BTGmoderatorDC wrote: ↑

Mon Jun 21, 2021 3:44 pm

Suppose x is the product of all the primes less than or equal to 59. How many primes appear in the set {x + 2, x + 3, x + 4, …, x + 59}?

A. 0
B. 17
C. 18
D. 23
E. 24



OA A

Source: Veritas Prep

Since \(2 \cdot 3 \cdot 5 \cdot \ldots \cdot 59\) is even, \(x\) is even.

As \(x\) is even, so will be \(x+2, x+4, x+6, \ldots, x+58,\) so these numbers cannot be prime.

Now, \(x+3\) is nothing but \((2\cdot 3\cdot \ldots \cdot 59) + 3.\) Rewriting this as \(\{3 \cdot (2 \cdot 5\cdot \ldots \cdot 59)\} +3.\)

One can observe here that the term in the curly brace is a multiple of \(3\) and that term plus \(3\) will definitely be a multiple of \(3.\)

(Multiple of \(3 + 3\) is a multiple of \(3,\) Multiple of \(59 + 59\) is a multiple of \(59\))

As \(x+3\) is a multiple of \(3\) it cannot be a prime.

Similarly, \(x+5\) can be written as \(\{5 \cdot (2 \cdot 3 \cdot 7 \cdot \ldots \cdot 59)\} +5\) which is a multiple of \(5.\)

Like wise, we can see that all the terms in the set are multiples of prime numbers.

Therefore, the correct answer is A