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M16-11

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Official Solution:

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

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


Total 3-digit numbers: \(999-100+1=900\);

# of multiples of 3 in the given range: \(\frac{\text {last-first}}{multiple}+1=\frac{999-102}{3}+1=300\);

{Total} - {# of multiples of 3} = {# of not multiples of 3}, hence the answer is \(900-300=600\).


Answer: B
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Re: M16-11 [#permalink]

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Hi,

Total 3-digit numbers: 999 - 100 + 1 = 900

We know that the range goes from 100 to 999

There will always be a 2:1 ratio between numbers not divisible by 3 and numbers divisible by 3.
For example: 100, 101, 102... 103, 104, 105... 997, 998, 999

We can infer that the 900 integers in the desired range will also fulfil this 2:1 ratio. Thus, 600 integers are not divisible by 3 and 300 are divisible by 3.

Is my apporach correct?
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New post 31 Oct 2014, 04:58
minwoswoh wrote:
Hi,

Total 3-digit numbers: 999 - 100 + 1 = 900

We know that the range goes from 100 to 999

There will always be a 2:1 ratio between numbers not divisible by 3 and numbers divisible by 3.
For example: 100, 101, 102... 103, 104, 105... 997, 998, 999

We can infer that the 900 integers in the desired range will also fulfil this 2:1 ratio. Thus, 600 integers are not divisible by 3 and 300 are divisible by 3.

Is my apporach correct?


Yes, that's correct.
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Re: M16-11 [#permalink]

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New post 13 Oct 2015, 07:21
Hi , the questions says three digit integers why should we not consider -100 to -999 ?

There can be another 600 -ve integers from -100 to -999 which are not divisible by 3.

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New post 18 Oct 2015, 13:47

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New post 02 Jun 2016, 20:31
Are you subtracting 102 from 999 because 102 is the first three-digit multiple of 3, and 999 is the last three-digit multiple of 3?

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New post 02 Jun 2016, 20:59
glochou wrote:
Are you subtracting 102 from 999 because 102 is the first three-digit multiple of 3, and 999 is the last three-digit multiple of 3?


Yes this gives us the range of multiples of 3, and in this range every THIRD number will be div by 3....
so we divide the range by 3..
And add 1 as both the first and last integers are div by 3 BUT in our calculation we take ONLY one of them..
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Re: M16-11 [#permalink]

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New post 11 Jul 2016, 15:01
I did it this way:

We have an arithmetic sequence with 1st term 102 (smallest 3 digit multiple of 3) and 999 (last term), d=3. Using the formula, we have 999 = 102 + 3*(n1) <=> n = 300. Total number of possible 3 digit numbers is 999-100+1 = 900. Deduct those which are divisible by 3 from all possible numbers, 900-300 = 600 are NOT divisible by 3.

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New post 16 Oct 2016, 13:27
Still dont understand why we are adding 1 in the equation. I got 601. Help!

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New post 23 Aug 2017, 05:01
Its the same formula as the one for A.P.

Nth term= a+ (n-1)d

Nth term = last term

(Last term- first term)/difference +1= n
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Re: M16-11   [#permalink] 23 Aug 2017, 05:01
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