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

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Joined: 26 Feb 2016
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For the integer n, the function f(n) is defined as the product of all  [#permalink]

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12 Sep 2018, 02:11
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35% (medium)

Question Stats:

48% (01:13) correct 52% (01:23) wrong based on 46 sessions

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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

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Re: For the integer n, the function f(n) is defined as the product of all  [#permalink]

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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

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]

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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]

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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  [#permalink]

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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.

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