Math, asked by aditynitu2018, 2 months ago

Find the number of divisors of 42336 excluding 1 and the number itself. Find the sum of these divisors.

Answers

Answered by santhoshsm2008
0

Answer:

1678

Step-by-step explanation:

Given number is 38808

It can be expressed as

(2)

3

×(3)

2

×(7)

2

×(11)

⇒ Total number of factors

=(4+1)(2+1)(3+1)(3+1)(6+1)−2

=5×3×4×4×7−2

=1880−2

On subtracting we get 1678

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Answered by mathdude500
1

\large\underline{\sf{Solution-}}

Given number is 42336

Let first find the prime factorization of 42336.

\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:42336 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:21168 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:10584\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:5292 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:2646 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:1323 \:\:}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:441 \:\:}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:147 \:\:}} \\ {\underline{\sf{7}}}& \underline{\sf{\:\:49 \:\:}} \\ {\underline{\sf{7}}}& \underline{\sf{\:\:7\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}

Thus, Prime factorization of

\rm :\longmapsto\:42336 =  {2}^{5} \times  {3}^{3} \times  {7}^{2}

We know,

If any natural number 'n' is represented as its prime factorization as

\rm :\longmapsto\:n =  {x}^{y}

\rm :\implies\:n \: is \: divisible \: by \: 1,x, {x}^{2}, {x}^{3}, -  -  -  - , {x}^{y}

\rm :\implies\:n \: has \: y + 1 \: divisors

So,

Here ,

\rm :\longmapsto\:42336 =  {2}^{5} \times  {3}^{3} \times  {7}^{2}

It implies number of divisors of 42236 is given by

\rm \:  =  \:  \:(5 + 1)(3 + 1)(2 + 1)

\rm \:  =  \:  \:6 \times 4 \times 3

\rm \:  =  \:  72

And to find the number of divisors exclusive of 1 and number itself,

Then,

Number of divisors = 72 - 2 = 70.

Now, To find the sum of divisors.

Again, Prime factorization of 42336 is

\rm :\longmapsto\:42336 =  {2}^{5} \times  {3}^{3} \times  {7}^{2}

So, sum of divisors is

\rm \:  =  \:  \:\dfrac{ {2}^{5 + 1}  - 1}{2 - 1} \times \dfrac{ {3}^{3 + 1}  - 1}{3 - 1} \times \dfrac{ {7}^{2 + 1}  - 1}{7 - 1}

\rm \:  =  \:  \:\dfrac{ {2}^{6}  - 1}{1} \times \dfrac{ {3}^{4}  - 1}{2} \times \dfrac{ {7}^{3}  - 1}{6}

\rm \:  =  \:  \:(64 - 1) \times \dfrac{81 - 1}{2}  \times \dfrac{343 - 1}{6}

\rm \:  =  \:  \:63 \times \dfrac{80}{2}  \times \dfrac{342}{6}

\rm \:  =  \:  \:63 \times 40  \times 57

\rm \:  =  \:  \:143640

So,

Sum of proper divisors = 143640 - 1 - 42336 = 101303

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