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The sum of the prime factors of \[1729\] is:
A) \[13\]
B) \[19\]
C) \[32\]
D) \[39\]

Answer
VerifiedVerified
454.8k+ views
Hint:
In the given question, we have been given a number. We have to find the sum of its prime factors. To do that, we are going to first find the prime factorization of the number using the factorization table. If a number can divide another number, then the first number is called a factor of the second number. And the second number is called a multiple of the first number. A prime number only has two factors – one and the number itself. While a composite number always has more than two factors. Two is the only even prime. So, if we have a number which is even but not two, then it is for sure composite.

Complete step by step solution:
We have to find the sum of prime factors of \[1729\]. To do that, we first find the prime factors of the number using prime factorization.
\[\begin{array}{l}{\rm{ }}7\left| \!{\underline {\,
  {1729} \,}} \right. \\13\left| \!{\underline {\,
  {247} \,}} \right. \\19\left| \!{\underline {\,
  {19} \,}} \right. \\{\rm{ }}1\left| \!{\underline {\,
  1 \,}} \right. \end{array}\]
So, \[1729 = 7 \times 13 \times 19\]
Thus, the sum of the prime factors is \[7 + 13 + 19 = 39\]

Hence, the correct option is D.

Additional Information:
While the number of factors of a number is limited, i.e., at one point, the list of factors ends, or we can say, the list of factors is exhaustive. But the number of multiples of a number is infinite. This is because the counting never ends, and by multiplying any number, we get one number more in the set of multiples.

Note:
In the given question, we had to find the sum of the prime factors of the number. To do that, we first found the prime factorization of the given number. We found the prime factorization of the number using the prime factorization table. We solve it by dividing the number by the smallest prime factor which divides the number. Then we again check if it is divisible by the prime factor. We do that until we reach the only number \[1\], and then we stop.
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