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If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)
- a)252
- b)254
- c)225
- d)224
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q...
∵ r, s, t are prime numbers,
∴ Section of (p, q) can be done as follows
∴ Section of (p, q) can be done as follows
∴ r can be selected 1 + 1 + 3 = 5 ways
Similarly s and t can be selected in 9 and 5 ways respectivley.
∴ Total ways = 5 × 9 × 5 = 225
∴ Total ways = 5 × 9 × 5 = 225
Most Upvoted Answer
If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q...
Where M is the number of distinct prime factors of r2s2t4.
First, we can write the prime factorization of p and q as:
p = r^a * s^b * t^c * x
q = r^d * s^e * t^f * y
where x and y are coprime to r, s, and t.
The LCM of p and q is then:
LCM(p, q) = r^max(a,d) * s^max(b,e) * t^max(c,f) * xy
Since the LCM is r2t4s2, we have:
max(a,d) = 2, max(b,e) = 0, max(c,f) = 2
This means that for p and q, we have:
p = r^2 * t^2 * x
q = r^2 * t^2 * y
We also know that r, s, and t are prime. Therefore, there are only two possible values for each of a, b, c, d, e, and f: either 0 or 2. This gives us a total of 2^6 = 64 possible combinations of exponents for p and q.
However, we must also ensure that x and y are coprime to r, s, and t. Since x and y can be any positive integers, there are t-1 choices for x and s-1 choices for y (since they cannot be divisible by t and s, respectively). Therefore, the total number of ordered pairs (p, q) is:
64 * (t-1) * (s-1)
Finally, we can simplify this expression using the fact that r2s2t4 has 3 distinct prime factors: r, s, and t. Therefore, M = 3 and the number of ordered pairs is:
64 * (t-1) * (s-1) = 64 * (t*s - t - s + 1)
= 64 * ((r2s2t4)^(1/4) - r^(1/2) - s^(1/2) + 1)
= 64 * (2^(3/4) - 2 - 3 + 1)
= 64 * (2^(3/4) - 4)
= 64 * (2^(3/4) - 2^(2))
= 64 * (2^(3/4) - 2^(4/4))
= 64 * (2^(3/4) - 2^(1))
= 64 * (2^(3/4) - 2)
= 64 * (1.682 - 2)
= 64 * (-0.318)
= -20.192
Rounding down to the nearest integer, we get:
-20, so the answer is 2006 - 3M - 20 = 1956.
First, we can write the prime factorization of p and q as:
p = r^a * s^b * t^c * x
q = r^d * s^e * t^f * y
where x and y are coprime to r, s, and t.
The LCM of p and q is then:
LCM(p, q) = r^max(a,d) * s^max(b,e) * t^max(c,f) * xy
Since the LCM is r2t4s2, we have:
max(a,d) = 2, max(b,e) = 0, max(c,f) = 2
This means that for p and q, we have:
p = r^2 * t^2 * x
q = r^2 * t^2 * y
We also know that r, s, and t are prime. Therefore, there are only two possible values for each of a, b, c, d, e, and f: either 0 or 2. This gives us a total of 2^6 = 64 possible combinations of exponents for p and q.
However, we must also ensure that x and y are coprime to r, s, and t. Since x and y can be any positive integers, there are t-1 choices for x and s-1 choices for y (since they cannot be divisible by t and s, respectively). Therefore, the total number of ordered pairs (p, q) is:
64 * (t-1) * (s-1)
Finally, we can simplify this expression using the fact that r2s2t4 has 3 distinct prime factors: r, s, and t. Therefore, M = 3 and the number of ordered pairs is:
64 * (t-1) * (s-1) = 64 * (t*s - t - s + 1)
= 64 * ((r2s2t4)^(1/4) - r^(1/2) - s^(1/2) + 1)
= 64 * (2^(3/4) - 2 - 3 + 1)
= 64 * (2^(3/4) - 4)
= 64 * (2^(3/4) - 2^(2))
= 64 * (2^(3/4) - 2^(4/4))
= 64 * (2^(3/4) - 2^(1))
= 64 * (2^(3/4) - 2)
= 64 * (1.682 - 2)
= 64 * (-0.318)
= -20.192
Rounding down to the nearest integer, we get:
-20, so the answer is 2006 - 3M - 20 = 1956.
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If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer?.
If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the LCM of p, q is r2t4s2, where r, s, t are prime numbers and p, q are the positive integers then the number of ordered pair (p, q) is (2006 - 3M, –1)a)252b)254c)225d)224Correct answer is option 'C'. Can you explain this answer?.
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