Remainder when (1! + 2! + 3! + ... + 4000!) is divided by 7
options
a)7 ,
b)1,
c)5,
d)none
Answers
Given :- Remainder when (1! + 2! + 3! + ... + 4000!) is divided by 7
options
a)7 ,
b)1,
c)5,
d)none
Solution :-
we know that,
→ 1! = 1
→ 2! = 1 * 2 = 2
→ 3! = 1 * 2 * 3 = 6
→ 4! = 1 * 2 * 3 * 4 = 24
→ 5! = 1 * 2 * 3 * 4 * 5 = 120
→ 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720
→ 7! = (1 * 2 * 3 * 4 * 5 * 6) * 7 = multiple of 7 .
and, next all factorial greater than 7 will be multiple of 7 . { As they have 7 in multiply . }
then,
→ (1! + 2! + 3! + _____+ 4000!) ÷ 7
→ (1!/7) + (2!/7) + (3!/7) + _______ (4000!/7)
→ 1/7 + 2/7 + 6/7 + 24/7 + 120/7 + 720/7 ____ next all gives remainder as 0 .
→ (1 + 2 + 6 + 24 + 120 + 720)/7
→ 873/7
→ 124 quotient and 5 remainder .
therefore, we can conclude that,
→ (1! + 2! + 3! + _____+ 4000!) ÷ 7 = Remainder 5 (Option C) (Ans.)
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SOLUTION
TO CHOOSE THE CORRECT OPTION
Remainder when ( 1! + 2! + 3! + ... + 4000! ) is divided by 7
a) 7
b) 1
c) 5
d) None
EVALUATION
Here the given expression is
1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! + 7 ! +... + 4000 !
We see that
Starting from 7 ! upto 4000 ! each terms contain the number 7
So each of 7 ! , 8 ! , 9 ! , ... , 4000 ! is divisible by 7
Now we have to check the expression
1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 !
Here
5 ! + 6 ! = ( 1 + 6 ) 5 ! = 7 × 5 !
So 5 ! + 6 ! is divisible by 7
1 ! + 3 ! = 1 + 6 = 7
So 1 ! + 3 ! is divisible by 7
Now
2 ! + 4 ! = 2 + 24 = 26
26 = ( 3 × 7 ) + 5
By Division algorithm the Remainder when 26 is divided by 7 is 5
Hence the required Remainder when ( 1! + 2! + 3! + ... + 4000! ) is divided by 7 is 5
FINAL ANSWER
Hence the correct option is c) 5
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