Question

Find the number of factors 14,98,176 and also find their sum.

Answer 1

Hint: We solve this problem by using the prime factorization method.
The prime factorization method is a method of representing the given number as the product of prime numbers. We represent the given number 1498176 as a product of the prime number.
We have the formula that is if a number represented in prime factorisation as
\[N={{x}^{a}}\times {{y}^{b}}\times {{z}^{c}}\]
Where \[x,y,z\] are prime numbers, then
\[\text{number of factors}=\left( a+1 \right)\left( b+1 \right)\left( c+1 \right)\]
\[\text{Sum of factors}=\left( \dfrac{{{x}^{a+1}}-1}{x-1} \right)\left( \dfrac{{{y}^{b+1}}-1}{y-1} \right)\left( \dfrac{{{z}^{c+1}}-1}{z-1} \right)\]

Complete step by step answer:
We are given that the number 1498176
Let us assume that the given number as
\[\Rightarrow N=1498176\]
Now, let us use the prime factorization method.
We know that the prime factorization method is a method of representing the given number as the product of prime numbers
Now, let us use the first prime number 2 that is dividing the given number with 2 then we get
\[\Rightarrow N=2\times 749088\]
Here we can see that the number 749088 can be further reduced.
Now, by dividing the number 749088 with 2 we get
\[\Rightarrow N={{2}^{2}}\times 374544\]
Now, by dividing the number 374544 by 2 in above equation then we get
\[\Rightarrow N={{2}^{3}}\times 187272\]
We know that all the even numbers are divided by 2
Now, let us use the prime factorisation with 2 until we get odd number in above equation then we get
\[\begin{align}
  & \Rightarrow N={{2}^{4}}\times 93636 \\
 & \Rightarrow N={{2}^{5}}\times 46818 \\
 & \Rightarrow N={{2}^{6}}\times 23409 \\
\end{align}\]
Here, we can see that we got a odd number that is 23409.
So, let us use the next prime number that is 3
Now, by dividing the number 23409 with 3 we get
\[\Rightarrow N={{2}^{6}}\times 3\times 7803\]
Now, again by dividing the number 7803 by 3 we get
\[\Rightarrow N={{2}^{6}}\times {{3}^{2}}\times 2601\]
We know that if the sum of digits is divisible by 3 then the whole number is divisible by 3
By using the above rule let us use the prime factorisation method until we get a number that is not divisible by 3 then we get
\[\begin{align}
  & \Rightarrow N={{2}^{6}}\times {{3}^{3}}\times 867 \\
 & \Rightarrow N={{2}^{6}}\times {{3}^{4}}\times 289 \\
\end{align}\]
Here, we can see that the number 289 is not divisible by 3 and can be represented as square of 17 that is
\[289={{17}^{2}}\]
By using this result in above equation we get
\[\Rightarrow N={{2}^{6}}\times {{3}^{4}}\times {{17}^{2}}\]
We know that the formula that is if a number represented in prime factorisation as
\[N={{x}^{a}}\times {{y}^{b}}\times {{z}^{c}}\]
Where \[x,y,z\] are prime numbers, then
\[\text{number of factors}=\left( a+1 \right)\left( b+1 \right)\left( c+1 \right)\]
Let us assume that the number of factors of 1498176 as \[n\]
By using the above formula to given number then we get the number of factors as
\[\begin{align}
  & \Rightarrow n=\left( 6+1 \right)\left( 4+1 \right)\left( 2+1 \right) \\
 & \Rightarrow n=7\times 5\times 3 \\
 & \Rightarrow n=105 \\
\end{align}\]
We also know that the formula that is if a number represented in prime factorisation as
\[N={{x}^{a}}\times {{y}^{b}}\times {{z}^{c}}\]
Where \[x,y,z\] are prime numbers, then
\[\text{Sum of factors}=\left( \dfrac{{{x}^{a+1}}-1}{x-1} \right)\left( \dfrac{{{y}^{b+1}}-1}{y-1} \right)\left( \dfrac{{{z}^{c+1}}-1}{z-1} \right)\]
Now, let us assume that the sum of factors of 1498176 as \[S\]
By using the above formula to given number then we get the sum of factors as
\[\begin{align}
  & \Rightarrow S=\left( \dfrac{{{2}^{7}}-1}{2-1} \right)\left( \dfrac{{{3}^{5}}-1}{3-1} \right)\left( \dfrac{{{17}^{3}}-1}{17-1} \right) \\
 & \Rightarrow S=\dfrac{127}{1}\times \dfrac{242}{2}\times \dfrac{4912}{16} \\
 & \Rightarrow S=4026154 \\
\end{align}\]
Therefore we can conclude that the number of factors of 1498176 is 105 and the sum of factors of 1498176 is 4026154.

Note:
 Students may do mistakes in the number of factors formula.
We have the formula that is if a number represented in prime factorisation as
\[N={{x}^{a}}\times {{y}^{b}}\times {{z}^{c}}\]
Where \[x,y,z\] are prime numbers, then
\[\text{number of factors}=\left( a+1 \right)\left( b+1 \right)\left( c+1 \right)\]
Here the factors 1 and itself are also included and gives the total number of factors.
Nut some students may misunderstand and assume that factors 1 and itself are not included and add 2 after finding the number of factors using the formula. But there is no need to add 2 to the number of factors that we get from the formula.