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For an integer n, the function f(n) is defined as the product of all
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30 Oct 2017, 22:14
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47% (01:34) correct 53% (01:36) wrong based on 134 sessions
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For an integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n)+1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II and III
Re: For an integer n, the function f(n) is defined as the product of all
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30 Oct 2017, 23:08
nkmungila wrote:
For an integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n)+1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II and III
f(n) = n!, where n > 10. So, f(n) is even, a multiple of 3 and a multiple of 10.
Therefore, f(n) + 1 is odd, 1 more than a multiple of 3 and 1 more than a multiple of 10, which means that f(n) + 1 is not divisible by 2, 3, or 10.
Re: For an integer n, the function f(n) is defined as the product of all
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30 Oct 2017, 23:10
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Bunuel wrote:
nkmungila wrote:
For an integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n)+1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II and III
f(n) = n!, where n > 10. So, f(n) is even, a multiple of 3 and a multiple of 10.
Therefore, f(n) + 1 is odd, 1 more than a multiple of 3 and 1 more than a multiple of 10, which means that f(n) + 1 is not divisible by 2, 3, or 10.
Re: For the integer n, the function f(n) is defined as the product of all
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12 Sep 2018, 05:01
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pushpitkc wrote:
For the integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n) + 1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II, and III
F(n)=1*2*3*...n where n>10 So f(n)=1*2*3*....*10*..*n Thus f(n) is product of all numbers till n and at least till 1 f(n) and f(n)+1 are co-prime, as they are consecutive terms.
Therefore f(n)+1 will not have any common factor that f(n) has.. So all numbers till 10 will NOT be factors
Re: For the integer n, the function f(n) is defined as the product of all
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12 Sep 2018, 09:27
urvashis09 wrote:
f(n) can be expressed as n! hence n! + 1 will be a prime integer and none of the integers smaller than n can be a factor of the same. Hence, E.
Although the second part that none of the integers smaller than n can be a factor of n!+1 is correct..
But that n!+1 will be prime integer may not be correct every time.. Example 3!+1=6+1=7 yes.. 4!+1=24+1=25....Not a prime 5!+1=120+1=121=11^2... Again not aprime integer..
So the point is that n!+1 will not be a multiple of any prime number <n
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Re: For the integer n, the function f(n) is defined as the product of all
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14 Sep 2018, 17:40
pushpitkc wrote:
For the integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n) + 1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II, and III
Notice that f(n) = n!. Since two consecutive integers cannot share any of the same prime factors, f(n) and f(n) + 1 cannot share any of the same prime factors. Since f(n) is greater than 10!, and 10! has prime factors of 2, 3, 5, and 7, we see that 2, 3, and 2 x 5 = 10 cannot be primes of f(n) + 1.
Re: For an integer n, the function f(n) is defined as the product of all
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06 Feb 2019, 19:38
nkmungila wrote:
For an integer n, the function f(n) is defined as the product of all integers from 1 to n, where n is greater than 10. Which of the following is NOT a factor of f(n)+1?
I. 2 II. 3 III. 10
A. None B. II only C. I and II only D. I and III only E. I, II and III
We must remember that f(n) and f(n) + 1 WON’T SHARE ANY OF THE SAME PRIME FACTORS because they are consecutive integers.
So, since n is greater than 10, we see that f(n) will have prime factors of at least 2, 3, 5, and 7. Thus, 2, 3, and 10 won’t be factors of f(n) + 1.
Re: For an integer n, the function f(n) is defined as the product of all
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08 Sep 2019, 22:39
US09 wrote:
f(n) can be expressed as n! hence n! + 1 will be a prime integer and none of the integers smaller than n can be a factor of the same. Hence, E.
Is this a applicable rule for other problems or just "made up" for this problem? Because if I try it wih 4! => 4*3*2*1 = 24 + 1 will result in 25 which is not a prime integer.
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47% (01:34) correct 
