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How do we use Venn Diagram approach to this question

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by [email protected] » Sun Oct 13, 2013 11:48 pm
Hi shibsriz,

While it is possible to use a 3-circle Venn Diagram to answer this question, it's not the easiest way for most Test Takers to answer this question. With some basic notes and minor calculations, you can answer this question.

First, the range of numbers. 200 to 300, inclusive is 300 - 200 + 1 = 101 numbers

Now we need to eliminate the numbers that are divisible by 2, 3, 5. There will be some numbers that are divisible by more than one of those values, but those won't be hard to find.

First the 2s. We need to eliminate ALL the even values. From 201 to 300, we'd remove half the numbers = 50. Add in the "200" and that's 51 numbers
Next the 5s. About half of those 5s have been removed already (200, 210, etc.), so let's list the ones that haven't: 205, 215, 225...295 = 10 numbers
Finally, the 3s. We've already accounted for all the even multiples and all the multiples that end in 5, so focus on the ones that we haven't hit yet:

201, 207, 213, 219 (notice how we're increasing by 6? Just make sure that we aren't hitting anything that ends in a 5)......
231, 237, 243, 249, 261, 267, 273, 279, 291, 297. That is = 14 numbers

51 + 10 + 14 = 75 numbers eliminated.

[spoiler]101 - 75 = 26 numbers[/spoiler]

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by [email protected] » Mon Oct 14, 2013 3:04 am
Hi Rich,

While I understand your solution, my question is if this approach can be utilised for all ranges for example 200 to 600??

[email protected] wrote:Hi shibsriz,

While it is possible to use a 3-circle Venn Diagram to answer this question, it's not the easiest way for most Test Takers to answer this question. With some basic notes and minor calculations, you can answer this question.

First, the range of numbers. 200 to 300, inclusive is 300 - 200 + 1 = 101 numbers

Now we need to eliminate the numbers that are divisible by 2, 3, 5. There will be some numbers that are divisible by more than one of those values, but those won't be hard to find.

First the 2s. We need to eliminate ALL the even values. From 201 to 300, we'd remove half the numbers = 50. Add in the "200" and that's 51 numbers
Next the 5s. About half of those 5s have been removed already (200, 210, etc.), so let's list the ones that haven't: 205, 215, 225...295 = 10 numbers
Finally, the 3s. We've already accounted for all the even multiples and all the multiples that end in 5, so focus on the ones that we haven't hit yet:

201, 207, 213, 219 (notice how we're increasing by 6? Just make sure that we aren't hitting anything that ends in a 5)......
231, 237, 243, 249, 261, 267, 273, 279, 291, 297. That is = 14 numbers

51 + 10 + 14 = 75 numbers eliminated.

[spoiler]101 - 75 = 26 numbers[/spoiler]

GMAT assassins aren't born, they're made,
Rich
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by GMATGuruNY » Mon Oct 14, 2013 5:12 am
[email protected] wrote:How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3
B. 16
C. 75
D. 24
E. 26

Answer is 26


The LCM of 2, 3 and 5 is 30.
Multiples are CYCLICAL:
Between successive multiples of 30, there will always be the SAME NUMBER of values that are divisible by 2, 3, or 5.
By implication, there will also always be the same number of values that are NOT divisible by 2, 3, or 5.

Consider an EASY CASE.
Count how many integers between 1 and 30 are NOT divisible by 2, 3 or 5:
1, 3, 7, 11, 13, 17, 19, 23, 29.
Total options = 8.
Implication:
Between successive multiples of 30, there will always be 8 integers that are not divisible by 2, 3 or 5.

Thus, between 200 and 300, we get:
211-240 --> 8 options
241-270 --> 8 options
271-300 --> 8 options.

Now we need to count the integers between 200 and 210 that are not divisible by 2, 3 or 5.
If the sum of the digits of an integer is a multiple of 3, then the integer itself is a multiple of 3.
Thus, between 200 and 210, an integer is viable only if the sum of its digits is NOT a multiple of 3 and if its units digit is an odd number OTHER than 5:
203, 209.

Thus:
Total good options = 8+8+8+2 = 26.

The correct answer is E.
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by s.vishnu » Mon Oct 14, 2013 9:20 am
Hi Experts,
I approach this problem using Arithmetic Progression to find the number of terms and then take the lcm of 2,3 , 2,5 and 3,5 to get the common terms counted.However,i am not able to get to the result.

Can you please let me know if this approach is wrong and if not, how the terms/numbers counted twice can be eliminated to reach 26 as the answer.

Thank you.
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by sanjoy18 » Mon Oct 14, 2013 10:04 am
total number 101

Number divisible by 2: 51
Number divisible by 3: 34
Number divisible by 5: 21
Number divisible by 6: 17
Number divisible by 10: 11
Number divisible by 15: 7
Number divisible by 30: 4

hence total number divisible by either 2 ,3 or 5
51+34+21-17-11-7+4=75

hence not divisible by either 2 ,3 or 5
101 -75 =26
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