Question

Find the remainder of (1!) + (2!)2 + (3!)2 + (4!) + ... + (50!)? divided by 10, .
Your answer​

Answer 1

Answer:

n=1!+2!+3!+4!+5!+…+50!

=> n=1+2+6+24+5!+…+50!

=>n=33+5!+…+50!

Now, if we observe terms other than 33, we find an interesting property

5!=5×4×3×2×1 ,divisible by 10

6!=6×5! ,divisible by 10

7!=7×6!=7×6×5! ,divisible by 10

In a this manner, every term from 5! to 50! is divisible by 10

Therefore their unit digit is 0

Now ,

n=30+ (a large no. with unit place 0)+3

This means that the unit digit of this large no. Is 3

We know , the remainder obtained when a no. is divided by 10 is the unit digit of that no.

Therefore , we can conclude that the remainder when n is divided by 10 is 3

Problem solved