Latest TCS Aptitude Question SOLUTION: Find d max value of n such that 157! is perfectly divisible by 18^n...can sm 1 solve dis wid xplanation..plz

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Find d max value of n such that 157! is perfectly divisible by 18^n...can sm 1 solve dis wid xplanation..plz

Asked In TCS SHAIK NAYAB RASOOL (12 years ago)

Unsolved Read Solution (8)

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vivek singh  (12 years ago)

52+17+5+1=75
so ans=3^75=3(3^74)
=3(9^37)
ans=37

Is this solution Helpfull? Yes (27) | No (4)  

pawan  (12 years ago)

Value of n will depend on value of power of 9 in 157!.

number of powers of 3 in 157! = 157/3 +157/9+ 157/27 + 157/81
= 52+ 17 + 5 + 1= 79
so 157! = k* 3^79= 3k*(3^78) = 3k*9^39

so max value of n= 39

Is this solution Helpfull? Yes (17) | No (16)  

shresth pal  (11 years ago)

18 = 3*3*2
We need to find the power of 3 in 157!
= [157/3] + [157/9] + [157/27] + [157/81]
where [x] represents highest integer less than or equal to x.
= 52 + 17 + 5 + 1
= 75

So, we have 37 pairs of 3.

The number of 2s will definitely more than 37 so we do not need to calculate.

Hence, ans will be 37

Is this solution Helpfull? Yes (7) | No (1)  

akhil kogta  (12 years ago)

n=18
18 can be written as 9*2
157/9=17 & 81=9*9
so total 9 used in 157! are 18

Is this solution Helpfull? Yes (4) | No (9)  

snehil  (12 years ago)

there are 8, 18 in 157
and some more can be found by
9*2 ,6*3,
so total number of 8's will be 8+2=10
so n max=10

Is this solution Helpfull? Yes (2) | No (6)  

pihu  (12 years ago)

akhil is right ans is 18

Is this solution Helpfull? Yes (1) | No (5)  

rocks  (12 years ago)

how 79???????

Is this solution Helpfull? Yes (1) | No (5)  

Pawan Ahuja  (12 years ago)

use formula
[n/p]+[n/P^2]+[n/P^3]+[n/P^4]+[n/P^5]........ soo on
n is given no with factorial take on n and [2.33]=2 greatest integer
p is prime no
so 18=9*2=3^2*2

Is this solution Helpfull? Yes (1) | No (2)  

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