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What is the number of multiples of 6 between 60 and 100

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What is the number of multiples of 6 between 60 and 100 [#permalink] New post 12 Sep 2006, 11:19
What is the number of multiples of 6 between 60 and 100 inclusive and exclusive?

Please show your work.
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 [#permalink] New post 12 Sep 2006, 11:59
Inclusive:
(96-60)/6 + 1
= 7

Exclusive:
(96-66)/6 + 1
= 6
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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 [#permalink] New post 12 Sep 2006, 14:47
Follow up:

Number of multiples of 7 between 237 and 489 inclusive?
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 [#permalink] New post 12 Sep 2006, 17:33
haas_mba07 wrote:
Follow up:

Number of multiples of 7 between 237 and 489 inclusive?


can we do (489-237)/7+1=37?

BTW why do we add 1 at the end?
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 [#permalink] New post 12 Sep 2006, 17:44
Yes 37 is correct.

You add the 1 when stated for the inclusive to include the first term, because after you've substracted, the first term effectively becomes 0.

So multiples of 6 between 60 and 100:

100-60/6 = 40/6 6r4 However, this does not include the original term of 60 which is divisible by 6. Hence 6+1
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 [#permalink] New post 12 Sep 2006, 17:54
But note that neither of 489 or 237 are divisible by 7.

The smallest number greater than 237 divisible by 7, is 238.
The greatest number less than 437 divisible by 7 is, 434.

Is the answer still 37? Or is it 36? Or is it something else?

MBAlad wrote:
Yes 37 is correct.

You add the 1 when stated for the inclusive to include the first term, because after you've substracted, the first term effectively becomes 0.

So multiples of 6 between 60 and 100:

100-60/6 = 40/6 6r4 However, this does not include the original term of 60 which is divisible by 6. Hence 6+1
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 [#permalink] New post 12 Sep 2006, 18:05
Actually, I stand corrected - it's 36 because the first term isn't divisible by 7.

Correct?
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 [#permalink] New post 12 Sep 2006, 18:05
What about the last one? Its not divisible by 7 either...

MBAlad wrote:
Actually, I stand corrected - it's 36 because the first term isn't divisible by 7.

Correct?
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 [#permalink] New post 12 Sep 2006, 19:07
it is 7.

60 is divisible by 6.
96 is divisible by 6.

so just take these 2 numbers, because 100 is not divisible by 6 and it is excluded anyway.

(96-60)/6 + 1 = 7.

so it is 7.
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 [#permalink] New post 12 Sep 2006, 20:39
haas_mba07 wrote:
What about the last one? Its not divisible by 7 either...

MBAlad wrote:
Actually, I stand corrected - it's 36 because the first term isn't divisible by 7.

Correct?


There are 252 numbers from 237 - 489
489-237 = 252
252/7 = 36.

So there are 36 numbers divisible by 7
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 [#permalink] New post 13 Sep 2006, 03:21
Number of multiples of 7 between 237 and 489 inclusive?

(489-237)/7 could be rewritten as 489/7 - 237/7 = 69 - 33 = 36

We add the "1" if the number excluded by the subtraction is divisible by n.

ie if it had been (490-237))/7 = 36 + 1 (490 is divisible by 7)

or (489-231)/7 = 36 +1 (231 is divisible by 7)

If both upper and lower terms are divisible eg multiples of 7 between 7-28 inclusive then you still add "1" because you only effectively exclude one of the terms:

(28-14)/7 = 2 But there are 3 numbers divisible by 7; 7,14,21 so "+1"
= 2 +1

I'm sure this is a long winded explanation and that one of the more math fluent members can do a better job of explaining this.[/quote]
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 [#permalink] New post 13 Sep 2006, 04:56
Great stuff! Thank you.
  [#permalink] 13 Sep 2006, 04:56
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