The number of even prime factor(s) in $1955$ is/are

5

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0

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3

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2

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Solution

The correct option is B $0$
Factor of $1955 = 5 \times 17 \times 23$
$1, 5, 17, 23$ are not even, so there are no even factors.
Therefore, B is the correct answer.

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Similar questions

1955

State whether the following statements are True or False:

(a) The sum of three odd numbers is even.

(b) The sum of two odd numbers and one even number is even.

(d) If an even number is divided by 2, the quotient is always odd.

(e) All prime numbers are odd.

(f) Prime numbers do not have any factors.

(g) Sum of two prime numbers is always even.

(h) 2 is the only even prime number.

(i) All even numbers are composite numbers.

(j) The product of two even numbers is always even.

The number of distinct prime factors in (216) $^{3 / 5}$ (2500) $^{2 / 5}$ (300) $^{1 / 5}$ is

N = 2^{5} \times 3^{7} \times 9^{2} \times 11^{4} \times 13^{3}

(216)^{\frac{3}{5}} \times (2500)^{\frac{2}{5}} \times (300)^{\frac{1}{5}}

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