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tejal777
Integer x is equal to the product of all even numbers from 2 to 60. If y is the smallest prime no. that is also a factor of x-1, then which of the following expressions must be true?

y lies bet. 0 and 4
y lies bet. 4 and 10
y lies bet. 10 and 20
y lies bet. 20 and 30
y is greater than 30

Guys please help me understand the underlying principles..

This is a replication of one of the most discussed and hardest question (I guess the question is from retired gmat question bank).

Per the question, x = the product of all even numbers from 2 to n, where n is any even integer value from 2 to 60 including.

If x = 2, x - 1 = 2! -1
If x = 2x4, x - 1 = 2! (2^2) - 1
If x = 2x4x6, x - 1 = 3! (2^3) - 1
If x = 2x4x6x8, x - 1 = 4! (2^4) - 1
.
.
.
.
If x = 2x4x6x.......x60, x - 1 = 30! (2^30) - 1



Lets find y, a smallest prime factor of (x-1):

If y were 2, [(30!) (2^30) -(1)] would be divisible by 2 however thats not the case.
If y were 3, [(30!) (2^30) -(1)] would be divisible by 3 however thats not the case.
If y were 5, [(30!) (2^30) -(1)] would be divisible by 5 however thats not the case.
.
.
.
.
.
If y were 29, [(30!) (2^30) -(1)] would be divisible by 29 however thats not the case.

If none of the primes under 31 is a factor of (x-1), (x-1) must have a prime factore above 30.

Thats E.



Previous discussions are found here:

gmatprep-question-need-solution-74516.html#p556411
ps-function-74417.html#p556064



dear gmatiger,

i was able to understand the x part...but hw can u say the y=2 , 3 ,5 etc etc be not a factor of (x-1).I mean how can u be so sure that it wont be a factor..

please explain in brief.
thanks in advance ,
kyle
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dear gmatiger,

i was able to understand the x part...but hw can u say the y=2 , 3 ,5 etc etc be not a factor of (x-1).I mean how can u be so sure that it wont be a factor..

please explain in brief.
thanks in advance ,
kyle


Lets find y, a smallest prime factor of (x-1):

If y were 2, [30! (2^30) - 1]/2 must be divisible by 2 and result in an integer. Lets do little work on it:

= [30! (2^30) -1] / 2
= [30! (2^30)/2 - 1/2]
= [15 x 29! (2^30) - 1/2] ......... is it an integer? No. If it were, then 2 would be a smallest prime.

Similarly:

= [30! (2^30) - 1] / 3
= [30! (2^30) / 3 - 1/3]
= [10 x 29! (2^30) - 1/3] ......... is it an integer? No. If it were, then 3 would be a smallest prime.

Also

= [30! (2^30) - 1] / 29
= [30! (2^30)/29 - 1/29]
= [30 x 28! (2^30) - 1/29] ......... is it an integer? No. If it were, then 3 would be a smallest prime.


So what would be y, the smallest prime factor? It must be greater than 29 i.e. greater than 30 too.

Hope it is clear now.
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X can be written as = 2.4.6.8....60

or X = 2.(2.2)(2.3)(2.4)....(2.30)
i.e. X = 2^30(1.2.3.4.5...30)

so, X-1 = 2^30(1.2.3....30)-1

Now take a small example, prime factorization of a number 10 = 1*2*5, if i add or subtract 1 from the number then the prime factors involved (i.e. 2 and 5) will no longer divide 9 or 11.

Going by the same principle, if X-1 contains all prime numbers between 1 and 30, X-1 is bound to be divisible by a prime number that is not already present in X.

Greater than 30 is the only possible choice. Hence E.
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tejal777
Integer x is equal to the product of all even numbers from 2 to 60, inclusive. If y is the smallest prime number that is also a factor of x-1, then which of the following expressions must be true?

(A) 0<y<4
(B) 4<y<10
(C) 10<y<20
(D) 20<y<30
(E) y>30

so the sum is 29 and x = 870, x-1 = 869 = 79x11 => y =11 ? smallest prime? so answer is C? no ?

Bunuel ? what do you think?
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tejal777
Integer x is equal to the product of all even numbers from 2 to 60. If y is the smallest prime no. that is also a factor of x-1, then which of the following expressions must be true?

y lies bet. 0 and 4
y lies bet. 4 and 10
y lies bet. 10 and 20
y lies bet. 20 and 30
y is greater than 30

Guys please help me understand the underlying principles..

This is a replication of one of the most discussed and hardest question (I guess the question is from retired gmat question bank).

Per the question, x = the product of all even numbers from 2 to n, where n is any even integer value from 2 to 60 including.

If x = 2, x - 1 = 2! -1
If x = 2x4, x - 1 = 2! (2^2) - 1
If x = 2x4x6, x - 1 = 3! (2^3) - 1
If x = 2x4x6x8, x - 1 = 4! (2^4) - 1
.
.
.
.
If x = 2x4x6x.......x60, x - 1 = 30! (2^30) - 1



Lets find y, a smallest prime factor of (x-1):

If y were 2, [(30!) (2^30) -(1)] would be divisible by 2 however thats not the case.
If y were 3, [(30!) (2^30) -(1)] would be divisible by 3 however thats not the case.
If y were 5, [(30!) (2^30) -(1)] would be divisible by 5 however thats not the case.
.
.
.
.
.
If y were 29, [(30!) (2^30) -(1)] would be divisible by 29 however thats not the case.

If none of the primes under 31 is a factor of (x-1), (x-1) must have a prime factore above 30.

Thats E.



Previous discussions are found here:

gmatprep-question-need-solution-74516.html#p556411
ps-function-74417.html#p556064
If y were 29 then x-1 would be divisible by 29 however that is not the case is not proved
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Bunuel
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tejal777
Integer x is equal to the product of all even numbers from 2 to 60, inclusive. If y is the smallest prime number that is also a factor of x-1, then which of the following expressions must be true?

(A) 0<y<4
(B) 4<y<10
(C) 10<y<20
(D) 20<y<30
(E) y>30

so the sum is 29 and x = 870, x-1 = 869 = 79x11 => y =11 ? smallest prime? so answer is C? no ?

Bunuel ? what do you think?

x is equal to the product of all even numbers from 2 to 60, inclusive: \(x = 2*4*6*...*58*60=(2*1)*(2*2)*(2*3)*...*(2*29)*(2*30)=2^{30}*30!\).

y is the smallest prime number that is also a factor of \(x-1= 2^{30}*30!-1\).

Now, two numbers \(x-1= 2^{30}*30!-1\) and \(x=2^{30}*30!\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since \(x=2^{30}*30!\) has all prime numbers from 1 to 30 as its factors, then according to the above \(x-1= 2^{30}*30!-1\) won't have ANY prime factor from 1 to 30. Hence y, the smallest prime factor of \(x-1= 2^{30}*30!-1\) must be more than 30.

Answer: E.

Similar questions to practice:
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Hope it helps.
If x=21 and x-1=20
factors of 21=1, 3, 7, 21; and 20 = 1, 2, 4, 5, 10, 20
x-1 has a factor which is less than 21 which is a factor of x
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Bunuel
jimwild
Integer x is equal to the product of all even numbers from 2 to 60, inclusive. If y is the smallest prime number that is also a factor of x-1, then which of the following expressions must be true?

(A) 0<y<4
(B) 4<y<10
(C) 10<y<20
(D) 20<y<30
(E) y>30

x is equal to the product of all even numbers from 2 to 60, inclusive: \(x = 2*4*6*...*58*60=(2*1)*(2*2)*(2*3)*...*(2*29)*(2*30)=2^{30}*30!\).

y is the smallest prime number that is also a factor of \(x-1= 2^{30}*30!-1\).

Now, two numbers \(x-1= 2^{30}*30!-1\) and \(x=2^{30}*30!\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since \(x=2^{30}*30!\) has all prime numbers from 1 to 30 as its factors, then according to the above \(x-1= 2^{30}*30!-1\) won't have ANY prime factor from 1 to 30. Hence y, the smallest prime factor of \(x-1= 2^{30}*30!-1\) must be more than 30.

Answer: E.

Similar questions to practice:
for-every-positive-even-integer-n-the-function-h-n-is-126691.html
for-every-positive-even-integer-n-the-function-h-n-149722.html
x-is-the-product-of-all-even-numbers-from-2-to-50-inclusive-156545.html
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does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

Hope it helps.
If x=21 and x-1=20
factors of 21=1, 3, 7, 21; and 20 = 1, 2, 4, 5, 10, 20
x-1 has a factor which is less than 21 which is a factor of x

The question talks about prime factors, not just factors.
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Bunuel
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tejal777
Integer x is equal to the product of all even numbers from 2 to 60, inclusive. If y is the smallest prime number that is also a factor of x-1, then which of the following expressions must be true?

(A) 0<y<4
(B) 4<y<10
(C) 10<y<20
(D) 20<y<30
(E) y>30

so the sum is 29 and x = 870, x-1 = 869 = 79x11 => y =11 ? smallest prime? so answer is C? no ?

Bunuel ? what do you think?

x is equal to the product of all even numbers from 2 to 60, inclusive: \(x = 2*4*6*...*58*60=(2*1)*(2*2)*(2*3)*...*(2*29)*(2*30)=2^{30}*30!\).

y is the smallest prime number that is also a factor of \(x-1= 2^{30}*30!-1\).

Now, two numbers \(x-1= 2^{30}*30!-1\) and \(x=2^{30}*30!\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since \(x=2^{30}*30!\) has all prime numbers from 1 to 30 as its factors, then according to the above \(x-1= 2^{30}*30!-1\) won't have ANY prime factor from 1 to 30. Hence y, the smallest prime factor of \(x-1= 2^{30}*30!-1\) must be more than 30.

Answer: E.

Similar questions to practice:
for-every-positive-even-integer-n-the-function-h-n-is-126691.html
for-every-positive-even-integer-n-the-function-h-n-149722.html
x-is-the-product-of-all-even-numbers-from-2-to-50-inclusive-156545.html
if-n-is-a-positive-integer-greater-than-1-then-p-n-represe-144553.html
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Hope it helps.

Many thanks Bunuel !! :wink:
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Bunuel
Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1.

This is a beautiful and quick way to solve the problem. Thanks. =)
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Here is how I approached this question -

x-1 = 2*4*6*8*10......*60

Let's take the example of 29 -
x-1 = 29*integer - 1 [hence not divisible by 29]

Similarly, for all the prime numbers < 30 ====> No Solution is possible.

Answer = E
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Note : Taking y=31, x-1 will NOT have 31 or any of it's multiple 62,93, etc.. hence, it still has a chance of being a factor.
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x= 2x4x.....x60
= (2x1)x(2x2)x.....x(2x30)
= (2^30)x(1x2x...30)
= (2^30)x(30!)

Clearly, as 30! is a factor of x, so is every prime number less than or equal to 30.
Hence, 2, 3, 5,....,29 are all factors of x.

Consider 2. We know that among consecutive integers, every second integer has 2 as a factor (since every second integer is even).
This means that if x has 2 as a factor, x-1 CANNOT have 2 as a factor. (since they're consecutive, they cannot both be even)

Consider 3.We know that among consecutive integers, every third integer has 3 as a factor.
This means that if x has 3 as a factor, x-1 CANNOT have 3 as a factor.

Continuing this way, we can conclude that x-1 would not have any of the prime factors till 29.
Thus, the only prime factors it will have have to be greater than 29.
Since 30 is not prime, we can say that the only prime factors it will have have to be greater than 30.

Hence, the answer is E.
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