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x# is defined for every positive even integer x as the product of all
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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
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Updated on: 06 Sep 2015, 17:41
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
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06 Sep 2015, 17:23
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
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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
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14 Sep 2015, 23:34
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
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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
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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
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19 Dec 2016, 02:57
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
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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
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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
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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
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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
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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
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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
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16 Jul 2018, 04:00
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
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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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