It is currently 24 Jun 2017, 12:48

### 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

# M16#11

Author Message
Intern
Joined: 19 Jun 2008
Posts: 20

### Show Tags

20 Nov 2008, 11:25
5
This post was
BOOKMARKED
How many three-digit integers are not divisible by 3 ?

(A) 599
(B) 600
(C) 601
(D) 602
(E) 603

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Denote any number divisible by 3 $$D$$ and any number not divisible by 3 $$N$$ . The number range 100-999 starts with $$NNDNND$$ ... and ends with ... $$NNDNND$$ . In all, there are $$\frac{999 - 100 + 1}{3} = 300$$ $$NNDs$$ and thus there are $$2*300 = 600 Ns$$ .

Would somebody please explain the +1 in the numerator of the given fraction? I don't quite understand why 1 is added.
SVP
Joined: 29 Aug 2007
Posts: 2473

### Show Tags

20 Nov 2008, 22:08
snowy2009 wrote:
How many three-digit integers are not divisible by 3 ?

(C) 2008 GMAT Club - m16#11

* 599
* 600
* 601
* 602
* 603

Denote any number divisible by 3 $$D$$ and any number not divisible by 3 $$N$$ . The number range 100-999 starts with $$NNDNND$$ ... and ends with ... $$NNDNND$$ . In all, there are $$\frac{999 - 100 + 1}{3} = 300$$ $$NNDs$$ and thus there are $$2*300 = 600 Ns$$ .

Would somebody please explain the +1 in the numerator of the given fraction? I don't quite understand why 1 is added.

count even integers from 10 to 20 including.
= (20-10)/2 + 1 = 6

this is how counting works.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

VP
Joined: 18 May 2008
Posts: 1260

### Show Tags

02 Dec 2008, 01:49
How many three-digit integers are not divisible by 3 ?

(C) 2008 GMAT Club - m16#11

I could nt nderstand the solution given. Instead I followed an alternate approach:
Total 3 diit nos are 999-100 +1=900
3 digit nos divisible by 3= [(999-102)/3] +1=300
So 3 digit nos not divisible by 3= 900-300=600
Intern
Joined: 19 Dec 2008
Posts: 12

### Show Tags

20 Dec 2008, 15:32
3
KUDOS
1
This post was
BOOKMARKED
I think is B) 600

There are 6 one-digit numbers that are not divisible by 3, 60 two-digit numbers that are not divisible and so on.
Intern
Joined: 21 Nov 2008
Posts: 13

### Show Tags

20 Dec 2008, 19:17
1
KUDOS
jjhko wrote:
How many three-digit integers are not divisible by 3 ?

A) 599
B) 600
C) 601
D) 602
E) 603

the range is from 100 to 999. Find the difference between the lowest and the highest three digit number and divide by 3, then add 1.

((999-102)/3)+1=300 - are divisible by three

900 - 300 = 600 are not divisible
Manager
Joined: 09 Dec 2009
Posts: 122

### Show Tags

20 Mar 2010, 11:08
I did 999-102 divided by 3 = 299

999-299 = 600.
_________________

G.T.L. - GMAT, Tanning, Laundry

Round 2: 07/10/10 - This time it's personal.

Intern
Joined: 08 Jun 2010
Posts: 12

### Show Tags

18 Jun 2010, 06:48
1
KUDOS
All above solution implicitly using AP. (Arithmetic Progression)

Nth number in AP = a+(n-1)d
a=> first number
d=> common difference.

so ;

equation => 999 = 102 + (n-1)3

999=> last number divisible by 3
102 => first number to be divisible by 3

solving: n = 300 ... Hence 102 to 999 there 300 numbers divisble by 3
now even between 100 - 999, 300 numbers are divisible by 3;

Hence 900 - 300 = 600.

--------------------
Ans : B = 600
Intern
Joined: 16 Jun 2010
Posts: 2

### Show Tags

18 Jun 2010, 09:20
TheSituation wrote:
I did 999-102 divided by 3 = 299

999-299 = 600.

999-299 is 700 not 600 ....careless....
Math Expert
Joined: 02 Sep 2009
Posts: 39660

### Show Tags

18 Jun 2010, 11:08
ZMAT wrote:
Can some body pls. explain clearly the strategy involved?

How many three-digit integers are not divisible by 3 ?

Total 3 digit numbers: $$999-100+1=900$$.
Multiples of 3 in the range 100-999: $$\frac{999-102}{3}+1=300$$ (check this: totally-basic-94862.html#p730075).

{Total} - {# multiples of 3} = {# of not multiples of 3} --> $$900-300=600$$.

Hope it helps.
_________________
Intern
Joined: 18 Jun 2010
Posts: 11

### Show Tags

18 Jun 2010, 11:43
Thanks Brunel for the clear explanation!
I forgot the tip to calculate the multiples in sequential order number..This was referred in one of the MGMAT guides.
Intern
Joined: 22 Jun 2010
Posts: 18

### Show Tags

22 Jun 2010, 04:16
1
KUDOS
Total number of numbers divisible by 3 between 1 and 999 -- 999/3 = 333.
Total number of numbers divisible by 3 between 1 and 99 -- 99/3 = 33
Total 3 digit numbers divisible by 3 = 333-33 = 300

Total 3 digit numbers not divisible by 3 are 900 - 300 = 600
Intern
Joined: 22 Jun 2010
Posts: 9

### Show Tags

23 Jun 2010, 19:24
i think its worth mentioning also that when you do the arithmatic progression equation and divide by 3 or whatever number, it is not necessary that the number is wholly divisible in this case....just round down.,...

so 998/3 means 32 divisors and then you add 1 from the other side....
Intern
Joined: 30 Aug 2009
Posts: 26

### Show Tags

14 Aug 2010, 00:19
Bunuel wrote:
ZMAT wrote:
Can some body pls. explain clearly the strategy involved?

How many three-digit integers are not divisible by 3 ?

Total 3 digit numbers: $$999-100+1=900$$.
Multiples of 3 in the range 100-999: $$\frac{999-102}{3}+1=300$$ (check this: totally-basic-94862.html#p730075).

{Total} - {# multiples of 3} = {# of not multiples of 3} --> $$900-300=600$$.

Hope it helps.

What's the difference between your approach and the arithmetic progression? Here, both get the same result, but for the example you have in the basic section ("How many numbers betw. 12 and 96 div. by 4?") the arith. progr. approach doesn't work!? Thx.
Math Expert
Joined: 02 Sep 2009
Posts: 39660

### Show Tags

14 Aug 2010, 11:14
1
KUDOS
Expert's post
thb wrote:
Bunuel wrote:
ZMAT wrote:
Can some body pls. explain clearly the strategy involved?

How many three-digit integers are not divisible by 3 ?

Total 3 digit numbers: $$999-100+1=900$$.
Multiples of 3 in the range 100-999: $$\frac{999-102}{3}+1=300$$ (check this: totally-basic-94862.html#p730075).

{Total} - {# multiples of 3} = {# of not multiples of 3} --> $$900-300=600$$.

Hope it helps.

What's the difference between your approach and the arithmetic progression? Here, both get the same result, but for the example you have in the basic section ("How many numbers betw. 12 and 96 div. by 4?") the arith. progr. approach doesn't work!? Thx.

The formula I wrote is basically the one counting # of terms in AP - $$# \ of \ terms=\frac{Last \ term -First \ term}{common \ difference}+1$$ ($$n=\frac{a_n-a_1}{d}+1$$ as $$a_n=a_1+d(n-1)$$), as multiples of some integers basically are AP.

So for the question: "How many multiples of 4 are there between 12 and 96, inclusive?" the formula will be the same and will give the same answer --> $$96=12+4(n-1)$$ --> $$n=22$$.

Hope it's clear.
_________________
Intern
Joined: 30 Aug 2009
Posts: 26

### Show Tags

14 Aug 2010, 12:10
Ah ok, problem solved. Thanks! (means kudos )
Manager
Joined: 09 Jun 2011
Posts: 103

### Show Tags

22 Jun 2011, 07:15
What is the sum of all the possible 3-digit numbers that can be constructed using the digit 3,4, and 5, if each digit can be used only once in each number?
Manager
Status: Married
Affiliations: MENSA India
Joined: 30 Apr 2011
Posts: 102
Location: India
GMAT 1: 630 Q42 V35
GMAT 2: 640 Q42 V35
GMAT 3: 640 Q44 V32
GMAT 4: 660 Q47 V34
GMAT 5: 680 Q46 V37
GPA: 3.6
WE: Sales (Computer Software)

### Show Tags

22 Jun 2011, 09:37
To solve the problem we need to find out total numbers with 3 digits and then subtract 3 digit multiples of 3 from it.

Total 3 digit numbers = Maximum number - Minimum number + 1 (dont't forget that the first minimum number also counts. Like 1 to 5 has 5 numbers)
So,
999 - 100 + 1 = 899 + 1 = 900

I took a different approach after this.
1 in every 3 numbers is a multiple of 3.
So 1/3 of 900 numbers will be multiples of 3.
So Multiples of 3 = 900/3 = 300

Subtract 3 Digit Multiples of 3 from Non-multiples

900 - 300 = 600

So IMHO the answer is B.
_________________

Always aim at the moon. Never mind if you miss, you will fall in the stars

Intern
Joined: 15 Jul 2010
Posts: 6
Location: Bangkok, Thailand

### Show Tags

22 Jun 2011, 18:04
I think B.
3-digit integers 100, 101,102, ..., 999
you can see that every 3 3-digit integers, there will be 1 3-digit integer that is divisible by 3 and number of 3-digit integer = 999-99 = 900
As a result, number of 3-digit integers that is divisible by 3 = 900/3 = 300
number of 3-digit integers that is NOT divisible by 3 = 900-300 =600
_________________

"You can do it if you believe you can!" - Napoleon Hill
"Insanity: doing the same thing over and over again and expecting different results." - Albert Einstein

Intern
Joined: 10 May 2011
Posts: 1

### Show Tags

23 Jun 2011, 16:19
May be its a stupid question - but why are we not considering negative integers here? The questions says three digit integers!
Senior Manager
Joined: 15 Sep 2009
Posts: 265

### Show Tags

26 Jun 2012, 06:49
A good question that illustrates the need to know certain simple but important formulae when it comes to topics such as Counting and Probability.

I solved using Bunuel's mathematical approach as well. Nothing more to add.

Cheers,
Der alte Fritz

Der
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Re: M16#11   [#permalink] 26 Jun 2012, 06:49

Go to page    1   2    Next  [ 27 posts ]

Display posts from previous: Sort by

# M16#11

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO 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®.