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 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
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 2512:30 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 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.
06 Mar 2007, 05:02
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
100%
(03:02)
wrong
based on 2
sessions
History
Date
Time
Result
Not Attempted Yet
How many sets of two or more consecutive integers are there such that no element is negative and the sum of the elements is 210?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
06 Mar 2007, 06:09
Kudos
Bookmarks
(A), which is 4
I do not have any formula but 210 has to be divisible by odd numbers.
For ex.
210/3 = 70, so 69+70+71 = 210.
similarly, 210/5 = 52, so 50+51+52+53+54 = 210
But if it is divisible by even number such as 2, then summation of 2 consecutive numbers can not be 210.
I do not have any formula but 210 has to be divisible by odd numbers.
For ex.
210/3 = 70, so 69+70+71 = 210.
similarly, 210/5 = 52, so 50+51+52+53+54 = 210
But if it is divisible by even number such as 2, then summation of 2 consecutive numbers can not be 210.
06 Mar 2007, 07:40
Kudos
Bookmarks
answer is 8 (D).
for sequences with odd count, rdg's method is very good:
the middle number is always the average, so you need to find a all odd divisors of 210 - these are 3,5,7,15,21 (and 105... see below)
and you get:
3: 210/3 = 70 => 69+70+71=210
5: 210/5 = 42 => 40+41+42+43+44=210
7: 210/7 = 30 => 27+28+29+30+31+32+33=210
15: 210/15 = 14 => 7+8+...+14+..+21=210
21: 210/21 = 10 => 0+1+2+...+10+...+20=210
for 105 this doesn't work, as the resulting sequence will involve negative numbers
now we try to find sequences with even count: such sequences always have a k+0.5 average where k is an integer. so if 2n is te number of terms in the sequence, then 2n(k+0.5)=210 => n(2k+1)=210 since (2k+1) is odd, then n must be even. so the length of the sequences can be 2n, where n is an even divisor of 210 these are 2,6,10,14,20...
n=2 (length is 4) => k= 52 => 51+52+53+54=210
n=6 (length is 12) => k= 17 => 12+13+14+15+16+17+..+23=210
n=10 (length is 20) => k=10 => 1+2+3+..10+11+...20=210
for longer sequences you eventually need negative numbers...
all in all, we found 8 sequences of non-negative consecutive integers whose sum is 210.
for sequences with odd count, rdg's method is very good:
the middle number is always the average, so you need to find a all odd divisors of 210 - these are 3,5,7,15,21 (and 105... see below)
and you get:
3: 210/3 = 70 => 69+70+71=210
5: 210/5 = 42 => 40+41+42+43+44=210
7: 210/7 = 30 => 27+28+29+30+31+32+33=210
15: 210/15 = 14 => 7+8+...+14+..+21=210
21: 210/21 = 10 => 0+1+2+...+10+...+20=210
for 105 this doesn't work, as the resulting sequence will involve negative numbers
now we try to find sequences with even count: such sequences always have a k+0.5 average where k is an integer. so if 2n is te number of terms in the sequence, then 2n(k+0.5)=210 => n(2k+1)=210 since (2k+1) is odd, then n must be even. so the length of the sequences can be 2n, where n is an even divisor of 210 these are 2,6,10,14,20...
n=2 (length is 4) => k= 52 => 51+52+53+54=210
n=6 (length is 12) => k= 17 => 12+13+14+15+16+17+..+23=210
n=10 (length is 20) => k=10 => 1+2+3+..10+11+...20=210
for longer sequences you eventually need negative numbers...
all in all, we found 8 sequences of non-negative consecutive integers whose sum is 210.
