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M16#11

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M16#11 [#permalink] New post 20 Nov 2008, 10:25
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How many three-digit integers are not divisible by 3 ?

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

[Reveal] Spoiler: OA
B

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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.
[Reveal] Spoiler: OA
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Re: m16 #11 [#permalink] New post 20 Dec 2008, 14:32
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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.
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Re: m16 #11 [#permalink] New post 20 Dec 2008, 18:17
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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
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Re: M16#11 [#permalink] New post 18 Jun 2010, 05:48
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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
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Re: M16#11 [#permalink] New post 22 Jun 2010, 03:16
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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
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Re: M16#11 [#permalink] New post 14 Aug 2010, 10:14
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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.

Answer: B.

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.
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Re: M16#11 [#permalink] New post 26 Jun 2013, 18:27
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100-999 = # of 3 digit integers.
999-100+1 = 900 (# of different 3 digit integers)

of these, we know that 2 of every 3 integers are not going to be divisible by 3.
so, 900 x 2 / 3 = 600
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Re: M16#11 [#permalink] New post 20 Nov 2008, 21:08
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.
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m16, #11 [#permalink] New post 02 Dec 2008, 00:49
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
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Re: m16 #11 [#permalink] New post 20 Mar 2010, 10:08
I did 999-102 divided by 3 = 299

999-299 = 600.
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Re: m16 #11 [#permalink] New post 18 Jun 2010, 08:20
TheSituation wrote:
I did 999-102 divided by 3 = 299

999-299 = 600.



999-299 is 700 not 600 ....careless....
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Re: M16#11 [#permalink] New post 18 Jun 2010, 10:08
Expert's post
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.

Answer: B.

Hope it helps.
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Re: M16#11 [#permalink] New post 18 Jun 2010, 10:43
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.
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Re: M16#11 [#permalink] New post 23 Jun 2010, 18:24
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....
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Re: M16#11 [#permalink] New post 13 Aug 2010, 23:19
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.

Answer: B.

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.
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Re: M16#11 [#permalink] New post 14 Aug 2010, 11:10
Ah ok, problem solved. Thanks! (means kudos ;-))
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Re: M16#11 [#permalink] New post 22 Jun 2011, 06:15
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?
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Re: M16#11 [#permalink] New post 22 Jun 2011, 08:37
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.
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Re: M16#11 [#permalink] New post 22 Jun 2011, 17:04
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
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Re: M16#11 [#permalink] New post 23 Jun 2011, 15:19
May be its a stupid question - but why are we not considering negative integers here? The questions says three digit integers!
Re: M16#11   [#permalink] 23 Jun 2011, 15:19
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