It is currently 21 Nov 2017, 21:20

### 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

we will pick new questions that match your level based on your Timer History

Track

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M01 Question 18

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Math Expert
Joined: 02 Sep 2009
Posts: 42284

Kudos [?]: 132984 [0], given: 12399

Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]

### Show Tags

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.
_________________

Kudos [?]: 132984 [0], given: 12399

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7743

Kudos [?]: 17839 [0], given: 235

Location: Pune, India
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]

### Show Tags

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.html
Check out the subtraction part.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17839 [0], given: 235

Manager
Joined: 18 Aug 2010
Posts: 88

Kudos [?]: 20 [0], given: 22

Re: M01 Question 18 [#permalink]

### Show Tags

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 $$\frac{994-105}{7}+1=128$$.

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

Kudos [?]: 20 [0], given: 22

Math Expert
Joined: 02 Sep 2009
Posts: 42284

Kudos [?]: 132984 [1], given: 12399

Re: M01 Question 18 [#permalink]

### Show Tags

04 Feb 2011, 11:46
1
KUDOS
Expert's post
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.html
_________________

Kudos [?]: 132984 [1], given: 12399

Manager
Joined: 18 Aug 2010
Posts: 88

Kudos [?]: 20 [0], given: 22

Re: M01 Question 18 [#permalink]

### Show Tags

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.html

i 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

Kudos [?]: 20 [0], given: 22

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1965

Kudos [?]: 2097 [1], given: 376

Re: M01 Question 18 [#permalink]

### Show Tags

10 May 2011, 11:11
1
KUDOS
gmatprep09 wrote:
How many of the three-digit numbers are divisible by 7?

(A) 105
(B) 106
(C) 127
(D) 128
(E) 142

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Description of the technique:

Q: How many 3 digits numbers are divisible by 7
OR
How many multiples of 7 are there between 100 and 999, inclusive?

Choose the first number greater than equal to 100 that is divisible by 7.
Remainder of 100/7 is 2. 100-2+7=105
"105" is the first number greater than equal to 100 that is divisible by 7.

Choose the first number less than equal to 999 that is divisible by 7.
Remainder of 999/7 is 5. 999-5=994
"994" is the first number less than equal to 999 that is divisible by 7.

$$n=\frac{994-105}{7}+1=127+1=128$$

Ans: "D"

************************
This rules hold good for all numbers.

Find the numbers divisible by 3 between 1 and 18, inclusive.

First Number greater than equal to 1 that is divisible by 3 = 3
First Number less than equal to 18 that is divisible by 3 = 18

$$Number of multiples = \frac{18-3}{3}+1=5+1=6$$

Validate:
Count of {3,6,9,12,15,18} = 6

Try it with any number and you should get the count.
_________________

Kudos [?]: 2097 [1], given: 376

Intern
Affiliations: Phi Theta Kappa, Dean's list, Chancelors list, Who's Who award, ASCPA
Joined: 16 May 2011
Posts: 25

Kudos [?]: 4 [0], given: 7

WE 1: Tax
Re: M01 Question 18 [#permalink]

### Show Tags

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.

Kudos [?]: 4 [0], given: 7

Intern
Joined: 24 Jul 2011
Posts: 4

Kudos [?]: 1 [0], given: 5

Re: M01 Question 18 [#permalink]

### Show Tags

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

Kudos [?]: 1 [0], given: 5

Manager
Affiliations: University of Tehran
Joined: 06 Feb 2011
Posts: 200

Kudos [?]: 38 [0], given: 57

Location: Iran (Islamic Republic of)
Concentration: Marketing
Schools: Wharton
GMAT 1: 680 Q45 V38
GPA: 4
WE: Marketing (Retail)
Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]

### Show Tags

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.

Kudos [?]: 38 [0], given: 57

Intern
Status: Preparing for GMAT
Joined: 19 Sep 2012
Posts: 19

Kudos [?]: 10 [0], given: 8

Location: India
GMAT Date: 01-31-2013
WE: Information Technology (Computer Software)
Re: M01 Question 18 [#permalink]

### Show Tags

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.

Kudos [?]: 10 [0], given: 8

Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 534

Kudos [?]: 361 [0], given: 75

Location: India
GPA: 3.82
WE: Account Management (Retail Banking)
Re: M01 Question 18 [#permalink]

### Show Tags

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

Kudos [?]: 361 [0], given: 75

Manager
Joined: 18 Jan 2012
Posts: 51

Kudos [?]: 106 [0], given: 26

Location: United States
Schools: IIM A '15 (A)
Re: M01 Question 18 [#permalink]

### Show Tags

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

-----------------------------------------------------------------------------------------------------

Kudos [?]: 106 [0], given: 26

Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 506

Kudos [?]: 72 [0], given: 562

Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Re: M01 Question 18 [#permalink]

### Show Tags

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 : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992

Kudos [?]: 72 [0], given: 562

Manager
Joined: 20 Dec 2011
Posts: 86

Kudos [?]: 113 [0], given: 31

Re: M01 Question 18 [#permalink]

### Show Tags

23 Sep 2013, 08:34
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:
$$\frac{994-105}{7}+1=128$$

Be careful using approach one because it can lead to wrong answers. It is best used as an approximation because it will be within 1 of the correct answer.

How many two-digit numbers that start with 3 are divisible by 7: 39 - 30 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Correct because we have 35 only.)
How many two-digit numbers that start with 4 are divisible by 7: 49 - 40 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Incorrect because we have 42 and 49.)

Kudos [?]: 113 [0], given: 31

Intern
Joined: 31 Aug 2013
Posts: 15

Kudos [?]: 2 [0], given: 1

Re: M01 Question 18 [#permalink]

### Show Tags

23 Sep 2013, 20:05
How many of the three-digit numbers are divisible by 7?

(A) 105
(B) 106
(C) 127
(D) 128
(E) 142

Firstly,find the first 3 digit number that is divisible by 7 ie 105 and the last 3 digit number that is divisible by 7 is 994.

so, series become like this:

105,112,.....................994.

l = a + (n-1) d

994 = 105+(n-1)7
994-105+7=7n
n=896/7=128

so, right answer is

128.

Kudos [?]: 2 [0], given: 1

Re: M01 Question 18   [#permalink] 23 Sep 2013, 20:05

Go to page   Previous    1   2   [ 35 posts ]

Display posts from previous: Sort by

# M01 Question 18

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.