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For any integer m greater than 1, $m denotes the product of all the in
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10 Sep 2021, 09:44
10 Sep 2021, 09:44
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For any integer m greater than 1, $m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between $7 + 2 and $7 + 10, inclusive?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: For any integer m greater than 1, $m denotes the product of all the in
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10 Sep 2021, 10:31
10 Sep 2021, 10:31
Question can be restated as "How many prime numbers are there between 7! + 2 and 7! + 10, inclusive?"
All factors are present in 7! (including those above number 7, i.e. 2*4, 3*3, 2*5). Hence, answer is none and A
All factors are present in 7! (including those above number 7, i.e. 2*4, 3*3, 2*5). Hence, answer is none and A
Carcass wrote:
For any integer m greater than 1, $m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between $7 + 2 and $7 + 10, inclusive?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Moderator
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Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
For any integer m greater than 1, $m denotes the product of all the in
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02 Sep 2022, 20:49
02 Sep 2022, 20:49
Given that $m denotes the product of all the integers from 1 to m, inclusive and we need to find how many prime numbers are there between $7 + 2 and $7 + 10, inclusive
$7 = Product of all the integers from 1 to 7, inclusive = 1*2*3*4*5*6*7 = 7!
$7 + 2 = 7! + 2 = 1*2*3*4*5*6*7 + 2 = 2*(1*3*4*5*6*7 + 1) = a Multiple of 2 => NOT a Prime Number
Similarly, 7! is also a multiple of all numbers from 3 to 10
=> all of the numbers from 7! + 2 to 7! + 10 will be Non-Prime numbers.
So, Answer will be A.
Hope it helps!
(Working below)
$7 + 3 = 7! + 3 = 1*2*3*4*5*6*7 + 3 = a Multiple of 3 => NOT a Prime Number
$7 + 4 = 7! + 4 = 1*2*3*4*5*6*7 + 4 = a Multiple of 4 => NOT a Prime Number
$7 + 5 = 7! + 5 = 1*2*3*4*5*6*7 + 5 = a Multiple of 5 => NOT a Prime Number
$7 + 6 = 7! + 6 = 1*2*3*4*5*6*7 + 6 = a Multiple of 6 => NOT a Prime Number
$7 + 7 = 7! + 7 = 1*2*3*4*5*6*7 + 7 = a Multiple of 7 => NOT a Prime Number
$7 + 8 = 7! + 8 = 1*2*3*4*5*6*7 + 8 = a Multiple of 8 => NOT a Prime Number
$7 + 9 = 7! + 9 = 1*2*3*4*5*6*7 + 9 = a Multiple of 9 => NOT a Prime Number
$7 + 10 = 7! + 10 = 1*2*3*4*5*6*7 + 10 = a Multiple of 10 => NOT a Prime Number
Watch the following video to learn the Basics of Functions and Custom Characters
$7 = Product of all the integers from 1 to 7, inclusive = 1*2*3*4*5*6*7 = 7!
$7 + 2 = 7! + 2 = 1*2*3*4*5*6*7 + 2 = 2*(1*3*4*5*6*7 + 1) = a Multiple of 2 => NOT a Prime Number
Similarly, 7! is also a multiple of all numbers from 3 to 10
=> all of the numbers from 7! + 2 to 7! + 10 will be Non-Prime numbers.
So, Answer will be A.
Hope it helps!
(Working below)
$7 + 3 = 7! + 3 = 1*2*3*4*5*6*7 + 3 = a Multiple of 3 => NOT a Prime Number
$7 + 4 = 7! + 4 = 1*2*3*4*5*6*7 + 4 = a Multiple of 4 => NOT a Prime Number
$7 + 5 = 7! + 5 = 1*2*3*4*5*6*7 + 5 = a Multiple of 5 => NOT a Prime Number
$7 + 6 = 7! + 6 = 1*2*3*4*5*6*7 + 6 = a Multiple of 6 => NOT a Prime Number
$7 + 7 = 7! + 7 = 1*2*3*4*5*6*7 + 7 = a Multiple of 7 => NOT a Prime Number
$7 + 8 = 7! + 8 = 1*2*3*4*5*6*7 + 8 = a Multiple of 8 => NOT a Prime Number
$7 + 9 = 7! + 9 = 1*2*3*4*5*6*7 + 9 = a Multiple of 9 => NOT a Prime Number
$7 + 10 = 7! + 10 = 1*2*3*4*5*6*7 + 10 = a Multiple of 10 => NOT a Prime Number
Watch the following video to learn the Basics of Functions and Custom Characters
