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Last updated date: 13th Jun 2024
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Hint: In this question, we need to find the factors of 150. The factors of 150 are the numbers that divide 150. Multiplying a pair of factors produces the result to 150. So, we will find such pairs of 150 which when multiplied, produce the result as 150. These factor pairs will be whole numbers only and not negative, fraction or decimal numbers. We will find such pairs by starting from 1 and dividing by 150. Quotient and divisor will both be factors of 150. We will stop when factors will start repeating.

Complete step-by-step answer:
Here we are given numbers as 150 and we need to find its factors.
First of all, 1 and 150 are definitely the factors of 150 because both divide 150 and $1\times 150=150$ (Product of factor pair is the number).
Now let us divide 150 by 2. We get $\dfrac{150}{2}=75$. Since 2 divides 150, so 2 is the factor of 150. Also 75 is the quotient so 75 is also the factor of 150.
Now let us divide 150 by 3. We get $\dfrac{150}{3}=50$. Since 3 divides 150, so 3 is the factor of 150. Also 50 is the quotient, so 50 is also the factor of 150.
Now divide the number by 4, we get $\dfrac{150}{4}=37.5$. Since 37.5 is not the whole number hence 4 is not a factor of 150.
Now dividing the number by 5, we get $\dfrac{150}{5}=30$. Since 5 divides 150, so 5 is a factor of 150. Also, 30 is the quotient, so 30 is also a factor of 150.
Now dividing the number by 6, we get $\dfrac{150}{6}=25$. So 6 and 25 are also the factors of 150.
Dividing by 7, we get $\dfrac{150}{7}=21.4$. Hence, 7 is not a factor of 150.
Since the number does not get divided by 4 so it will not get divided by 8 also.
Dividing by 9, we get $\dfrac{150}{9}=16.67$ so 9 is not a factor of 150.
Dividing by 10, we get $\dfrac{150}{10}=15$ so 10 and 15 are factors of 150.
As we can see, there will be no more factors because 150 will not divide 11 $\left( \dfrac{150}{11}=13.6 \right)$, 12 (not divisible by 4 or 8), 13 $\left( \dfrac{150}{13}=11.538 \right)$ and 14 $\left( \dfrac{150}{14}=10.71 \right)$. So, 15 factors will repeat.
Hence, obtained factors are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Note: To simplify calculations, students can first check divisibility using the divisibility rule. Some of the divisibility rules are:
For 2: at one place there should be 0, 2, 4, 6 or 8.
For 3 and 9: sum of digits should be divisible by 3 or 9 respectively.
For 5: at one place, there should be 0 or 5.
For 6: number should be divisible by 2 and 3 both.
For 11: sum of difference of digits at odd places and even places should be either zero or divisible by 11.