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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.

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

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

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} --> \(900-300=600\).

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.

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

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....

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.

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 --> \(96=12+4(n-1)\) --> \(n=22\).

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?

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

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