The "length" of a positive integer greater than 1

[BTGmoderatorDC]

The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because 2×2×5=202×2×5=20. What is the length of 5,950?

A. 1
B. 2
C. 3
D. 4
E. 5

OA E

Source: Manhattan Prep

[Ash Mo]

The prime factorization of the given number is
5950 = 5*5*2*17*7

Since there are 5 prime factors( including the repeating ones) ,
the length of the number is 5.

[fskilnik@GMATH]

The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because 2×2×5=20. What is the length of 5,950?

A. 1
B. 2
C. 3
D. 4
E. 5
Source: Manhattan Prep

$$?\,\,\, = \,\,\,\# \,\,{\rm{prime}}\,\,{\rm{factors}}\,\,\,\left( {{\rm{not}}\,\,{\rm{necessarily}}\,\,{\rm{distincts}}} \right)$$
$$5,950 = 10 \cdot 595\,\,\mathop = \limits^{\left( * \right)} \,\,10 \cdot 5 \cdot \underline {119} \,\,\mathop = \limits^{\left( {**} \right)} \,\,2 \cdot {5^2} \cdot \underline {7 \cdot 17} $$
$$\left( * \right)\,\,{{550 + 45} \over 5} = 110 + 9\,\,\,\,\, \Rightarrow \,\,\,\,595 = 5 \cdot 119$$
$$\left( {**} \right)\,{{70 + 49} \over 7} = 10 + 7\,\,\,\,\, \Rightarrow \,\,\,\,119 = 7 \cdot 17$$
$$?\,\,:\,\,\,2,5,5,7,17\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 5$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

[Scott@TargetTestPrep]

The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because 2×2×5=202×2×5=20. What is the length of 5,950?

A. 1
B. 2
C. 3
D. 4
E. 5

5,950 = 595 x 10 = 119 x 5 x 2 x 5 = 7 x 17 x 5 x 2 x 5

Since 5,950 has 5 prime factors (including repeats), the length of 5,950 is 5.

Answer: E