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# The number of prime factors in 1955 areA) 5B) 8C) 3D) 2

Last updated date: 16th Jun 2024
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Hint: Here we will be using the concept of prime factorization of numbers, and thereby count them.
Prime factorisation is the method of finding the prime factors of the given number.

We know that in the number system, prime numbers are the numbers which have only two factors, they are unity and the number itself. Hence, it is obvious that they will be divisible by unity and the number itself.

Now, the given number is 1955. So, let us find its divisibility by prime divisors, starting with the smallest prime number, which is 2.
We observe that the number 1955 is an odd number and hence it is not divisible by 2.
Next, we sum up the digits of the given number.
Hence, we get $1+9+5+5=20$.
Thus, we see that the sum of the digits of the given number is not divisible by 3.
So, it is obvious that it is not divisible by 3.

Now, the last digit of the given number 1955 is 5.
Hence, we can conclude that it is divisible by 5.
Thus, dividing by 5, we get
$\dfrac{1955}{5}=391$.
Now, we observe that 391 is perfectly divisible by 17. So, dividing it by 17, we get
$\dfrac{391}{17}=23$
Hence, resolving into factors, we get that 1955 can be expressed as
$1955=5\times 17\times 23$.

So, the total number of prime factors of 1955 is 3.

Note: In these types of questions, it should always be remembered that it is important to know the rules of divisibility to find the prime factors of a given number. For checking divisibility by smaller prime numbers, in most cases, it is not required to actually divide the number by the prime number, but we can apply the rules of divisibility effectively.