Question

If \[r,\,s,\,t\] are prime numbers and \[p,\,q\] are the positive integers such that the LCM of \[p,\,q\] is \[{{r}^{2}}{{s}^{4}}{{t}^{2}}\], then the ordered pair \[(p,q)\] is
A. 252
B. 254
C. 225
D. 224

Answer 1

Hint: It is given that LCM \[(p,q)={{r}^{2}}{{s}^{4}}{{t}^{2}}\] that means it is at least one of the p and q must have \[{{r}^{2}},\,\,{{s}^{4}}\] and \[\,{{t}^{2}}\] in their prime factorization. Now, we have to consider each case for selecting the number of ways for power of \[r,\,\,s\,\,\,\text{ and }\,\,t\]. So, in this way we have to approach the problems.

Complete step by step answer:
According to the given condition that LCM \[(p,q)={{r}^{2}}{{s}^{4}}{{t}^{2}}\]
That is at least one of the p and q should have \[{{r}^{2}},\,\,{{s}^{4}}\] and \[\,{{t}^{2}}\] in primary factorization.
Now, we will be considering each case for selecting the power of r, s and t.
First we consider the power of r
It can be done by
Case 1: p contains \[{{r}^{2}}\] then q has \[{{r}^{k}}\] with \[k=(0,1)\]
That is, the number of ways should be\[=2\].
Case 2: q contains \[{{r}^{2}}\] then q has \[{{r}^{k}}\] with \[k=(0,1)\]
That is the number of ways\[=2\].
Case 3: Both p and q contain\[{{r}^{2}}\].
Then, number of ways\[=1\]
Therefore, exponent of r may be chosen in \[2+2+1=5\]

Similarly we can also perform for selection of s.
Case 1: p contains \[{{s}^{4}}\] then $q$ has \[{{s}^{k}}\] with \[k=(0,1,2,3)\].
That is, the number of ways should be\[=4\].
Case 2: q contains \[{{s}^{4}}\] then q has \[{{s}^{k}}\] with \[k=(0,1,2,3)\]
That is the number of ways\[=4\].
Case 3: Both p and q contain \[{{s}^{4}}\].
Then, number of ways\[=1\].
Therefore, the exponent of s may be chosen in \[4+4+1=9\].

Similarly we can also perform for selection of t.
Case 1: p contains \[{{t}^{2}}\] then q has \[{{t}^{k}}\] with \[k=(0,1)\]
That is, the number of ways should be\[=2\].
Case 2: q contains\[{{t}^{2}}\] then q has \[{{t}^{k}}\] with \[k=(0,1)\]
That is the number of ways\[=2\].
Case 3: Both p and q contain \[{{t}^{2}}\].
Then, number of ways\[=1\]
Therefore, the exponent of r may be chosen in \[2+2+1=5\].
Thus, the total number of ways is:
\[5\times 9\times 5=225\]
Hence the number of ordered pairs\[(p,q)\] is \[225\].

So, the correct answer is “Option C”.

Note: Be careful about selecting the r s and t for this particular problem. Because number ways for each selection of r, s and t there are 3 cases. Because there are two values given\[p,\,q\]. Hence to calculate the total number of ways is to multiply the number of ways for r, number of ways for s and number of ways for t. So, in this way to solve this particular type of problem.