|
Author |
Message |
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11506
Followers: 1791
Kudos [?]:
9526
[0], given: 826
|
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
 29 Dec 2010, 01:40
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?
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
|
Senior Manager
Status: ready to boMBArd
Joined: 31 Oct 2010
Posts: 493
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 710 Q48 V40
WE: Project Management (Manufacturing)
Followers: 10
Kudos [?]:
31
[0], given: 67
|
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
02 Jan 2011, 01:13
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.
_________________
My GMAT debrief: from-620-to-710-my-gmat-journey-114437.html
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11506
Followers: 1791
Kudos [?]:
9526
[0], given: 826
|
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
02 Jan 2011, 02:38
gmatpapa wrote: 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. You should read the solution carefully: it doesn't matter whether the the range is inclusive or exclusive, the formula gives you the correct answer in any case. How many multiples of 5 are there between 5 and 35, not inclusive? Last multiple of 5 IN the range is 30 (not 35 but 30 as 35 is not IN the range); First multiple of 5 IN the range is 10 (not 5 but 10 as 5 is not IN the range); \frac{30-10}{5}+1=5: 10, 15, 20, 25 and 30. How many multiples of 5 are there between 5 and 35, inclusive? Last multiple of 5 IN the range is 35; First multiple of 5 IN the range is 5; \frac{35-5}{5}+1=7: 5, 10, 15, 20, 25, 30 and 35.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3104
Location: Pune, India
Followers: 567
Kudos [?]:
1994
[0], given: 92
|
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
02 Jan 2011, 11:17
A quick demonstration of the formula and why it is intuitive: How many numbers are there from 3 to 7 (inclusive)? 3, 4, 5, 6, 7 i.e. 5 numbers When we do (7 - 3 = 4), we remove 3 from the list. I I I I I I I : 7 sticks [strike] I I I[/strike] I I I I : (7 - 3 = 4) sticks The third stick has been removed but we need it to be included too. So we add 1 to the result 4 + 1 = 5 sticks These 5 sticks are: I I I I I Now, how many numbers are there from 3 to 7 (exclusive)? 4, 5, 6 i.e. 3 numbers When we do (7 - 3 = 4), we still have 7 in the list. I I I I I I I : 7 sticks [strike] I I I[/strike] I I I I : (7 - 3 = 4) sticks The seventh stick is still there which has to be removed. So we subtract 1 from the result 4 - 1 = 3 sticks These 3 sticks are: I I I or you can think of the exclusive case as (inclusive case - 2) You say in inclusive case: (Last term - First term + 1): Both end terms are included here. Then you subtract 2 to remove both the end terms and you get: (Last term - First term + 1) - 2 = Last term - First term - 1 I have also explained this in the post: http://gmatquant.blogspot.com/2010/11/four-prongs.htmlCheck out the subtraction part.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Manager
Joined: 18 Aug 2010
Posts: 92
Followers: 1
Kudos [?]:
3
[0], given: 22
|
Re: M01 Question 18 [#permalink]
04 Feb 2011, 11:27
Bunuel wrote: jayasimhaperformer wrote: 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. Answer: D. Hope it helps. Hello Bunuel, nice explanation as usual  is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7? or we have to use the long method of division? thx 4 response
|
|
|
|
|
|
Manager
Joined: 18 Aug 2010
Posts: 92
Followers: 1
Kudos [?]:
3
[0], given: 22
|
Re: M01 Question 18 [#permalink]
04 Feb 2011, 12:03
Bunuel wrote: tinki wrote: Hello Bunuel, nice explanation as usual is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7? or we have to use the long method of division? thx 4 response Well, it depends multiple of which number you want to find and the range you are looking in. Common sense, and divisibility rules should help you in this but sometimes 'trial and error' is also a good method. Check for example divisibility rules here: math-number-theory-88376.htmli did not know about 7 and 11 . nowhere seen the rule before. GREAT HELP . THANKS A LOT A LOT A LOT FOR SWIFT RESPONSES +kudo from me
|
|
|
|
|
|
Intern
Affiliations: Phi Theta Kappa, Dean's list, Chancelors list, Who's Who award, ASCPA
Joined: 16 May 2011
Posts: 26
WE 1: Tax
Followers: 1
Kudos [?]:
3
[0], given: 7
|
Re: M01 Question 18 [#permalink]
22 Sep 2011, 06:27
hockeyplaya182 wrote: 1) Take all numbers between 1 and 999 divisible by 7: 999/7=142
2) Subtract the number that are not three digits: 99/7=14
142-14=128 Great explanation. Thanks.
|
|
|
|
|
|
Intern
Joined: 24 Jul 2011
Posts: 4
Followers: 0
Kudos [?]:
0
[0], given: 5
|
Re: M01 Question 18 [#permalink]
22 Sep 2011, 10:07
Here is a simple Solution :
The count of numbers<=999 divisible by seven = rounded[999/7] = 142
The number of 2 digit numbers divesible by 7 = rounded[99/7] = 14
The number of 3 digit numbers divisible by 7 = 142 - 14 = 128
|
|
|
|
|
|
Manager
Affiliations: University of Tehran
Joined: 06 Feb 2011
Posts: 204
Location: Iran (Islamic Republic of)
Grad GPA: 4
Concentration: Marketing
GMAT 1: 680 Q45 V38
GPA: 4
WE: Marketing (Retail)
Followers: 5
Kudos [?]:
32
[0], given: 57
|
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
22 Sep 2011, 10:30
james12345 wrote: 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. The key point is that in a simple subtraction, we are acting inclusive in one side and exclusive in the other side. For example, when we say 10-6=4, we have not considered 6 itself (exclusive in one side) but we have considered 10 (inclusive in the other side). This is the rule. So, if you figure that out, you see that when GMAT says exclusive, we have to just exclude the side which was not originally excluded. Since one side is not excluded in a normal subtraction, we just subtract 1, not 2. On the other hand, when GMAT says inclusive, we have already included one side, so we just add 1, not 2, to include the other side. Hope it helps
_________________
Ambition, Motivation and Determination: the three "tion"s that lead to PERFECTION.
World! Respect Iran and Iranians as they respect you! Leave the governments with their own.
|
|
|
|
|
|
Intern
Status: Preparing for GMAT
Joined: 19 Sep 2012
Posts: 19
Location: India
GMAT Date: 01-31-2013
WE: Information Technology (Computer Software)
Followers: 0
Kudos [?]:
7
[0], given: 8
|
Re: M01 Question 18 [#permalink]
24 Sep 2012, 05:18
Ans is 128. 14*7 = 98 next will be a 3 digit factor.. and 994 /7 = 142 next will be a 4 digit.. so 142 -14 = 128
_________________
Rajeev Nambyar
Chennai, India.
|
|
|
|
|
|
Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 566
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)
Followers: 10
Kudos [?]:
68
[0], given: 75
|
Re: M01 Question 18 [#permalink]
24 Sep 2012, 05:22
Total 3 digit nos 100 to 999 among them total digits div.by 7 ={(last digit div.by 7 Minus 1st digit div by 7)/7 }+1 {(994-105)/7}+1 = 128
_________________
" Make more efforts "
Press Kudos if you liked my post
|
|
|
|
|
|
Intern
Joined: 18 Jan 2012
Posts: 46
Location: United States
Followers: 1
Kudos [?]:
36
[0], given: 21
|
Re: M01 Question 18 [#permalink]
24 Sep 2012, 08:47
T How do we find out the largest 3 digit number that is divisible by 7 ? Well one way is trial and error, but that is not too efficient. Rather , let's divide 999 by 7 to find the remainder. Subtract the remainder from 999 and voila ! For eg : When 999 is divided by , the remainder is 5. Subtract from 999, the answer is 994. If m/n leaves "r" as the remainder, then m-r is divisible by n. What is the logic behind this ? Well when you divided a number by another divisor, the remainder represents the piece that is not divisible by the divisor. Remove the offending remainder from the quotient and the new quotient will be divisible by the divisor
_________________
-----------------------------------------------------------------------------------------------------
IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES.
YOUR KUDOS IS VERY MUCH APPRECIATED
-----------------------------------------------------------------------------------------------------
|
|
|
|
|
|
Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 551
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 1
Kudos [?]:
13
[0], given: 560
|
Re: M01 Question 18 [#permalink]
27 Feb 2013, 03:54
bb wrote: 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 Hi, I don't understand how and why approach 1 works. Please help explain why it works and provide some examples.
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595
My GMAT Journey : end-of-my-gmat-journey-149328.html#p1198742
|
|
|
|
|
|
|
Re: M01 Question 18
[#permalink]
27 Feb 2013, 03:54
|
|
|
|
|
|
|
|
|
|
|
close
