In a more general way, $900 = 2^a \times 3^b \times 5^c$, where $(a,b,c) \in (0,1,2)$.

With this, you can get started and modify the values of $a$, $b$ and $c$ according to the constraints.

0 votes

In $900$ how many

$(A)$ Number of factors(divisors) are possible$?$

$(B)$ Number of Even factors are possible$?$

$(C)$ Number of Odd-factors are possible$?$

$(D)$ If the number is divisible by $25$ then a number of factors are possible$?$

$(E)$ Sum of factors$?$

$(F)$ Product of factors$?$

$(A)$ Number of factors(divisors) are possible$?$

$(B)$ Number of Even factors are possible$?$

$(C)$ Number of Odd-factors are possible$?$

$(D)$ If the number is divisible by $25$ then a number of factors are possible$?$

$(E)$ Sum of factors$?$

$(F)$ Product of factors$?$

0

You can write it as $900 = 2^2\times3^2\times5^2$.

In a more general way, $900 = 2^a \times 3^b \times 5^c$, where $(a,b,c) \in (0,1,2)$.

With this, you can get started and modify the values of $a$, $b$ and $c$ according to the constraints.

In a more general way, $900 = 2^a \times 3^b \times 5^c$, where $(a,b,c) \in (0,1,2)$.

With this, you can get started and modify the values of $a$, $b$ and $c$ according to the constraints.

0

the answer is $(A)27$ $(B)18$ $(C)9$ $(D)9$ $(E)2821$ $(F)(900)^{\frac{27}{2}}$

please correct me if i'm wrong?

please correct me if i'm wrong?

0

$(D)$ If the number is divisible by $25$ then a number of factors are possible?

It means if $900$ divisible by $25$ then the number of factors are possible.

0

@goxul

i think for D as , 900 = 25*36 and it is given that it is divisible by 25 so we have to consider only prime factors of 36 i,e 2^2*3^2 = (2+1)*(2+1) = 9

corrct me if wrong

i think for D as , 900 = 25*36 and it is given that it is divisible by 25 so we have to consider only prime factors of 36 i,e 2^2*3^2 = (2+1)*(2+1) = 9

corrct me if wrong

1

- Product of factors of N = N
^{No. of factors/2} - Number of factors of N = (a+1)(b+1)(c+1)
- Sum of factors: ( p
^{0}+p^{1}+...+p^{a}) ( q^{0}+ q^{1}+....+q^{b}) (r^{0}+r^{1}+...+r^{c})/ (p^{a}-1)(q^{b}-1)(r^{c}-1)

1 vote

Best answer

In 900

900 = 2^{2}×3^{2}×5^{2}.

**( A) Number of factors(divisors) are possible?**

** ** (2+1) * (2+1) * (2+1) = 27

**( B) Number of Even factors are possible?**

you need even divisor means you must included '2' and any no multiply with the 2 given even number

(2)*(2+1)*(2+1) = 18

**( C) Number of Odd-factors are possible?**

"didn't include '2' "

(2+1)*(2+1) = 9

**( D) If the number is divisible by 25 then a number of factors are possible?**

25 = 5^{ 2}

Number must have the multiple of 5^{2} then only it's divisible by 25 right ??

**(2+1)*(2+1)*(1) = 9**

**( E) Sum of factors? **

= (2^{ 0 }+ 2^{1 }+ 2^{2}) x (3 ^{0} + 3^{1} + 3 ^{2}) x (5 ^{0 }+ 5 ^{1 }+ 5 ^{2})

= $(\frac{2^{3}-1}{2-1}) X (\frac{3^{3}-1}{3-1}) X( \frac{5^{3}-1}{5-1})$

**F) Product of factors** = (Number)^{ no of factors /2}

^{ }= (900) ^{27/2}