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x# is defined for every positive even integer x as the product of all

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x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 06 Sep 2015, 10:28
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x# is defined for every positive even integer x as the product of all even integers from 2 to x. What is the smallest possible prime factor of (x#+7)?

A. 2
B. 3
C. 5
D. 7
E. 11

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x# is defined for every positive even integer x as the product of all  [#permalink]

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New post Updated on: 06 Sep 2015, 17:41
1
We can resolve this problem by picking some smart numbers.

First of all, x# is an even number, accordingly x#+7 is odd, so the answer cannot be 2 even if it's tempting at the first sight !
Now we can proceed by taking some numbers, if we consider


X=4, then x#+7= 15 = 3 x 5. we got it, 3 is the smallest prime factor that we can obtain.

Originally posted by issamL on 06 Sep 2015, 14:28.
Last edited by issamL on 06 Sep 2015, 17:41, edited 1 time in total.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 06 Sep 2015, 17:23
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IMO: B

Since smallest possible prime factor is required.
Let x = 2 then --> 2# = 2
x#+7 ==> 2#+7 = 2+7 = 9 = 3*3.
Thus 3 is the smallest possible Prime Factor

2 cannot be the possible prime factor because --> x# is always even since it is product of even numbers
Even +7 = Odd. Hence 2 cannot be prime factor of this result.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 14 Sep 2015, 21:32
issamL wrote:
We can resolve this problem by picking some smart numbers.

First of all, x# is an even number, accordingly x#+7 is odd, so the answer cannot be 2 even if it's tempting at the first sight !
Now we can proceed by taking some numbers, if we consider


X=4, then x#+7= 15 = 3 x 5. we got it, 3 is the smallest prime factor that we can obtain.


If x=4, should x#+7 be 31?
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 14 Sep 2015, 23:34
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Aves wrote:
issamL wrote:
We can resolve this problem by picking some smart numbers.

First of all, x# is an even number, accordingly x#+7 is odd, so the answer cannot be 2 even if it's tempting at the first sight !
Now we can proceed by taking some numbers, if we consider


X=4, then x#+7= 15 = 3 x 5. we got it, 3 is the smallest prime factor that we can obtain.


If x=4, should x#+7 be 31?


No. # is a user defined operator i.e. the question defines for what # stand.

x# = 2*4*..*x

So if x = 2, x# = 2
If x = 4, x# = 2*4 = 8
If x = 6, x# = 2*4*6 = 48
and so on...

So x# + 7 when x = 4 will be 8 + 7 = 15 because when x is 4, x# is 8.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 14 Sep 2015, 23:45
VeritasPrepKarishma wrote:
Aves wrote:
issamL wrote:
We can resolve this problem by picking some smart numbers.

First of all, x# is an even number, accordingly x#+7 is odd, so the answer cannot be 2 even if it's tempting at the first sight !
Now we can proceed by taking some numbers, if we consider


X=4, then x#+7= 15 = 3 x 5. we got it, 3 is the smallest prime factor that we can obtain.


If x=4, should x#+7 be 31?


No. # is a user defined operator i.e. the question defines for what # stand.

x# = 2*4*..*x

So if x = 2, x# = 2
If x = 4, x# = 2*4 = 8
If x = 6, x# = 2*4*6 = 48
and so on...

So x# + 7 when x = 4 will be 8 + 7 = 15 because when x is 4, x# is 8.


Oh, it was only the product of all even integers not all integers.

I made a silly mistake.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 18 Dec 2016, 20:50
When x is taken as 6,X#+7 will be 55.
In this case ,the smallest prime factor will be 5 right?
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 19 Dec 2016, 02:57
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reto wrote:
x# is defined for every positive even integer x as the product of all even integers from 2 to x. What is the smallest possible prime factor of (x#+7)?

A. 2
B. 3
C. 5
D. 7
E. 11


We have

\(2*4*6*8*10* … * x = 2^{\frac{x}{2}}*1*2*3*4*5* … * \frac{x}{2} = 2^{\frac{x}{2}}*(\frac{x}{2})!\)

If we need this expression to be an integer, min value that \(x\) can take is \(2\).

\(x# + 7 = 2^1*1! + 7 = 9 = 3^2\)

Min prime factor is \(3\)

Answer B
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 03 Apr 2017, 23:58
x# is the product of all even nos. from 2 to x. Thus the answer would be an even no.

x#+7= Even +Odd= Odd

Therefore the smallest possible prime factor would be 3 and not 2 because the final answer is ODD and not even.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 10 Apr 2017, 05:43
Yes it will have to be odd and thus 3 is the small prime factor
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 13 Apr 2017, 09:35
VeritasPrepKarishma wrote:
Aves wrote:
issamL wrote:
We can resolve this problem by picking some smart numbers.

First of all, x# is an even number, accordingly x#+7 is odd, so the answer cannot be 2 even if it's tempting at the first sight !
Now we can proceed by taking some numbers, if we consider


X=4, then x#+7= 15 = 3 x 5. we got it, 3 is the smallest prime factor that we can obtain.


If x=4, should x#+7 be 31?


No. # is a user defined operator i.e. the question defines for what # stand.

x# = 2*4*..*x

So if x = 2, x# = 2
If x = 4, x# = 2*4 = 8
If x = 6, x# = 2*4*6 = 48
and so on...

So x# + 7 when x = 4 will be 8 + 7 = 15 because when x is 4, x# is 8.


But karishma (48+7= 55) which is not divisible by 3.
So it is not true in all cases.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 13 Apr 2017, 22:59
Shiv2016 wrote:

But karishma (48+7= 55) which is not divisible by 3.
So it is not true in all cases.


We need the smallest "possible" prime factor which CAN be a factor of a number of the form x# + 7.

There are some prime factors which CANNOT be factors of a number of this form. One such prime factor is 2 because numbers of the form x# + 7 will always be odd.

The next smallest prime factor is 3. Since there is a number of the form x# + 7 for which 3 is a factor (it needn't be a factor of EVERY such number), 3 is the smallest such prime factor.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 13 Apr 2017, 23:36
VeritasPrepKarishma wrote:
Shiv2016 wrote:

But karishma (48+7= 55) which is not divisible by 3.
So it is not true in all cases.


We need the smallest "possible" prime factor which CAN be a factor of a number of the form x# + 7.

There are some prime factors which CANNOT be factors of a number of this form. One such prime factor is 2 because numbers of the form x# + 7 will always be odd.

The next smallest prime factor is 3. Since there is a number of the form x# + 7 for which 3 is a factor (it needn't be a factor of EVERY such number), 3 is the smallest such prime factor.



Okay. Got it. Thanks!! :)
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 15 Jul 2018, 04:46
Any conceptual approach to solving this problem?
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 16 Jul 2018, 04:00
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agarwalaayush2007 wrote:
Any conceptual approach to solving this problem?



x# is defined for every positive even integer x as the product of all even integers from 2 to x.
implies
2# = 2
4# = 2*4 = 8
6# = 2*4*6 = 48
and so on...

What is the smallest possible prime factor of (x#+7)?
2# + 7 will be 9
4# + 7 will be 15
6# + 7 will be 55
and so on...

Note that x# will always be even so if you add 7 to it, it will become odd. So 2 will never be a factor of x# + 7.
Next we see that 3 CAN be a factor of x# + 7 (9 and 15 have 3 as a factor) so the smallest such prime number must be 3.
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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New post 17 Jul 2018, 07:08
KarishmaB wrote:
agarwalaayush2007 wrote:
Any conceptual approach to solving this problem?



x# is defined for every positive even integer x as the product of all even integers from 2 to x.
implies
2# = 2
4# = 2*4 = 8
6# = 2*4*6 = 48
and so on...

What is the smallest possible prime factor of (x#+7)?
2# + 7 will be 9
4# + 7 will be 15
6# + 7 will be 55
and so on...

Note that x# will always be even so if you add 7 to it, it will become odd. So 2 will never be a factor of x# + 7.
Next we see that 3 CAN be a factor of x# + 7 (9 and 15 have 3 as a factor) so the smallest such prime number must be 3.


Thank you Karishma. Great help!
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Re: x# is defined for every positive even integer x as the product of all   [#permalink] 17 Jul 2018, 07:08
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