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M21#29

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M21#29 [#permalink] New post 23 Nov 2008, 12:43
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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

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B

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Re: M21#29 [#permalink] New post 23 Nov 2008, 14:05
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ventivish wrote:
How many integers between 18 and 70 inclusive are divisible by either 7 or 9 but not both?
# 11
# 12
# 13
# 14
# 15


divisible by 7 = 21 to 70 i.e. 8 but take out 63. so 7.
divisible by 9 = 18 to 63 i.e. 6 but take out 63. so 5.
= 7 + 5
= 12
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Re: M21#29 [#permalink] New post 23 Nov 2008, 14:20
Thank You GMATTIGER.
I do not understand why we subtract 63 twice although its part of the same series?
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Re: M21#29 [#permalink] New post 23 Nov 2008, 19:15
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ventivish wrote:
How many integers between 18 and 70 inclusive are divisible by either 7 or 9 but not both?
# 11
# 12
# 13
# 14
# 15


Note here: divisible by either 7 or 9 but not both?

ventivish wrote:
Thank You GMATTIGER.
I do not understand why we subtract 63 twice although its part of the same series?

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Re: M21#29 [#permalink] New post 22 Jun 2010, 08:06
I was thrown off by "between 18 and 70". To me, that didn't include 18 or 70, so my initial answer was 10. Does "between" generally include the first and last numbers in a series?
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Re: M21#29 [#permalink] New post 22 Jun 2010, 08:07
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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)

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Re: M21#29 [#permalink] New post 22 Jun 2010, 08:09
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michigancat wrote:
I was thrown off by "between 18 and 70". To me, that didn't include 18 or 70, so my initial answer was 10. Does "between" generally include the first and last numbers in a series?


The question mentions the term "inclusive" which means first and last terms included.
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Re: M21#29 [#permalink] New post 22 Jun 2010, 21:39
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counting would be difficult if the numbers are large.

One can try following method:

70/7 = 10
70/9 = 7

Divisible by both = 70/(7*9) = 1
Therefore divisible by 7 & 9, but not both = 10+7-1 = 16

Similary for 18
18/7 = 2
18/9 = 2

divisible by 7 & 9, but not both = 2+2 - 0 = 4

between 18&70 is 16-4 = 12

Therefore answer is 12
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Re: M21#29 [#permalink] New post 22 Jun 2010, 23:33
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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...
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Re: M21#29 [#permalink] New post 06 Aug 2010, 05:10
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Basic Mathematical Tables & Counting:
i) For 7, the numbers lying between 18 and 70 (inclusive) are from:
7*3=21 to 7*10=70
Therefore 10-3 = 7

ii) For 9, the numbers lying between 18 and 70 (inclusive) are from:
9*2=18 to 9*7=63
Therefore 7-2 = 5

Now we simply add them up;

7+5 = 12
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Re: M21#29 [#permalink] New post 24 Jun 2011, 08:39
My Approach,
considering multiple of 7 and 9 as two arithmetic progressions,
general formula
an(last term of the series)= a1(first term of the series) + (n-1) d (distance btw terms)
applied:

1) (multiple of 7)
70=21+(n-1)7 (diving by 7) 10=3+n-1
n= 8

2) (multiple of 9)
63=18+(n-1)9 (diving by 9) 7=n+1
n=6

Note: 63 is counted twice and we need to exclude it since the question stem is telling us # of integer divisible by either 7 or 9 (BUT NO BY BOTH)

N=14-2=12 ANSWER IS B

P.S. this approach might not be as useful as other approaches (posted here) on this question considering the small range of values. However, it might be good to keep in mind when you need to consider a larger series of integers btw i.e 322 to 1345 ....
my 2 cents,
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Re: M21#29 [#permalink] New post 24 Jun 2011, 10:30
Hello Everyone!

I had the same approach as iPinnacle.

These are my steps:

1) How many numbers are divisible by 7 in the range?

(70 - 21)/7 + 1
= 8

2) (63 - 18)/9 + 1
= 6

3) How many are divisible by both?
Ans from #1 - Ans from #2
= 8 - 6
= 2

Therefore 2 numbers in the series are divisible by both 7 and 9

4) Number of items divisible by 7 and 9 but not both
= Ans from #1 + Ans from #2 - Ans from #3
= 8 + 6 - 2

Therefore answer is 12 ie B

Hope that helps!
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Re: M21#29 [#permalink] New post 26 Jun 2011, 06:55
intelindahouse wrote:
Basic Mathematical Tables & Counting:
i) For 7, the numbers lying between 18 and 70 (inclusive) are from:
7*3=21 to 7*10=70
Therefore 10-3 = 7

ii) For 9, the numbers lying between 18 and 70 (inclusive) are from:
9*2=18 to 9*7=63
Therefore 7-2 = 5

Now we simply add them up;

7+5 = 12



@intelindahouse,
Not sure if i quite understand your process, when you say 10-3= 7 you are excluding one integer on the table count ...
we have 21 28 35 42 49 56 63 70 .... it will be good to say that you are already not considering 63 which is divisible by both (and we need to exclude as the question stem is telling us)
same process for the series of 9,

when counting range of integers (inclusive of both start and end integer) always remember to add 1,
my two cents, if I misunderstood you I am sorry,
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Re: M21#29 [#permalink] New post 27 Jun 2011, 04:48
70-18=52

52/9+1=6
52/7+1=8

total=14 sabstract the common 63 from both of them=12
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Re: M21#29 [#permalink] New post 28 Jun 2012, 04:25
Ans 11

9,{18,27,36,45,54,63},72
14,{21,28,35,42,49,56,63,70}

So total 11 excluding 63 hope i am rgt A
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Re: M21#29 [#permalink] New post 28 Jun 2012, 05:02
Use counting principles, i.e. {[(Last #)-(First number)]/multiple}+1

Don't forget to take out 1 (representing 63) from final answer. This yields 12 (B).

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Re: M21#29 [#permalink] New post 28 Jun 2012, 05:07
answer is B

to find total nos between 18 and 70 that which are divisible by 7 :

first count the total nos between the 21 and 70 (including both).
It comes out to be 50.
so, total nos divisible by 7 = 50/7 + 1= 8 ( adding 1 because 70 gets left out while dividing the range by 7).
but we cannot count 63 ( because it is divisible by both 7 and 9)
hence total nos divisible by 7 = 7.

similarly for the total nos divisible by 9 :

we select the range 18 to 63.
But since 63 will not be counted anyways, so we can just count the nos between 18 and 54.

so the total nos divisible by 9 comes out as (54-18)/9 + 1 = 5

the answer is 7+5 = 12.

it looks lengthy but once you get the idea it can be done really fast!
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Re: M21#29 [#permalink] New post 04 Jul 2012, 04:21
but not both?
missed this part initially and got wrong answer easily.
so need to be very careful from next time.
thanks for nice question.
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Re: M21#29 [#permalink] New post 04 Jul 2012, 04:26
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ventivish 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

[Reveal] Spoiler: OA
B

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# of multiples of 7 in the given range is (last-first)/multiple+1=(70-21)/7+1=8;
# of multiples of 9 in the given range is (last-first)/multiple+1=(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] New post 28 Jun 2013, 06:52
I just quickly wrote out the multiples of 7 and 9

7=21,28,35,42,49,56,63,70
9=18,27,36,45,54,63

Then I counted them (minus the two "63"s) and came up with 12.
Re: M21#29   [#permalink] 28 Jun 2013, 06:52
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