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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
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Hello,

Calculate the no. of terms from 1 to 1000 (inclusive) that are divisible by 11 or 35 or both.

1. No of terms divisible by 11 -> 1000/11 = 90
2. No of terms divisible by 35 -> 1000/35 = 28
3. No of terms divisible by 11 and 35 -> 1000/(11*35) = 2

Answer = 1000- (90+28-2) = 884.
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
manalq8 wrote:
What is the number of integers from 1 to 1000 (inclusive) that are divisible by neither 11 nor by 35?

A. 884
B. 890
C. 892
D. 910
E. 945


Number divisible by 11:-
1000/11= 90

numbers divisible by 35:-
1000/35= 28

Numbers divisible by both 11 and 35= 2

Total numbers divisible by both 11 and 28= 90+28-2= 116 (because we counted 2 in both 90 and 28)

Total numbers not divisible by 11 or 35= 1000-116= 884

A is the answer
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
Numbers divisible by 11: 1001/11 = 91
Numbers divisible by 35: 1001/35 = 28
Numbers divisible by 11x35: 1001/385 = 2

1001-(91+28)+2 = 884

A.
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
law258 wrote:
Numbers divisible by 11: 1001/11 = 91
Numbers divisible by 35: 1001/35 = 28
Numbers divisible by 11x35: 1001/385 = 2

1001-(91+28)+2 = 884

A.

­Can you explain why you check for numbers dividible by 11x35?

Thank
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
Expert Reply
gmatmartell wrote:
law258 wrote:
Numbers divisible by 11: 1001/11 = 91
Numbers divisible by 35: 1001/35 = 28
Numbers divisible by 11x35: 1001/385 = 2

1001-(91+28)+2 = 884

A.

­Can you explain why you check for numbers dividible by 11x35?

Thank

­
I updted the solution HERE. Hope now it's clear.
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
Bunuel wrote:
Official Solution:

How many integers are there between 1 and 1000, inclusive, that are not divisible by either 11 or 35?

A. 884
B. 890
C. 892
D. 910
E. 945


Let's determine the number of multiples of 11 or 35 between 1 and 1000, inclusive, and subtract that number from 1000.

The number of multiples of an integer within a range can be calculated using the following formula:


\(\frac{\text{last multiple in the range - first multiple in the range} }{\text{multiple} }+1\)

Thus:


• The number of multiples of 11 in the given range is \(\frac{ last - first}{multiple}+1=\frac{990-11}{11}+1=90\);

• The number of multiples of 35 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{980-35}{35}+1=28\);

• The number of multiples of both 11 and 35 is 2 (since \(11*35=385\) and \(385*2=770\));

Observe that the two numbers 385 and 770 are included in the count of multiples of 11 and the count of multiples of 35. To avoid double-counting, we need to subtract these numbers once from the total count of multiples of 11 and 35. Therefore, the number of multiples of either 11 or 35 in the given range is \(90+28-2=116\).

Consequently, the count of numbers that are not divisible by either 11 or 35 is \(1000-116=884\).


Answer: A­

­Hello Bunuel. Could you please help me understand a quick way to find out the greatest multiple of 35? How did you got within less than 2 minutes to 980?
It would help me a lot when using this method with big numbers like these.
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
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ruis wrote:
Bunuel wrote:
Official Solution:

How many integers are there between 1 and 1000, inclusive, that are not divisible by either 11 or 35?

A. 884
B. 890
C. 892
D. 910
E. 945


Let's determine the number of multiples of 11 or 35 between 1 and 1000, inclusive, and subtract that number from 1000.

The number of multiples of an integer within a range can be calculated using the following formula:




\(\frac{\text{last multiple in the range - first multiple in the range} }{\text{multiple} }+1\)

Thus:




• The number of multiples of 11 in the given range is \(\frac{ last - first}{multiple}+1=\frac{990-11}{11}+1=90\);

• The number of multiples of 35 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{980-35}{35}+1=28\);

• The number of multiples of both 11 and 35 is 2 (since \(11*35=385\) and \(385*2=770\));

Observe that the two numbers 385 and 770 are included in the count of multiples of 11 and the count of multiples of 35. To avoid double-counting, we need to subtract these numbers once from the total count of multiples of 11 and 35. Therefore, the number of multiples of either 11 or 35 in the given range is \(90+28-2=116\).

Consequently, the count of numbers that are not divisible by either 11 or 35 is \(1000-116=884\).


Answer: A­

­Hello Bunuel. Could you please help me understand a quick way to find out the greatest multiple of 35? How did you got within less than 2 minutes to 980?
It would help me a lot when using this method with big numbers like these.

­ruis You can divide 1000 by 35, and then subtract the remainder, 20, from 1000:



 ­
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Re: How many integers are there between 1 and 1000, inclusive, that are [#permalink]
GMATCoachBen wrote:
ruis wrote:
Bunuel wrote:
Official Solution:

How many integers are there between 1 and 1000, inclusive, that are not divisible by either 11 or 35?

A. 884
B. 890
C. 892
D. 910
E. 945


Let's determine the number of multiples of 11 or 35 between 1 and 1000, inclusive, and subtract that number from 1000.

The number of multiples of an integer within a range can be calculated using the following formula:




\(\frac{\text{last multiple in the range - first multiple in the range} }{\text{multiple} }+1\)

Thus:




• The number of multiples of 11 in the given range is \(\frac{ last - first}{multiple}+1=\frac{990-11}{11}+1=90\);

• The number of multiples of 35 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{980-35}{35}+1=28\);

• The number of multiples of both 11 and 35 is 2 (since \(11*35=385\) and \(385*2=770\));

Observe that the two numbers 385 and 770 are included in the count of multiples of 11 and the count of multiples of 35. To avoid double-counting, we need to subtract these numbers once from the total count of multiples of 11 and 35. Therefore, the number of multiples of either 11 or 35 in the given range is \(90+28-2=116\).

Consequently, the count of numbers that are not divisible by either 11 or 35 is \(1000-116=884\).


Answer: A­

­Hello Bunuel. Could you please help me understand a quick way to find out the greatest multiple of 35? How did you got within less than 2 minutes to 980?
It would help me a lot when using this method with big numbers like these.

­ruis You can divide 1000 by 35, and then subtract the remainder, 20, from 1000:



 ­


Thats awesome! Thanks :)

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