# solve these questions

solve these questions......wth full procedure plzzzz

Q1..a person add first N natural no. and find the sum to be 1850 but actually one number was added twice by mistake .find the difference b/w N and that no.?

a)40 b)33 c)60 d)17

Q2....what is the unit digit of 1!+2!+3!.....99!+100!?

a) 3 b)1 c)5 d)6

Q3...the no.444444........999times is definately divisible by

a)22 b)44 c)222 d)all of these

Q4...N=7!^3 how many factors of N are multiples of 10?

a)736 b)1008 c)1352 d)894

Q5....find the unit digit of N=27!^37!?

a)1 b)3 c)7 d)9

Q6...whts the unit digit of 2^3^3^4?

a)2 b)4 c)8 d)6

Q7...a no N has odd no. of divisors ,,which is true abt N?

a)all divisors are odd

b)there will be atleast N-11 prime divisors

c)N will be perfect square

d)atleast one divisr should be odd

Q8...how many zeroes will be there at the end of 1003*1001*999*....123?

Q9...whts the remainder when 1!+2!+3!.......1000! divided by 5 ?

a)1 b)2 c)3 d)4

Q10.......plzz tell me where i can get the fundas of unit digit questions in this site....

Q2....what is the unit digit of 1!+2!+3!.....99!+100!

Soln.:

Unit digit =(1!+2!+3!+...+100!)%10

As 5!+..+100! is divisible by 10

So unit digit=(1!+..+4!)%10=(1+2+6+24)%10=33%10=3

Unit digit is 3..

Option (a) is correct...

Giv ur best to d world.

Nd d best will cm back to u...

Regards,

Dipanjan....

n/a

Q3...the no.444444........999times is definately divisible by

Soln. 222=111*2

So option (c) is correct...

Giv ur best to d world.

Nd d best will cm back to u...

Regards,

Dipanjan......

n/a

Q4...N=7!^3 how many factors of N are multiples of 10?

Soln.:

7!^3=(1*2*3*4*5*6*7)^3=(2*5)*(2^2*5^2*3^3*4^3*6^3*7^3)

Total factors:

3^2*4^4=9*256=2304

Don't know why 2304 is missed in the option..

Giv ur best to d world.

Nd d best will cm back to u...

Regards,

Dipanjan....

n/a

Q5....find the unit digit of N=27!^37!?

Soln.:

Unit digit: 27!^37!%10

See as 27 and 10 are co-primes so we can apply Euler Theorem.

Euler of 10 is 4

So 27!^4 =1(mod 10)

27!^37!=1(mod 10)

Option (a) is correct..

Giv ur best to d world.

Nd d best will cm back to u..

Regards,

Dipanjan....

n/a

Q6...whts the unit digit of 2^3^3^4?

Soln.

2^3^3^4%10

=[2^(3^3^4-1)%5]*2

=[2^(..03)%5]*2

=3*2

=6

Giv ur best to d world.

Nd d best will cm back to u...

Regards,

Dipanjan....

n/a

Q7...a no N has odd no. of divisors ,,which is true abt N?

Soln.

N should be a PERFECT SQUARE..

Like 4,9...All have 3 factors.

So option (c) is correct...

Giv ur best to the world.

Nd d best will cm back to u...

Regards,

Dipanjan....

n/a

Q8...how many zeroes will be there at the end of 1003*1001*999*.

Soln.:

As all no.s are odd so no. of zeros at the end is 0.

Giv ur best to d world.

Nd d best will cm back to u...

Regards,

Dipanjan...

n/a

Q9...whts the remainder when 1!+2!+3!.......1000! divided by 5

Soln.:

(1!+2!+3!+..+1000!)%5=(1!+2!+3!+4!)%5=3

Option (c) is correct...

Giv ur best to the world.

Nd d best will cm back to u...

Regards,

Dipanjan...

n/a

Q4...N=7!^3 how many factors

Q4...N=7!^3 how many factors of N are multiples of 10?

a)736 b)1008 c)1352 d)894

ans is 1008

Little Star

n/a

Soln.:

As the person adds a no. twice within N so we can write the actual sum shud be less than 1850.

N(N+1)/2<1850

If N=60 then N(N+1)/2=1830. So he added 20 twice.

Diff :(60-20)=40

Option (a) is correct...

Giv ur best to the world.

Nd d best will cm back to u..

Regards,

Dipanjan....

n/a

n/a