Latest TCS Aptitude Question SOLUTION: Find the minimum value of n such that 50! is perfectly divisible by 2520^n.

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(#M40037133) TCS Numerical Ability question Keep an EYE

Find the minimum value of n such that 50! is perfectly divisible by 2520^n.

Asked In TCS makthabai akash (10 years ago)

Solved Saraswathy Read Solution (18)

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Read Solution (18) :

Saraswathy  (10 years ago)

50!/2520^n
2520 -> 2^3 * 3^2 * 5^1 * 7^1

Here 7 is the largest prime factor...
So in order to find the minimum value of "n", it is enough to find the minimum power of "7"... nd for maximum value of "n", find max power of 7...

For max. value of n, find
50/7^1 + 50/7^2 = 7 + 1 = 8 [quotient only]
Max. value of n which is perfectly divisible by 2520^n is (8)
Min. value is 1

Max value : 8
Min Value : 1

Is this solution Helpfull? Yes (72) | No (9)  

vicknesh  (10 years ago)

any way 50! gonna end in 0
similarly, 2520^1 will end in 0
so n =1

Is this solution Helpfull? Yes (9) | No (2)  

Avi  (10 years ago)

50!
---
(2520)^n
Factorize 2520-2*2*3*3*7
Take the biggest no-7.Every 2520 has a 7
now 1st find the no of 7's in 50!.
way is-50/7=quotient-7;again divide 7 by 7=quotient is 1.so stop.So it has 7+1=8 7's.Therefore,2520 must be 8 times to cancel all 7's.

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Krishnendu Lakshman  (10 years ago)

2520 = 2^3 * 3^2 * 5^1 * 7^1 = 8 * 9 * 5 * 7

These are co-prime factors of 2520 and the minimum value of n depends on the max power of 9 (and not 7)...

max power of 9 = floor(50/9)+floor(50/81) = 5 + 0 = 5

therefore, n = 5

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saurabh kumar  (10 years ago)

ans is 12. As in 50! the no. of zeros will be 50/5+50/25+50/125+.....=12.(in factorial of any no there will be so many zeros as many no. of 5 so we are calculating no. of 5 using above formula). So in 2520^n (10) ki powers same honi chahiye...that's why ans is 12.

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Sanjeevan DEY  (10 years ago)

for ur kind information,,,distance between delhi and Mumbai is 1384 km by train,,,!!!

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Anant Shaurya Gaur  (10 years ago)

1 because 2520=3.5*6! so any any factorial greater than 6! will be purely divisble by it.

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RUPESH KUMAR  (10 years ago)

n=0 then it wud become 1 den it easily divides it

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Mayuri  (10 years ago)

@ SARASWATHY: Why u took quotient only?

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jayakarthi K R  (10 years ago)

how did u tell the minimum value is 1.pls explain

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LOKESH RANJAN  (10 years ago)

1, 2520->2^3*3^2*5^1*7^1

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Priya Singh  (10 years ago)

min value of n is 1 only

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sahitya  (10 years ago)

can anyone please explain it more clearly??

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NADEEM AHMAD  (10 years ago)

2520=2^3*3^2*5*7
7 is the highest prime no.among all the factors 2520
highest power of 7 in 50! =50/7+50/7^2+50/7^3......
we get highest power of 7 as 8
which is lowest of all the highest powers of prime factors of 2520
so 8 will be the answer

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arjit  (10 years ago)

ans is 25

we have to divide 50! by 2520^n

now 50! contains 25s 5 in its factors
50=5*5*5
45=5*9

and so on...

now 2520 contains only one 5 in its prime factorization so we need 25 5 so value of n will be 25

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anviti nigam  (10 years ago)

guys....it is too simple.....see how

we are given the fac. of 50 and results always end with 0,and also 2520^n last with 0 ..
so any value of n is possible such that n>=1.....
thankyou

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joydeb  (9 years ago)

11..
2520=2^3*7*3^2
heighest power 9
so 22/2=11

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Kiran Namdev Derle  (8 years ago)

If you want to divide 50! by 2520, every factor of 2520 must divide 50!.
This is the basic logic.
Now, Factors of 2520 = 2^3 * 3^2 * 5 * 7.
It is obvious that no of 2 will be more in the expression 50!.
Check of No. 3 = 50/3 + 50/9 + 50/27 = 16 + 5 +1 = 22 (Only consider Quotient)
Check of No. 5 = 50/5 + 50/25 = 10 + 2 = 12
Check of No. 7 = 50/7 + 50/49 = 7 + 1 = 8.
To divide 50! by 2520 we need Three 2s, Two 3s, One 5 and One 7. We have to see the minimum number of occurrences to make the pairs of these. 7 has minimum number of occurrences, which is 8.
So 8 is the answer source:targetcampus(com).
Note: you have to use same logic as you do it for finding out number of zeros in an expression!

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