SOLUTION: a) What is the remainder left after dividing
1! + 2! + 3! +…………………+ 100! by 7?
b) If n >=4 then find whether 2.n is less than 4.n or not.
Algebra ->
Divisibility and Prime Numbers
-> SOLUTION: a) What is the remainder left after dividing
1! + 2! + 3! +…………………+ 100! by 7?
b) If n >=4 then find whether 2.n is less than 4.n or not.
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You can put this solution on YOUR website! a) What is the remainder left after dividing
1! + 2! + 3! +…………………+ 100! by 7?
GROUP AS FOLLOWS
=(1!+3!)+(2!+4!)+(5!+6!)+(7!+8!+9!+..........+100!)
=(7)+(26)+5!(1+6)+7!(1+8+8.9+............+8.9.10......100)...WE FIND THAT EXCEPT 26 THE SECOND TERM, ALL THE OTHER TERMS ARE DIVISIBLE BY 7.HENCE THE REMAINDER WE GET ON DIVIDING WITH 7 IS THE REMAINDER WE GET WHEN WE DO 26/7...THAT IS ...5 IS THE REMAINDER .
b) If n >=4 then find whether 2.n is less than 4.n or not.
CHECK THE QUESTION 2.N MEAns 2*n ..if n=5 say 2*5=10.....it is definitely less than 4*n=4*5=20
what is there to prove in that?
