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M12-36

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Manager
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Joined: 23 Jun 2009
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Re: M12-36 [#permalink]

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New post 26 Sep 2016, 09:00
Put every number from 1 - 200 in an excel spreadsheet and divide them by 12 and by 13.

You will see that 156 is counted twice

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Re: M12-36 [#permalink]

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New post 16 Jan 2017, 02:40
Hi Bunuel,
I believe that to count the number of multiples of 12 in 200 , we can simply divide 200 by 12 and leave the remainder. it is true for other numbers. is it correct way?
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Re: M12-36 [#permalink]

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New post 18 Jul 2017, 10:31
I think this is a high-quality question and I agree with explanation

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Re: M12-36 [#permalink]

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New post 11 Aug 2017, 07:27
Just a suggestion

to see how many multiples of 13, 12 or any number are there up to a particular number, say 200 in this case, all we need to do is divide 200 by 13 ( or 200 /12) and look at the quotient.
That number is the number of factors.

Hope it helps :)

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M12-36 [#permalink]

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New post 01 Sep 2017, 12:07
Dear Bunuel,

Is there a efficient way to find the last number in any given range that divides by an integer x.

In this question, for example, how do I figure out 195 is the last multiple of 13 when range is (0,200) faster?

Regards
Srinath

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Re: M12-36 [#permalink]

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New post 02 Sep 2017, 03:33
krsrinath wrote:
Dear Bunuel,

Is there a efficient way to find the last number in any given range that divides by an integer x.

In this question, for example, how do I figure out 195 is the last multiple of 13 when range is (0,200) faster?

Regards
Srinath


An in integer to be a multiple of 3, the sum of its digit must be a multiple of 3. For example, 315 is divisible by 3 because 3 + 1 + 5 = 9 and 9 is divisible by 3. In your example, 201 is a multiple of 3 (2 + 0 + 1 = 3), thus so must be 201 - 3 = 198.
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Re: M12-36   [#permalink] 02 Sep 2017, 03:33

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