solve these questions

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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....


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Soln.:As the person adds 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....

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CAT@IIMs's picture
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Q2....what is the unit digit

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....

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CAT@IIMs's picture
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Q3...the

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......

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CAT@IIMs's picture
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Q4...N=7!^3 how many factors

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....

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CAT@IIMs's picture
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Q5....find the unit digit of

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....

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Q6...whts the unit digit of

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....

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CAT@IIMs's picture
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Q7...a no N has odd no. of

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....

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CAT@IIMs's picture
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Q8...how many zeroes will be

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...

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Q9...whts the remainder when

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...

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jigar_er_civil's picture
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Q4...N=7!^3 how many

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

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