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How many three-digit integers are not divisible by 3 ? (A) 599 (B) 600 (C) 601 (D) 602 (E) 603 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.
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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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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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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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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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why are you subtracting from 900 and not 999?
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I did 999-102 divided by 3 = 299 999-299 = 600.
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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.
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Ans : B = 600
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TheSituation wrote: I did 999-102 divided by 3 = 299
999-299 = 600. 999-299 is 700 not 600 ....careless....
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Can some body pls. explain clearly the strategy involved?
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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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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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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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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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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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Ah ok, problem solved. Thanks! (means kudos  )
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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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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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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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