Latest TCS Aptitude Question SOLUTION: 1!+2!+3!+......+50!=? a) 3.1035*10^64 b) 2.1021*10^65 c) 3.1035*10^63 d) 3.1035*10^62

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(#M40038628) TCS Numerical Ability question Numbers Keep an EYE

1!+2!+3!+......+50!=?
a) 3.1035*10^64
b) 2.1021*10^65
c) 3.1035*10^63
d) 3.1035*10^62

Asked In TCS Satya Karteek (9 years ago)

Solved HARSHIT AGARWAL Read Solution (10)

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Read Solution (10) :

HARSHIT AGARWAL  (9 years ago)

Note that 50! ≈ 3.04 x 10^64 already, so this eliminates (C) and (D) as they are less than 50!. Intuitively, you would expect that 1! + 2! + ... + 50!'s value would be significantly greater than 50!, but not by a factor of about 10 as (B) expects, so you would expect that 1! + 2! + ... + 50! < 10^65. Thus, the only answer that makes sense is (A).

Therefore, the answer is (A).

I hope this helps!

Is this solution Helpfull? Yes (25) | No (15)  

maria  (9 years ago)

Please can u explain me clearly

Is this solution Helpfull? Yes (8) | No (2)  

miraj ansari  (9 years ago)

3.1035*10^64 should be the Ans.

Is this solution Helpfull? Yes (1) | No (19)  

rashmi lather  (9 years ago)

same que is in puzzle section..chk it from dere

Is this solution Helpfull? Yes (1) | No   

Mayank Sharma  (9 years ago)

Sum of n factorials has
first non-zero digit
with least significance
as 1
when n>4

Is this solution Helpfull? Yes (1) | No   

Sourav ganguly  (9 years ago)

Call the sum S, then 50! < S < 2*50! This is because 50! = 50*49! = 49! + 49! + 49! + ... +49!, which has 50 terms, each as large as the 49 other terms in 1!+2!+3!.......49!

30 414 093 201 713 378 043 612 608 166 064 768 844 377 641 568 960 512 000 000 000 000 < S <
60 828 186 403 426 756 087 225 216 332 129 537 688 755 283 137 921 024 000 000 000 000

It factors as (2^E)(3^F)(5^G)*K, with E=0, F=1, G=0

The least significant digits can be calculated, because they remain fixed! Starting at 1!+2!+3!+4! = 33, the units digit will remain 3 indefinitely. The sum at 9! will have the fixed last two digits (10! ends with two 0s).

The most significant digits can be found in a similar way, pretty much done above, with 50! having the same leading digit as S. So
S = 3. something...3 x 10^64

Is this solution Helpfull? Yes (1) | No (3)  

Rajdeep Sharma  (9 years ago)

calculator is allowed in tcs aptitude exam dont take tension

Is this solution Helpfull? Yes (1) | No (3)  

sahitya  (9 years ago)

can anyone please explain a good method to solve this.

Is this solution Helpfull? Yes | No   

Abra Dbra  (9 years ago)

Call the sum S, then 50! < S < 2*50! This is because 50! = 50*49! = 49! + 49! + 49! + ... +49!, which has 50 terms, each as large as the 49 other terms in 1!+2!+3!.......49!

30 414 093 201 713 378 043 612 608 166 064 768 844 377 641 568 960 512 000 000 000 000 < S <
60 828 186 403 426 756 087 225 216 332 129 537 688 755 283 137 921 024 000 000 000 000

It factors as (2^E)(3^F)(5^G)*K, with E=0, F=1, G=0

The least significant digits can be calculated, because they remain fixed! Starting at 1!+2!+3!+4! = 33, the units digit will remain 3 indefinitely. The sum at 9! will have the fixed last two digits (10! ends with two 0s).

The most significant digits can be found in a similar way, pretty much done above, with 50! having the same leading digit as S. So
S = 3. something...3 x 10^64

Is this solution Helpfull? Yes | No (7)  

SARAVANAN S K  (9 years ago)

please explain it. i cant understand

Is this solution Helpfull? Yes | No (3)  

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