Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2212:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
17 Dec 2025, 13:12
Kudos
Bookmarks
1<P<the number
An integer divided by 18 gives a remainder from 0-17 but not all remainders are possible for primes.
Prime are not divisible by 2 or 3 so they are; 1,5,7,11/13,17
so the total programs would be 6
The answer is D
An integer divided by 18 gives a remainder from 0-17 but not all remainders are possible for primes.
Prime are not divisible by 2 or 3 so they are; 1,5,7,11/13,17
so the total programs would be 6
The answer is D
17 Dec 2025, 13:57
Kudos
Bookmarks
Generated number = prime number (greater than 10) since it can't have a factor small than it and it's an integer.
On dividing the prime numbers greater than 10 by 18, we get following remainders
11/18 = rem 11
13/18 = rem 13
17/18 = rem 17
19/18 = rem 1
23/18 = rem 5
29/18 = rem 11
31/18 = rem 13
37/18 = rem 1
41/18 = rem 5
43/18 = rem 7
Looking at this, we realise, remainders are all prime number under 18 (and number 1), other than 2 & 3 since they are factors of 18.
= 1, 5, 7, 11, 13, 17
Answer = 6
On dividing the prime numbers greater than 10 by 18, we get following remainders
11/18 = rem 11
13/18 = rem 13
17/18 = rem 17
19/18 = rem 1
23/18 = rem 5
29/18 = rem 11
31/18 = rem 13
37/18 = rem 1
41/18 = rem 5
43/18 = rem 7
Looking at this, we realise, remainders are all prime number under 18 (and number 1), other than 2 & 3 since they are factors of 18.
= 1, 5, 7, 11, 13, 17
Answer = 6
Bunuel
17 Dec 2025, 17:57
Kudos
Bookmarks
Clearly, the question is asking if a prime number is divided by 18, how many distinct remainders are possible?
The number will be: \(18 * quotient + remainder\)
\(= 18 * q + r\)
if divisor = 18, possible remainders are {\(0, 1, 2, 3 .... 16, 17\)} i.e. 0 to divisor -1
for the number to be prime, r can't have any factors of 18, such as 3, 9, 2, and any even multiples. Otherwise, the number will be divisible
so first r has to be a set of odd numbers.
i.e.
{\(1, 3, 5, 7... 17\)}
among them, if we remove, multiples of 3, the leftovers are {\(1, 5, 7, 11, 13, 17\)}
\(Count = 6\)
Ans: Option D
The number will be: \(18 * quotient + remainder\)
\(= 18 * q + r\)
if divisor = 18, possible remainders are {\(0, 1, 2, 3 .... 16, 17\)} i.e. 0 to divisor -1
for the number to be prime, r can't have any factors of 18, such as 3, 9, 2, and any even multiples. Otherwise, the number will be divisible
so first r has to be a set of odd numbers.
i.e.
{\(1, 3, 5, 7... 17\)}
among them, if we remove, multiples of 3, the leftovers are {\(1, 5, 7, 11, 13, 17\)}
\(Count = 6\)
Ans: Option D
