Find the remainder when `sum_(r=1)^(n)r!` is divided by 15, if `n ge5`

1.	Find the remainder when `sum_(r=1)^(n)r!` is divided by 15, if `n ge5`.
Answer» Let `N=underset(r=1)overset(n)(sum)r!=1!+2!+3!+4!+5!+6!+7!+ . . .+n!` `=(1!+2!+3!+4!)+(5!+6!+7!+ . . .+n!)` `=33+(5!+6!+7!+ . . .+n!)` `implies(N)/(15)=(33)/(15)+((5!+6!+7!+ . . .+n!))/(15)` `=2+(3)/(15)+` interger [as `5!,6!`, . . . are divisible by 15] `=(3)/(15)+`Integer Hence, remainder is 3.

Discussion

No Comment Found

Related InterviewSolutions

The number of different numbers that can be formed by using all the digits 1, 2, 3, 4,3, 2,1 so that odd digits always occupy the odd places IS 1) 6 2)72 3) 60 4) 18A. 3!4!B. 34C. 18D. 12
There are 10 girls and 8 boys in a class room including Ms. Rani and Radha. A list of speakers consisting of 8 girls and 6 boys has to be prepared. Mr. Ravi refuses to speak if Ms. Rani is a speaker. Ms. Rani refuses to speak if Ms. Radha is a speaker . The number of ways the list can be prepared isA. 202B. 308C. 567D. 952
Ten persons numbered 1, ,2..,10 play a chess tournament, each player against every other player exactlyone game. It is known that no game ends in a draw. If `w_1, w_2, , w_(10)`are thenumber of games won by players 1, ,2 3, ...,10, respectively, and `l_1, l_2, , l_(10)`are thenumber of games lost by the players 1, 2, ...,10, respectively, thena. `sumw_1=suml_i=45`b. `w_1+1_i=9`c. `sumw l1 2()_=81+suml1 2`d. `sumw l i2()_=suml i2`A. `sumw_(i)^(2)+81-suml_(i)^(2)`B. `sumw_(i)^(2)+81=suml_(i)^(2)`C. `sumw_(i)^(2)=suml_(i)^(2)`D. None of these
In how many ways 5 different balls can be distributed into 3 boxes so that no box remains empty?
The total number of ways in which a beggar can be given atleast one rupee from four 25 paise coins, three 50 paise coins and 2 one rupee coins are
A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together, isA. 56B. 126C. 91D. none of these
There are 10 persons named `P_1, P_2, P_3 ..., P_10`. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement `P_1` must occur whereas `P_4` and `P_5` do not occur. Find the number of such possible arrangements.
There are 10 lamps in a hall.Each one of them can be switched on independently.Find the number of ways in which the hall can be illuminated.A. 102B. 1023C. 210D. 10!
There are 10 lamps in a hall.Each one of them can be switched on independently.Find the number of ways in which the hall can be illuminated.A. `10^(2)`B. 1023C. `2^(10)`D. 10!
A hall has 12 gates. In how many ways can a man enter the hall through one gate and come out through a different gate?

Find the remainder when `sum_(r=1)^(n)r!` is divided by 15, if `n ge5`.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment