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M21#29

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Joined: 20 Mar 2013
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GMAT Date: 07-25-2013
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Re: M21#29 [#permalink] New post 28 Jun 2013, 10:40
dagmat wrote:
My 2 cents:

Since 18 and 70 are included, the range of numbers allowed is (70-18) + 1. This becomes 53.
Next, divide 53/7, we get 7*7=49 as the closest. Hence only 7 numbers are present.

Similarly for 9, 53/9, we get 9*5=45 as the closest. Hence only 5 numbers are present.
63 is the first number divisible by both 9&7, but its outside the range. So total numbers = 7+5=12 (B). This method is similar to iPinnacle's method but I think its a bit faster.

Hope this helps...


this method would not work in all situations
eg : had the no. range been 19-69 inclusive then by this method the ans would still have been 12 whereas the correct ans is 10
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Re: M21#29 [#permalink] New post 29 Jun 2013, 14:00
Answer B.

I made a silly mistake, but this is how I worked the problem.

Number divisible by 7 inclusive 70 = 10, but under 18 there are two multiples - hence 8
Number divisible by 9 inclusive 70 = 7, but under 18 there is one multiple - hence 6

LCM (least common multiple) of 7 and 9 = 63
Number divisible by 63 inclusive 70 and under 18 = 1
that one is part of both 7 and 9 sets, hence remove the elements twice to count total = 8 + 6 - 2 = 12
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Re: M21#29 [#permalink] New post 30 Jun 2013, 13:21
iPinnacle wrote:
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

Ans is B

This is not hard counting. Here is the formula.
{( last divisible number - first divisible number)/divisor}+1

Here, between 18 and 70,
For 7,
last divisible number is 70
first divisible number is 21
so {(70-21)/7}+1 =9

similarly we can find out for 9, which is 5
So 9+5=14

But we have to exclude the number which are divisible by both. Here in this case that number is 63(here also you can use above formula by taking LCM of 7 and 9 as divisor). So remove this common number from both list.

so 14-2=12 (Ans)


I think there is a mistake here. {(70/21)/7}+1 = 8, not 9. Therefore, 8+5 =13 and you would only have to remove "1" to represent the 63.
Re: M21#29   [#permalink] 30 Jun 2013, 13:21
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