Difference between the number of even and prime divisors....

Q1) What is the difference between the numbers of even and prime divisor of 1020?

Q2)How many zeroes will be at the end of 7x14x21x.....x777 ?

Q3)The number 1 to 33 are wriiten ,what will be the remainder when it is divided by 9?

Q4)The number 4444...(999 times) will be definitely divided by?
a) 22   b) 44   c) 222   d) All of these.

Q5)How many divisor of N=420 will be in the form (4n+1)?

Q6)What is the unit digit of 2^(3)^(4)^(5)?

Q7)How many zeroes will be at the end of 1003x1001x999x....x123?Minimum value of HCF of [n^2+17] and[(n+1)^2+17]?

Q8)What is the sum of the least multiple of 13 which when didvided by 6,8,12 leaves 5,7 and 11 as the remainder?

Q9)What is the unit digit of 7^(11)^(22)^(33)?

Q10)What is the remainder when fact1+fact2+fact3+....+fact1000 will be divided by 5?

Q11)How many zero will be at the end of fact18 + fact 19?

Q12) How many integers N in the set of the integers {1,2,3,...,100} are such that N^2+N^3 is a perfect square?

Q13)How many prime numbers are there which when divided by another gives a quotient same as remaider?

Submitted by shivanibusiness on Fri, 2008-05-16 10:50.

CAT@IIMs's picture

CAT@IIMs points: 259
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Comments : 117

Soln.:1020=2^2*5*3*17
        No. of even divisors=24-8=16
        No. of prime divisors=4
        Difference=12......

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Nd d best will cm back to u....

Regards,
Dipanjan.......

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.7(1*2*3*...*111)=7*111!
       111! has [111/5]+[111/25]=22+4=26 no. of zeros at d end....
So 7X14X21X...X777 has 26 no. of zeros at d end....

Giv ur best to d world.
Nd d best will cm back to u.....

Regards,
Dipanjan...... 

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:123..33=123...27+2829...33=1234...27+2829303+1323+3
         First three parts are fully divisible by 9.So remainder is 3.

Giv ur best to d world.
Nd d best will cm back to u....

Regards,
Dipanjan.....

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:As 999 is divisible by 3.And 222 is a 3 digits no.
So option (c) is correct...

Giv ur best to d world.
Nd d best will cm back to u....

Regards,
Dipanjan.....

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:2^3^4^5=2^3^1024=2^(...1)
        As Cyclicity of 2 is (4n+1) & (..1) is always in 4n+1 format.
        2^(..1) gives unit digit 2.

Giv ur best to d world.
Nd d best will cm back to u...

Regards,
Dipanjan.....

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.As all are odd no.s so dr mulitiplication wud give odd no...
       Dr will b NO ZEROs at d end of dt no.

Giv ur best to d world.
Nd d best will cm back to u.....

Regards,
Dipanjan.....


CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:As 11^22^33=(.....1) can be xpressed in (4n+1) format...
        We know d cyclicity of 7 is (4n+1)....
So it gives 7 as unit digit....

Giv ur best to d world.
Nd d best will cm back to u....

Regards,
Dipanjan.....

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:All ongoing no. after 5! will b divisible by 5....
        R.O. 1!+2!+3!+4!+5!+..+1000!=1!+2!+3!+4!(mod 5)=33(mod 5)=3
So 3 is d remainder...

Giv ur best to d world.
Nd d best will cm back to u.....

Regards,
Dipanjan......

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.:As 18! has 3 zeros at d end of it.So 18!+19! has 3 zeros at d end...

Giv ur best to world.
Nd d best will cm back to u....

Regards,
Dipanajan......

CAT@IIMs's picture

CAT@IIMs points: 259
Location:
Comments : 117

Soln.: D no.s are 1,5,21,105.
         So 4no.s are possible...

Giv ur best to d world.
Nd d best will cm back to u....

Regards,
Dipanjan.....