Question

For an integer n, n! = n(n-1) (n-2)...3.2.1. Then 1! + 2! + 3! + ... + 100! When
divided by 5 leaves remainder

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Answer 1

Given:

Given that, n! = n (n-1) (n-2).....3 . 2 . 1

To find:

We need to find remainder when  1! + 2! + 3! + ... + 100! is divided by 5.

Solution:

We have to find the remainder when the expression is divided by 5.

So let us expand it a bit to understand it carefully,

1! + 2! + 3! + 4! + 5! + 6!  ... + 100!

If we observe all the elements from 5! contains 5 in it.

All the element from 5! have the factor 5 in it, so we can say that

(5! +  6!  ... + 100!) is a multiple of 5 ,hence the remainder is 0 if divided by 5.

so left behind is 1! + 2! + 3! + 4!

Now,

1! + 4! = (1 + 24) = 25 , which is again a multiple of 5.

So the remainder we get only depends on (2! + 3!), let us find it out

2! = 2

3! = 6

(2! + 3!) = 8

when 8 is divided by 5 we get our remainder 3.

So, combining all we can say the remainder is 3 when the expression is divided by 5.

Therefore, 3 is the remainder when we divide it.

Answer 2

$\sf { \underline{SOLUTION}}$

GIVEN

For an integer n, n! = n(n-1) (n-2)...3.2.1

TO DETERMINE

The remainder when

1! + 2! + 3! + ... + 100! is divided by 5

CONCEPT TO BE IMPLEMENTED

We write a ≡ b ( mod n )

When ( a - b) is divisible by n

EVALUATION

Here the given expression is

1! + 2! + 3! + ... + 100!

= 1! + 2! + 3! + 4! + 5! + 6!... + 100!

= ( 1! + 2! + 3! + 4! ) + ( 5! + 6!... + 100!)

Here

5! = 5 × 4 × 3 × 2 × 1

So 5! is divisible by 5

Similarly 6! , 7! , ..... 100! are divisible by 5

Now 1! = 1

∴ 1! ≡ 1 ( mod 5 )

Now 2! = 2 × 1 = 2

∴ 2! ≡ 2 ( mod 5 )

Now 3! = 3 × 2 × 1 = 6

∴ 3! ≡ 6 ( mod 5 )

Since 6 ≡ 1 ( mod 5 )

∴ 3! ≡ 1 ( mod 5 )

Now 4! = 4 × 3 × 2 × 1 = 24

∴ 4! ≡ 24 ( mod 5 )

Since 24 ≡ 4 ( mod 5 )

∴ 4! ≡ 4 ( mod 5 )

∴ 1! + 2! + 3! + 4! ≡ ( 1 + 2 + 1 + 4 ) ( mod 5 )

∴ 1! + 2! + 3! + 4! ≡ 8 ( mod 5 )

∵ 8 ≡ 3 ( mod 5 )

∴ 1! + 2! + 3! + 4! ≡ 3 ( mod 5 )

Hence from above

∴ 1! + 2! + 3! +...... + 100! ≡ 3 ( mod 5 )

Hence the required Remainder = 3

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