1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is

Question

1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is the remainder? 

A) 1 
B) 2 
C) 3 
D) 4 
E) 0

banerjeea_98 · Answer

"A". Will explain if correct.

toddmartin · Answer

The pattern seems to be as follows:


 sum of the numbers: 1 25 31

 remiainder /5: 1 0 1


 I stopped there because 16! would be too much to calculate. The 36th term would be (36^2)!, so the remainder would be 0 at this point based on the pattern, and the last term is 37!, so the remainder should be 1.

 So my answer is A

neelesh · Answer

The way I solved it: (probably similar to the previous post).

1! + (2^2)! + 3! + (4^2)! + ... + 37!

= 1 + 4! + 6 + 16! + ... + 37! 

= 25 + 6 + 16! + ... + 37!

Based on the above expression 25 would result in a remainder of 0, 6 would result in a remainder of 1 and every term after that all the way to 37! will always result in a remainder of 0, as 5 will be a part of each of them (due to the factorial). Hence remainder is 1.

twixt · Answer

I considered that any number equal or greater than 5! does not provide any remainder so I just focused on the smallest terms.

1! + (2^2)! + 3! + (4^2)! +5!+(6^2)!...+ 37!

(4^2)!>5! and includes 5 so only 1!, 4! and 3! contribute to the remainder calculation.

=1+24+6=31

So remainder is 1.

1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is

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