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M12-36

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Re: M12-36  [#permalink]

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New post 26 Sep 2016, 09:00
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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?

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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]

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New post 11 Mar 2018, 01:10
AdmitJA wrote:
A humble suggestions. :-)

30 would be a good trap answer here. I got 30 and then realised I'll have to reduce 1 twice. Had 30 been there, I'd got this one wrong.


This is so evil, i was trapped in the loop for quite awhile before I noticed the mistake too.

Really a good suggestion!

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M12-36  [#permalink]

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New post 27 Apr 2019, 06:29
Bunuel,

I have some basic doubt here, I know am missing something but not sure where am making the mistake.
On one other similar problem:
How many positive integers less than 100 are neither multiples of 2 nor multiples of 3 ?
We find # of multiples of 2, multiples of 3 and multiples of 6 which is 50,33 and 16 respectively.
100 - ((50 + 33) - 16) because P(Total) = P(A)+P(B)-P(Both)+P(Neither)

But here,
Here, we found # of multiples of 13, # of multiples of 16 which is 15,16 respectively
(15-1) + (16-1) since in explanation it was mentioned that 156 belonged to both 12 and 13..

on similar lines, each multiple of 6 would be there in multiples of 2 and 3 as well right,
so it should have been (50-16) + (33-16) right??
Why is this difference in the method..
I can see that question is slightly different, neither/or, but as I said am unable to connect the dots here and confused. Please help.

P(A or B) = P(A) + P(B) - P(Both A & B)
So what is the difference between multiples of A or multiples of B and multiples of A or multiples of B but not both
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Re M12-36  [#permalink]

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New post 10 Aug 2019, 09:44
I think this is a high-quality question and I agree with explanation.
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Re M12-36   [#permalink] 10 Aug 2019, 09:44

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