of ordered triplets \[(x,y,z)\] satisfying the given equation is
A) \[15\]
B) \[27\]
C) \[6\]
D) \[33\]

Answer 1

Hint: Here we are given to find the number of possible ordered triplets that can be formed such that product of those three values is equal to \[105\]. To do this we will use the method of prime numbers, where we write prime numbers with their powers, needed for the factorization of \[105\]. After finding the factors we find the possible triplets. Then we use those prime numbers to find other factors also and find the total possible triplets.
Formula used: Total numbers of possible ways to organize three things \[a,b,b\] where two things are common is given by \[\dfrac{{3!}}{{2!}}\].
Total numbers of possible ways to organize three things \[a,b,c\] are \[3!\].

Complete step-by-step solution:
Here we have to factorize \[105\] using prime numbers. We can see that we can write \[105\] as,
\[105 = {3^1}{5^1}{7^1}\]
Hence we get the factors as \[3,5,7\]. Now we have to find the possible triplets that can be formed by these numbers. We see that,
\[3\] can be written at any place be it \[x,y\,or\,z\]. Same way numbers \[5and7\] can also be written.
Hence, total possible triplets \[ = 3!\].
Same,
For \[105 = 35 \times 3 \times 1\], we get possible triplets as \[3! = 6\]
For \[105 = 21 \times 5 \times 1\], we get possible triplets as \[3! = 6\]
For \[105 = 7 \times 15 \times 1\], we get possible triplets as \[3! = 6\]
For \[105 = 105 \times 1 \times 1\], as two factors are same, we get possible triplets as \[\dfrac{{3!}}{{2!}} = 3\]
Hence we get possible triplets as, \[6 + 6 + 6 + 6 + 3 = 27\]
So answer is B).

Note: Whenever we have to find the number of factors of any big number, we should always try to find them through the prime number method and then find other factors after multiplying them with one another as we have done here.