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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$?$
in Numerical Ability by Veteran (54.7k points) | 108 views
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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.
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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?
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Seems correct to me - although I don't understand what D means.
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$(D)$ 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.

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@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
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 Manas Mishra

you are right.

check my answer for $(F)?$

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@lakshman

correct
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  • Product of factors of N = N No. of factors/2
  • Number of factors of N = (a+1)(b+1)(c+1)
  • Sum of factors: ( p0+p1+...+pa) ( q0+ q1+....+qb) (r0+r1+...+rc)/ (pa-1)(qb-1)(rc-1)
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$N=p^{a}\times q^{b}\times r^{c}....$ where $p,q,r$ are prime number

Sum of factors $S_{n}=\frac{(p^{a+1}-1).(q^{b+1}-1).(r^{c+1}-1)}{(p-1).(q-1).(r-1)}$

please correct me if i'm wrong?
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$1)$ will be $(2+1)(2+1)(2+1)=27$
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Yes ma'am

I write all the answers in the above comment, you can see and verify it.

1 Answer

+1 vote
Best answer

In 900 

900 = 22×32×52.

(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 52 then only it's divisible by 25 right ??

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

(E) Sum of factors?

= (2 0 + 21 + 22)  x (3 0 + 31 + 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

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