a little problem

montex · Joined: 18 Jun 2005 Posts: 23 Location: New Delhi

What is the least number that must be multiplied by 100! to make it perfectly divisible by 3 to the power of 50.

kindly give a simplified short and easy to understand solution.
thankyou so much Confused

addy91in · Joined: 30 Jun 2005 Posts: 1 Location: Mumbai

What is the least number that must be multiplied by 100! to make it perfectly divisible by 3 to the power of 50.

since 100 doesn't contain any 3s so the least multiplicant must be 3^50

any takers??

shaji

elan · Joined: 28 Jun 2005 Posts: 5

testing

vamsi-143 · Joined: 01 Jul 2005 Posts: 11

What is the least number that must be multiplied by 100! to make it perfectly divisible by 3 to the power of 50 ?

you hav 100!/(3^50)

but 100! = 1*2*3....*99*100

you got 33 numbers that are divisible by in 3 in 100!
now you 17 3's remaining in your denominator, but there are again 11 3's will cancel for 1*2*3....*32*33 and again there are three 3's and again you hav one 3 total 48 3's will e cancelled for 100! the denominator you got 9 hence the least that shuld be multiplied to 100! to make it perfectly divisible is 9

Hope you all got this solution Wink

montex · Joined: 18 Jun 2005 Posts: 23 Location: New Delhi

vamsi-143 · Joined: 01 Jul 2005 Posts: 11

100! = 100*99*98...*2*1

Here you will cancel 33 numbers from 1 to 100 but there are numbers 9,27... you hav to count the number of threes in 9 is not 1 but there are 2 and so on you cal. and include them also you will find the sol. bye Wink

neerajj · Joined: 29 Aug 2005 Posts: 7

I thk the number need to be multiplied by 27 as :

No of multiples of 3 in 100 is 33
No of multiples of 9 in 100 is 11
No of multiples of 27 in 100 is 3

Total number of 3's multiple available in 100! is 33+11+3 = 47

Remaining multiples of 3 reqd is 50-47 = 3 . Hence need to be multiplied by 27

baskar · Site Admin Joined: 11 Apr 2004 Posts: 41

jawla · Joined: 10 Sep 2005 Posts: 1

Hi Everybody,

the correct answer is 9.
Let me explain all of u.
To get the no. of powers of any number(say x) that can perfectly divide any other(say y) no's factorial.

divide the no y by x and get only integer part,
then divide y by x^2 and get only integer part,
then divide y by x^3 and get only integer part ans so on untill we exceeds or equal the no(y).

Take the sum of these integers value and that will be power of no(x) that can divide the no. y perfectly means no remainder left.

Now come to the problem here y is 100 and x is 3.
get first value 100/3 = 33.33333... but take integer value 33
then 100/3^2 = 11.1111.... but take integer value 11
then 100/3^3 = 3.6.. but integer value is 3
then 100/3^4 = 1.abc... but integer value is 1
then 100/3^5 = 0.xyz... but integer part will be zero here
and for next powers also integer part will be zero.

so now take sum 33 + 11 + 3 +1 = 48

Now it's clear that 100! can be divided by 3^48 perfectly so we need only 3^2 i.e 9 to be multiplied into 100! .

whitelotus1980 · Joined: 11 Jan 2006 Posts: 8