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gmatcracker2010
How many three-digit integers are not divisible by 3 ?

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

Total Single digit No. = 9 (1 to 9)
Total Two digit No. = 90 (10 to 99)
Total Single digit No. = 900 (100 to 999)

Total No. divisible by 3 from 1 through 999 = 999/3 = 333
Total No. divisible by 3 from 1 through 99 = 99/3 = 33

Total No. divisible by 3 from 100 through 999 = 333-33 = 300

So not divisible by 3 = 900-300 = 600

Answer: Option B
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gmatcracker2010
How many three-digit integers are not divisible by 3 ?

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

first, let's find how many ARE divisible...
102 is the minimum one - it's the 34th multiple of 3.
999 is the maximum one - it's the 333 multiple of 3.
333-34 +1 (inclusive counting) = 300 numbers are divisible by 3.
now...we have 999 total numbers. we exclude the non 3 digit ones (from 1 to 99)
999-99=900
900-300=600
answer is B.
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gmatcracker2010
How many three-digit integers are not divisible by 3 ?

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

Three-digit numbers range from 100 to 999

How many 3-digit numbers total?
Inclusive: (Greatest-Least) + 1
(999 - 100) = 899 + 1 = 900 numbers altogether

1) Find a pattern - divisible by 3? (digits must sum to 3 or a multiple of 3)
100: no
101: no
102: yes
103: no
104: no
105: yes

2 out of 3, \(\frac{2}{3}\), are NOT divisible by 3

\(\frac{2}{3}*900 = 600\)

Answer B

2) Use evenly spaced set's properties* to find how many numbers ARE divisible by 3, i.e. find how many are multiples of 3

Subtract those multiples of 3 from the total of 3-digit numbers

First and last multiples of 3 in this range?

The first multiple of 3 is 102
The last multiple of 3 is 999

Number of terms (multiples of 3) =

\(\frac{(Last Term-FirstTerm)}{Increment} + 1\)

\((\frac{999-102}{3}+1)=(\frac{897}{3}+1)=(299+1)=300\)


There are 900 numbers from 100 to 999

300 ARE divisible by 3

(900-300) = 600 are NOT divisible by 3

Answer B

*See benjiboo ,
PART 2:Find number of integers that are a multiple of a certain number in a set

The whole Guide to Series and Sequences is excellent
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