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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: M01 Question 18 [#permalink]
01 Jun 2009, 16:45
4
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
Is something not clear in the Official Explanation?
Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\) (Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128. Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1: \(\frac{994-105}{7}+1=128\) _________________
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
10 Jan 2010, 16:25
I'm kept my lame questions to a minimum. I see why plus 1 work for inclusive. I don't see for exclusive why minus 1 would work. If both of the end number "excluded" are actually multiples of the number in question it changes it. but that should be the case for inclusive as well. in other words, how many multiples of 7 between 8 and 22 exclusive means 9 to 21 which means the answer is 2. 21-9 is 12 which divided by 7 is 1 so plus 1 = 2. however 7 to 28 exclusive means 8-27 = 19 which is 2 7s which is actually correct.
It makes me very nervous to apply formulas without being crystal clear as to why they're right. Maybe I've misunderstood part of the premise of the question.
a 7*15=105 b 106/7=15.1 c 127/7=18.1 d 128/7=18.12 aprox e 142/7=20.--- aprox so a is right & a straight ans as its div by 7*15
Hi, and welcome to Gmat Club.
It seems that you misinterpreted the question.
The question is "How many of the three-digit numbers are divisible by 7?" It's not necessary the answer itself to be divisible by 7
GENERALLY: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;
\(\frac{30-(-5)}{5}+1=8\).
OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;
\(\frac{-7-(-21)}{7}+1=3\).
Back to the original question:
Last 3-digit multiple of 7 is 994; First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is \(\frac{994-105}{7}+1=128\).
The range of 3 digit numbers that are divisible by 7 are 105 to 994 inclusive. If we consider the set of these numbers as in Arithmetic Progression, {105, 112, 119 ... 994}, we can find the number of values in the set using the formula; Last term = First term + (n-1)(Common Difference), which implies, 994 = 105 + (n-1)(7). Solving n = 128 _________________
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
28 Dec 2010, 22:23
Can someone solve the same problem with some addtiional details How many of the 3 digit numbers are divisible by 7? Nos divisble by 7 = 128 Nos divisible by 5 = 180 Can we find nos divisible by 35 on the basis of above info ( not the traditional formula used for this problem , but using set theory ) _________________
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
28 Dec 2010, 23:59
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.
Also to figure out those 3 digit numbers divisible by 35 I would just use the same formula.
Thanks for your reply, i understand that i can find the nos with the same formula
But i wish to understand whethr the same is possible with set theory.. because the q could be asked as a DS question in some different format also _________________
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
29 Dec 2010, 00:40
Expert's post
How many of the three-digit numbers are divisible by 7? (A) 105 (B) 106 (C) 127 (D) 128 (E) 142
GENERALLY: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;
\(\frac{30-(-5)}{5}+1=8\).
OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;
\(\frac{-7-(-21)}{7}+1=3\).
Back to the original question:
Last 3-digit multiple of 7 is 994; First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is \(\frac{994-105}{7}+1=128\).
Answer: D.
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.
The above approach will work for both inclusive and exclusive cases, which is also.
hirendhanak wrote:
gettinit wrote:
this strategy should work for most of these types of problems. I have never seen an exclusive scenario.
Also to figure out those 3 digit numbers divisible by 35 I would just use the same formula.
Thanks for your reply, i understand that i can find the nos with the same formula
But i wish to understand whethr the same is possible with set theory.. because the q could be asked as a DS question in some different format also
What do you mean by the red part? _________________
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
02 Jan 2011, 00:13
Bunuel wrote:
How many of the three-digit numbers are divisible by 7? (A) 105 (B) 106 (C) 127 (D) 128 (E) 142
GENERALLY: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
For example: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;
\(\frac{30-(-5)}{5}+1=8\).
OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;
\(\frac{-7-(-21)}{7}+1=3\).
Back to the original question:
Last 3-digit multiple of 7 is 994; First 3-digit multiple of 7 is 105;
So # of 3-digt multiples of 7 is \(\frac{994-105}{7}+1=128\).
Answer: D.
Hi Bunuel,
I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?
For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above. _________________