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M21-29

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M21-29 [#permalink]

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New post 16 Sep 2014, 01:15
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Re M21-29 [#permalink]

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Official Solution:

How many integers between 18 and 70, inclusive are divisible by either 7 or 9 but not both?

A. 11
B. 12
C. 13
D. 14
E. 15


# of multiples of 7 in the given range is \(\frac{(last-first)}{multiple}+1=\frac{(70-21)}{7}+1=8\);

# of multiples of 9 in the given range is \(\frac{(last-first)}{multiple}+1=\frac{(63-18)}{9}+1=6\);

# of multiples of both 7 and 9 is 1: \(7*9=63\). Notice that 63 is counted both in 8 and 6;

So, # of multiples of either 7 or 9 but not both in the given range is \((8-1)+(6-1)=12\).


Answer: B
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Re: M21-29 [#permalink]

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New post 11 Oct 2015, 01:53
Bunuel wrote:
Official Solution:

How many integers between 18 and 70, inclusive are divisible by either 7 or 9 but not both?

A. 11
B. 12
C. 13
D. 14
E. 15


# of multiples of 7 in the given range is \(\frac{(last-first)}{multiple}+1=\frac{(70-21)}{7}+1=8\);

# of multiples of 9 in the given range is \(\frac{(last-first)}{multiple}+1=\frac{(63-18)}{9}+1=6\);

# of multiples of both 7 and 9 is 1: \(7*9=63\). Notice that 63 is counted both in 8 and 6;

So, # of multiples of either 7 or 9 but not both in the given range is \((8-1)+(6-1)=12\).


Answer: B



Hi,
Thanks for the post and knowing the fomular is very helpful.
But, just wondering why we need to add "1" after (last-first)/multiple?
# of multiples of 7 in the given range is \(\frac{(last-first)}{multiple}+1=\frac{(70-21)}{7}+1=8\);

Thanks!
Regarda
Andy
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M21-29 [#permalink]

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New post 12 Oct 2015, 16:19
Between 18 and 70 inclusive

    multiples of 7 consist {21,28,35,42,49,56,63,70}

    multiples of 9 {18,27,36,45,54,63}


since we need numbers that are divisible by either 7 or 9 but not both.

\(9*7=63\) is to be removed from both the sets.

final set {18,21,27,28,35,36,42,45,49,54,56,70}

no of terms in final set = 12
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Re: M21-29 [#permalink]

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New post 12 Dec 2017, 03:20
The answer is B. First find the number of multiples of 7 from 18 to 70. Then the number of multiples of 9 from 18 to 63. We need to exclude the common multiple from both. 63 has been counted two times The answer is 8+6-2 . i.e. 12. Hence B.
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Re: M21-29   [#permalink] 12 Dec 2017, 03:20
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