Question

Find the least no to be multiplied by 100! such that is its divisible by 3^50?

Answer 1

Heya User,

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

--> Now, let :->
-----> A = n{ x | 0<x<100 ; 3 divides x }
-----> B = n{ x | 0<x<100 ; 9 divides x }
-----> C = n{ x | 0<x<100 ; 27 divides x }
-----> D = n{ x | 0<x<100 ; 81 divides x }

NOW, A = [ 100 / 3 ] = 33 --> B = 11 --> C = 3 ---> D = 1;
--> A + B + C + D = 48 ;

Hence, '3' to the power '48' completely divides 100!.... Hence, 9 should be multiplied in 100! to make it divisible by 3^50...

CONCEPT APPLIED :->

--> 3 = 3*1;
--> 6 = 3*2 ; --> 9 = 3*3 ;

Hence, 33 times '3'

+++++++++++++++++++++++++

--> 9 = 3*3 --> repetition of 3 twice
--> 18 = 3*3*2 --> '3' twice

Hence, 11 times '3'

And similarly, ..... 48 times '3' altogether....

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Hope you understood the Answer xD ☺