Question

What is the remainder when 1!+2!+3!+4!+.... +100! divided by 5?
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Answer 1

Answer:

3

Step-by-step explanation:

See,

We know that 5! is 120

6!is 720

and on moving on the units place be zero

thats why don't require rest else as the unit's place be zero

so, considering 1!+2!+3!+4!

1+2+6+24

33

the total sum will something but we know that the units place is 3 so the remainder is 3

Answer 2

$\underline{\boxed{\bold{Question}}}$

• What is the remainder when E = 1!+2!+3!+4!+.... +100! divided by 5?

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$\underline{\boxed{\bold{Important\;concepts\;to\;be\;learned}}}$

• In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.

• n! = n × (n - 1) × (n - 2) × (n - 3) .................... × 2 × 1

            = n × (n - 1)!

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$\underline{\boxed{\bold{Answer}}}$

Let's start !

As from the above concept,

100! = 100 × 99 × 98 ........... × 5 × 4 × 3 × 2 × 1

99! = 99 × 98 × 97 ........... × 5 × 4 × 3 × 2 × 1

98! =  98 × 97 × 96 ........... × 5 × 4 × 3 × 2 × 1

.

.

.

.

5! = 5 × 4 × 3 × 2 × 1

4! =  4 × 3 × 2 × 1 = 24

3! = 3 × 2 × 1 = 6

2! = 2 × 1 = 2

1! =  1

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As as you can see from above expressions that from 100! to 5! there is multiple of 5 coming common.

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E = 1! + 2! + 3! + 4! + .... + 100!

E = 1! + 2! + 3! + 4! + 5(K)    →    {K is some number after taking 5 common}

E = 1 + 2 + 6 + 24 + 5K

E = 3 + 30 + 5K

E = 5(K + 6) + 3    

E = 5M + 3                      { K + 6 = M}

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So now, as we know

Dividend = Quotient × Divisor + Remainder.

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So By comparing above both equations,

Dividend = E

Quotient = M

Divisor = 5

Remainder = 3

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$\boxed{\boxed{\bold{\therefore Remainder\;of\;the\;Expression = 3 }}}$

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Hope this helps u.../

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