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If p is the product of the integers from 1 to 30 inclusive

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by bhumika.k.shah » Mon Jan 25, 2010 8:28 pm
If p is the product of the integers from 1 to 30 inclusive , what is the greatest integer k for which 3^k is a factor of p ?

A.10
B.12
C.14
D.16
E.18

OA C

How to solve such sums ? A detailed explanation of their basics would be good :-)
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by fibbonnaci » Mon Jan 25, 2010 8:49 pm
this is a simple question.
the number will have a power of 3 only when it is a multiple of 3.
from integers 1 to 30 we have 10 numbers that are multiples of 3.

in those numbers, 9= 3^2- so an extra 3 needs to be accounted.
similarly 18= 3*6=> 3*3*2- so again an extra 3 needs to be accounted.
27= 3*3*3 => 2 extra 3's need to be accounted.

so in total, 10 + 1+1+2 =>14 3's are there from 1-30 inclusive.

therefore the greatest integer k= 14.

hope this helps!
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by thephoenix » Mon Jan 25, 2010 8:51 pm
30/3=10------------>10
10/3=3.3333--------->3
3/3=1---------------->1

TOT=10+3+1=14
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by bhumika.k.shah » Mon Jan 25, 2010 9:12 pm
yes it did! thanks :D
fibbonnaci wrote:this is a simple question.
the number will have a power of 3 only when it is a multiple of 3.
from integers 1 to 30 we have 10 numbers that are multiples of 3.

in those numbers, 9= 3^2- so an extra 3 needs to be accounted.
similarly 18= 3*6=> 3*3*2- so again an extra 3 needs to be accounted.
27= 3*3*3 => 2 extra 3's need to be accounted.

so in total, 10 + 1+1+2 =>14 3's are there from 1-30 inclusive.

therefore the greatest integer k= 14.

hope this helps!
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by bhumika.k.shah » Mon Jan 25, 2010 9:13 pm
i agree this was a very simple sum!
because the #s asked were from 1 to 30..
what if its a huge 5 digit # ? then i cant sit in the exam and find all its multiples nah?
so then how would i go about with it ???

fibbonnaci wrote:this is a simple question.
the number will have a power of 3 only when it is a multiple of 3.
from integers 1 to 30 we have 10 numbers that are multiples of 3.

in those numbers, 9= 3^2- so an extra 3 needs to be accounted.
similarly 18= 3*6=> 3*3*2- so again an extra 3 needs to be accounted.
27= 3*3*3 => 2 extra 3's need to be accounted.

so in total, 10 + 1+1+2 =>14 3's are there from 1-30 inclusive.

therefore the greatest integer k= 14.

hope this helps!
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by bhumika.k.shah » Mon Jan 25, 2010 9:14 pm
Thanks phoenix! :)
But i would like to ask u the same question tht i asked fibbonaci...
what if this was a 5 digit #
then how would i go about with it ????

thephoenix wrote:30/3=10------------>10
10/3=3.3333--------->3
3/3=1---------------->1

TOT=10+3+1=14
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by fibbonnaci » Mon Jan 25, 2010 9:18 pm
yes thats why you have the factorization method. thephoenix has already articulated the method. i didnt want to repost the same thing but wanted to introduce you to another approach.

you need to decide on the approach based on the time you have at hand and the enormosity of the problem.
There is no point going all round the world with complex formulas for simple ones and vice versa.
so it is better that you know various approaches available and take a call!
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by bhumika.k.shah » Mon Jan 25, 2010 9:20 pm
true that!

thanks!! :)
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