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Retired Moderator Joined: 29 Apr 2015
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x# is defined for every positive even integer x as the product of all  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 55% (01:37) correct 45% (01:25) wrong based on 355 sessions

### HideShow timer Statistics 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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Intern  Joined: 16 Aug 2015
Posts: 18
x# is defined for every positive even integer x as the product of all  [#permalink]

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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.
Manager  B
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GMAT 1: 710 Q50 V34 Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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1
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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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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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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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.

Intern  S
Joined: 01 Sep 2016
Posts: 6
Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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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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Joined: 13 Oct 2016
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Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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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 $$3$$

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Joined: 02 Sep 2016
Posts: 657
Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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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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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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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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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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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!! Intern  B
Joined: 13 Feb 2018
Posts: 13
Re: x# is defined for every positive even integer x as the product of all  [#permalink]

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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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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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Karishma
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Intern  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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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! 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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