x is the product of all even numbers from 2 to 50, inclusive

BTGmoderatorDC wrote:x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10
(B) Between 11 and 15
(C) Between 15 and 20
(D) Between 20 and 25
(E) Greater than 25

OA E

Source: Veritas Prep

We have x = 2.4.6.8.....48.50 = (2.1).(2.2).(2.3).........(2.24).(2.25 ) = 2^(25)*(1.2.3........25) = 2^(25)∗25!

You may note that 2^(25)∗25! is divisible by all the prime numbers less than 25.

Again, note that two consecutive integers are co-prime to each other, i.e., they don't share any common factor except 1. For example, 14 and 15 are consecutive integers; thus, the only common factor they share is 1. We have x and (x + 1) two consecutive integers; thus, they don't share any common factor except 1.

Since x has all prime numbers from 1 to 25 as its factors, (x + 1) must not have any prime factors from 1 to 25. Thus, the smallest prime factor of (x + 1) will be greater than 25.

The correct answer: E

Hope this helps!

-Jay
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BTGmoderatorDC wrote:x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10
(B) Between 11 and 15
(C) Between 15 and 20
(D) Between 20 and 25
(E) Greater than 25

OA E

Source: Veritas Prep

Here's a very similar (official) question to practice with: https://www.beatthegmat.com/arthemetic- ... 75643.html

Cheers,
Brent

BTGmoderatorDC wrote:x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10
(B) Between 11 and 15
(C) Between 15 and 20
(D) Between 20 and 25
(E) Greater than 25

OA E

Source: Veritas Prep

We must recall the rule that two consecutive integers do not share the same prime factors.

We see that x breaks into all the prime factors from 2, 3, 5, …, 23

Since x + 1 won’t have any of the prime factors from 2 to 23, the smallest prime factor of x + 1 is greater than 25.

Alternate Solution:

Notice that x = 2 * 4 * 6 * … * 50 = (2^25)*25!.

Now, x + 1 will have a remainder of 1 when divided by all the prime numbers less than 25 (because all the prime numbers less than 25 is a factor of 25!). Thus, the smallest prime factor of x + 1 must be greater than 25.

Answer: E