First Naukri Number System Quiz -1

Solution:

Solution:

We can write 2700 as

2700=2^{2}*3^{3}*5^{2}

So, any factor of 2700 will be

2^{a}*3^{b}*5^{c}

[here 2>=a>=0, 3>=b>=0, 2>=c>=0]

For perfect square a can have 2 values 0/2

For perfect square b can have 2 values 0/2

For perfect square c can have 2 values 0/2

Hence the answer is (2*2*2)=8

Solution:

Any factor of 2^{3} * 5^{2} * 7^{8} can be written as

2^{a}*5^{b}*7^{c}

 

[here 3>=a>=0, 2>=b>=0, 8>=c>=0]

 

So, to become an odd factor a has to be 0

b can take any integer from 0 to 2 [3 values]

c can take any integer from 0 to 8 [9 values]

Hence the answer is 1*3*9=27

Solution:

10^{3}=2^{3}*5^{3}

So, 2^{3} * 5^{2} * 7^{8} * 10^{3}=2^{6} * 5^{5}*7^{8}

Any factor of  2^{6} * 5^{5}*7^{8} is

2^{a} * 5^{b}*7^{c}

[here 6>=a>=0, 5>=b>=0, 8>=c>=0]

For even factors

a can have any integer from 1 to 6 [6 values]

b can have any integer from 0 to 5 [6 values]

c can have any integer from 0 to 8 [9 values]

Hence the answer is = 6*6*9=324

Solution:

If we divide pqrpqr with pqr then we will get 1001

1001=7*11*13

Let pqr has no factor from 7,11,13 .

Then the number of factors in ‘pqrpqr‘ is = 2*2*2*4=32

For, having 4 factors pqr has to be in

x^{3} form where x is an prime number

If x = 7 or 11 or 13 then

‘pqrpqr’ will have = (4+1)*2*2=20 factors

Solution:

For a number p^{a}q^{b}r^{c},  the total number of factors are (1+a)(1+b)(1+c) [here p,q and r are prime numbers] for a perfect cube all of the a,b,c has to be divisible by 3.

So, if a number is a perfect cube and having k factors then k has to be divisible by (1+n) where n is divisible by 3.

As w having 49 factors,

49=(1+6)(1+6)

As 6 is divisible by 3 so w is a perfect cube

X having 7 factors

7=(1+6)

So, x is also a perfect cube

Y having 21 factors now 21 is divisible by

21=(1+6)(1+2)

Here 6 is divisible by 3 but 2 is not divisible by 3

Any combination which proves that 21 is a perfect cube is not possible.

So, Y is not a perfect cube

Z having 81 factors but there is no (1+n) [where n is divisible by 3] which can divide 81.

So, Z is also not a perfect cube

Hence the answer is W and X

Solution:

We know that, For a number p^{a}q^{b}r^{c} ,  the total number of factors are (1+a)(1+b)(1+c) [here p,q and r are prime numbers]

We can represent 30 as,

30=1*30

30=2*15

30=6*5

30=10*3

30=2*3*5

For

1*30 the smallest number is 2^{29}

2*15 the smallest number is 3^{1}*2^{14}=49152

6*5 the smallest number is 3^{4}*2^{5}=2592

10*3 the smallest number is 3^{2}*2^{9}=4608

2*3*5 the smallest number is 5^{1}*3^{2}*2^{4}=720

Solution:

We know that, For a number p^{a}q^{b}r^{c} ,  the total number of factors are (1+a)(1+b)(1+c) [here p,q and r are prime numbers]

As, x^{2} is a perfect square So, there is 2 ways possible for it having 21 factors

21=(1+20) here x^{2}=p^{20}

21=(1+2)(1+6) here x^{2}=p^{2}*q^{6}

For the first case x=p^{10} so, x has (1+10)=11 factors

For the second case x=p*q^{3} so, x has (1+1)(1+3)=8 factors

So, the answer is 8 or 11.

Solution:

The smallest number that is divisible by 3 and 5 is lcm(3,5)=15

The maximum integer below 250 which is divisible by 15 is = 240 [As if we divide 250 with 15 then will get 10 as a reminder so the number is 250-10=240]

So, there are (\frac{(240-15)}{15})+1=16 numbers which are divisible by 3 and 5

The smallest number that is divisible by 3 and 5 and 2 is lcm(2,3,5)=30

The maximum integer below 250 which is divisible by 30 is = 240 [As if we divide 250 with 30 then will get 10 as a reminder so the number is 250-10=240]

So, there are (\frac{(240-30)}{30})+1=8 numbers which are divisible by 2,3 and 5

So, there will be 16-8=8 numbers which are divisible by 3 and 5 but not by 2

Solution:

2^{160}=32^{32}

So, from the question we can say that a=32

Now,

32^{2}+2^{32}

= 2^{10}+2^{32}

= 2^{10}+2^{32}

= 2^{10}(1+2^{22})

So, the largest 2^{b} that can divide 32^{2}+2^{32} is 2^{10} as (1+2^{22}) is not divisible by 2

So, the largest possible b is 10