Number System: Basics of Factors Test-1

Step 1: Prime factorisation, so N=825= 3¹ 5² 11¹
Power of 3 as 3⁰  , 3¹ ( 1+1=2)ways ,
Power of 5 as 5⁰ , 5¹ ,5²  ( 2+1=3)ways
Power of 11 as 11⁰, 11¹ (1+1=2)ways
Step 2: Hence, the number of factors is (1+1)(2+1)(1+1)=2x3x2=12

Step 1: Prime factorisation, so N=3²5³7³11²13²17 ¹
Power of 3 as 3⁰  , 3¹ ,3²
Power of 5 as 5⁰ , 5¹ ,5²,5³
Power of 7 as 7⁰,7¹,7²,7³
Power of 11as 11⁰11¹11²
Power of 13 as 13⁰13¹13²
Power of 17 as 17⁰17¹
Step 2: Hence, the number of factors is(2+1)(3+1)(3+1)(2+1)(2+1)(1+1)=864

In this case we have to find number of even factors,an even factor is divisible by 2 or smallest power of 2 has to be 1 not 0.
Hence a factor must be 2^{(1 or 2 or 3 or 4)}3^{(0 or 1,2,3,4,5)}7^{(0 or 1 or 2)}
Hence total number of factors= (4)(5 + 1 )(2 + 1) = 72

To simply find odd factors, remove 2 from prime factors i.e from N remove ${{2}^{4}}$
Then we are left with $N={{3}^{5}}{{7}^{2}}$ Since these are two odd factors, they will have no even factors.
Therefore, the total number of factors is= (5+1)(2+1)=18

If a number is divisible by 10 then it must have minimum power of 2 and 5 as 1.
A factor divisible by 10 is 2^{(1 or 2 or 3or 4 or 5)}5^{(1 or 2 )}7^{(0 or 1 or 2 )}
Hence, total number of factors divisible by 10 is = (5)(2)(2 + 1) = 30