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For any integer n greater than 1, n! denotes the product of all intege
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11 Jul 2023, 00:12
11 Jul 2023, 00:12
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Question Stats:
35% (01:40) correct
64% (01:40) wrong based on 17 sessions
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For any integer n greater than 1, n! denotes the product of all integers from 1 to n, inclusive. How many prime numbers are there between 15! + 5 and 15! + 13?
A. Zero
B. One
C. Two
D. Three
E. Four
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
A. Zero
B. One
C. Two
D. Three
E. Four
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
Re: For any integer n greater than 1, n! denotes the product of all intege
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07 Aug 2023, 22:29
07 Aug 2023, 22:29
Expert Reply
OE
A prime number is a positive integer that has two distinct positive factors. Hence a
prime is divisible by only two positive integers 1 and itself.
If a number is divisible by anything else apart from 1 and itself then it will not be a
prime number.
Thus, considering the values between 15! + 5 and 15! + 13, we can see that;
(15! + 5)= (1 × 2 × 3 × 4 × 5 ×… × 15) + 5
Take 5 common,
= 5((1 × 2 × 3 × 4 × 6 × … × 15) + 1)
So, we can see it is divisible by 5.
Similarly, (15! + 6) will give 6 as common and so on.
Hence, none of the number will be prime.
Thus, among the values we don’t have any prime number.
Ans. (A)
A prime number is a positive integer that has two distinct positive factors. Hence a
prime is divisible by only two positive integers 1 and itself.
If a number is divisible by anything else apart from 1 and itself then it will not be a
prime number.
Thus, considering the values between 15! + 5 and 15! + 13, we can see that;
(15! + 5)= (1 × 2 × 3 × 4 × 5 ×… × 15) + 5
Take 5 common,
= 5((1 × 2 × 3 × 4 × 6 × … × 15) + 1)
So, we can see it is divisible by 5.
Similarly, (15! + 6) will give 6 as common and so on.
Hence, none of the number will be prime.
Thus, among the values we don’t have any prime number.
Ans. (A)
gmatclubot
Re: For any integer n greater than 1, n! denotes the product of all intege [#permalink]
07 Aug 2023, 22:29
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