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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?
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Verified 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.
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.
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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).
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