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For an integer n, the function f(n) is defined as the product of all

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nkmungila
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For an integer n, the function f(n) is defined as the product of all [#permalink] New post   30 Oct 2017, 22:14
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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
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E
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chetan2u
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Re: For the integer n, the function f(n) is defined as the product of all [#permalink] New post   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

Source: Experts Global


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

E
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Re: For the integer n, the function f(n) is defined as the product of all [#permalink] New post   12 Sep 2018, 09:27
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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 an integer n, the function f(n) is defined as the product of all [#permalink] New post   30 Oct 2017, 23:08
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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.

Answer: E.
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Re: For an integer n, the function f(n) is defined as the product of all [#permalink] New post   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.

Answer: E.


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Re: For the integer n, the function f(n) is defined as the product of all [#permalink] New post   12 Sep 2018, 09:16
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.
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Re: For the integer n, the function f(n) is defined as the product of all [#permalink] New post   14 Sep 2018, 17:40
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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


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.

Answer: E
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Re: For an integer n, the function f(n) is defined as the product of all [#permalink] New post   06 Feb 2019, 19:38
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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.

Answer: E
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Re: For an integer n, the function f(n) is defined as the product of all [#permalink] New post   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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Re: For an integer n, the function f(n) is defined as the product of all [#permalink] New post   16 Jul 2022, 12:57
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