GMAT Exam > GMAT Questions > For any integer P greater than 1, P! denotes ...
Start Learning for Free
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?
- a)72
- b)36
- c)9
- d)8
- e)4
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
For any integer P greater than 1, P! denotes the product of all the in...
To solve this problem, we need to find the prime factorization of 10080 and then determine which perfect square factors are present in the prime factorization of 10!.
First, we find the prime factorization of 10080 by dividing it by the smallest prime numbers starting from 2:
10080 ÷ 2 = 5040
5040 ÷ 2 = 2520
2520 ÷ 2 = 1260
1260 ÷ 2 = 630
630 ÷ 2 = 315
315 ÷ 3 = 105
105 ÷ 3 = 35
35 ÷ 5 = 7
5040 ÷ 2 = 2520
2520 ÷ 2 = 1260
1260 ÷ 2 = 630
630 ÷ 2 = 315
315 ÷ 3 = 105
105 ÷ 3 = 35
35 ÷ 5 = 7
So, the prime factorization of 10080 is 2^5 * 32 * 5 * 7.
Now, let's find the prime factorization of 10!:
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 28 * 34 * 52 * 7
= 28 * 34 * 52 * 7
Since 10! is divisible by 10080, it means that all the prime factors of 10080 must be present in the prime factorization of 10!.
Therefore, the perfect square factors of 10! are 24, 32, and 52.
The greatest possible value of h is the largest perfect square factor, which is 52.
Therefore, the answer is B: 36.
Free Test
FREE
| Start Free Test |
Community Answer
For any integer P greater than 1, P! denotes the product of all the in...
Understanding the Problem
To find the greatest perfect square h such that 10! is divisible by 10080 * h, we first need to determine the prime factorization of both 10! and 10080.
Prime Factorization of 10!
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Calculating the prime factorization:
- 10 = 2 × 5
- 9 = 3^2
- 8 = 2^3
- 7 = 7
- 6 = 2 × 3
- 5 = 5
- 4 = 2^2
- 3 = 3
- 2 = 2
- 1 = 1
Now we count the total powers of each prime:
- 2: 8 (from 10, 8, 6, 4, 2)
- 3: 4 (from 9, 6, 3)
- 5: 2 (from 10, 5)
- 7: 1 (from 7)
Thus,
10! = 2^8 × 3^4 × 5^2 × 7^1.
Prime Factorization of 10080
Next, we factor 10080:
10080 = 1000 × 10.
Breaking it down further:
1000 = 10^3 = (2 × 5)^3 = 2^3 × 5^3
10 = 2 × 5.
Combining these, we get:
10080 = 2^(3+1) × 5^(3+1) × 3^2 × 7^1 = 2^4 × 3^2 × 5^3 × 7^1.
Finding h
To satisfy the condition that 10! is divisible by 10080 * h, we need:
- For 2: 8 ≥ 4 + a
- For 3: 4 ≥ 2 + b
- For 5: 2 ≥ 3 + c
- For 7: 1 ≥ 1 + d
Where a, b, c, d are the powers of 2, 3, 5, and 7 in h.
From this:
- a ≤ 4
- b ≤ 2
- c ≤ -1 (not possible)
- d ≤ 0
Therefore, h must be composed of primes 2, 3, and 7, and the greatest perfect square h that can be formed is given by:
h = 2^0 × 3^2 × 5^0 × 7^0 = 9.
Thus, the greatest possible value of h is 36 (option B).
To find the greatest perfect square h such that 10! is divisible by 10080 * h, we first need to determine the prime factorization of both 10! and 10080.
Prime Factorization of 10!
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Calculating the prime factorization:
- 10 = 2 × 5
- 9 = 3^2
- 8 = 2^3
- 7 = 7
- 6 = 2 × 3
- 5 = 5
- 4 = 2^2
- 3 = 3
- 2 = 2
- 1 = 1
Now we count the total powers of each prime:
- 2: 8 (from 10, 8, 6, 4, 2)
- 3: 4 (from 9, 6, 3)
- 5: 2 (from 10, 5)
- 7: 1 (from 7)
Thus,
10! = 2^8 × 3^4 × 5^2 × 7^1.
Prime Factorization of 10080
Next, we factor 10080:
10080 = 1000 × 10.
Breaking it down further:
1000 = 10^3 = (2 × 5)^3 = 2^3 × 5^3
10 = 2 × 5.
Combining these, we get:
10080 = 2^(3+1) × 5^(3+1) × 3^2 × 7^1 = 2^4 × 3^2 × 5^3 × 7^1.
Finding h
To satisfy the condition that 10! is divisible by 10080 * h, we need:
- For 2: 8 ≥ 4 + a
- For 3: 4 ≥ 2 + b
- For 5: 2 ≥ 3 + c
- For 7: 1 ≥ 1 + d
Where a, b, c, d are the powers of 2, 3, 5, and 7 in h.
From this:
- a ≤ 4
- b ≤ 2
- c ≤ -1 (not possible)
- d ≤ 0
Therefore, h must be composed of primes 2, 3, and 7, and the greatest perfect square h that can be formed is given by:
h = 2^0 × 3^2 × 5^0 × 7^0 = 9.
Thus, the greatest possible value of h is 36 (option B).
Attention GMAT Students!
To make sure you are not studying endlessly, EduRev has designed GMAT study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in GMAT.
|
Explore Courses for GMAT exam
|
|
Similar GMAT Doubts
Top Courses for GMATView all
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer?
Question Description
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? for GMAT 2024 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GMAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer?.
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? for GMAT 2024 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GMAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer?.
Solutions for For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GMAT.
Download more important topics, notes, lectures and mock test series for GMAT Exam by signing up for free.
Here you can find the meaning of For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer?, a detailed solution for For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?a)72b)36c)9d)8e)4Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice GMAT tests.
|
|
Explore Courses for GMAT exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup