Number System: Basics of Factors Test-4

Factors for 1600 are= 2⁶5² =(6+1)(2+1)=21
So the number of factors which are perfect square , 4x2 =8
How did we decide that? We know that for a factor to be a perfect square,
powers of prime numbers should be even, so number of ways that happens 2⁰2²2⁴2⁶,5⁰,
5² i.e  (3+1)(1+1)=8
So the number of factors in 1600 which are not a perfect square are 21-8=13
Option b is the right answer .

We have number N= 2²3³5⁵7⁷11¹¹13¹³ 
Number of factors are (2+1)(3+1)(5+1)(7+1)(11+1)(13+1)= 96768
Now there are 6 prime numbers (that is the six prime factors) are used and there is also 1,
1 is not a prime nor composite so the number of composite factors are  96768-7=96761

Step 1: For even factors divide the number with 2 thus 420/2=210
Step 2:  210= 2x3x5x7 (if we multiply each factor 210 by 2 it will give the even factor of 420),
so will find the sum of even factors of 210 and then we will multiply the result with 2
Step 3:  {(2²-1)/(2-1)x (3²-1)/(3-1)x (5²-1)/(5-1)x(7²-1)/(7-1)}=576
(Use formula for sum of all factors)
Step 4: 576x2=1152
Hence option D is the right answer

Since this very simple problem if you know the formulae for this
Number of co- primes to the number N and also less than N is given by N (1-1/a)(1-1/b)(1-1/c)
Step 1:  Prime factorisation for 120= 2³
3¹5¹
Step2:    number of co-primes are = 120{(1-1/2)(1-1/3)(1-1/5)}=32.So option a is the right answer

Remember these two formulas:
Number of pairs which are co-prime to each other is given by the formula for a number having two prime factors (p,q,r) :  1+(p+q)+2(pq)
Number of pairs which are co-prime to each other is given by the formula for a number having three prime factors (p,q,r) :  1+(p+q+r)+2(pq+pr+qr)+4pqr
So step1:  900 = 2²3²5²
where p,q,r = 2,2,2, respectively.
Since 900 has 3 prime factors, we use the second formula.
Step 2: put the values in the above expression and get the value 63 , so the right option is c