When we are expressing the given number as product of primes factors, we have to be aware the divisibility rules for 2,3, 5.... etc.
Divisibility rule for 2 :
Every even number is divisible by 2. The number which ends with 0, 2 4, 6 or 8 are even numbers and it is divisible by 2.
Divisibility rule for 3 :
If the sum of the number is divisible by 3, then the given number is also divisible by 3.
Divisibility rule for 5 :
If the number ends with 0 or 5, then it is divisible by 5.
To express the given number as product of prime factors, we follow the steps:
Step 1 :
Divide the given number using prime numbers.
Step 2 :
Until we receive prime number, we have to continue the process.
Step 3 :
Write the factors as product.
Step 4 :
If we see the repeated factor, write the repeating factor in exponential form.
Express the following numbers as product of prime factors.
Problem 1 :
52
Solution :
In 52, the unit digit is 2. So, it is even number. Then it is divisible by 2.
52 = 2 x 2 x 13
Here 2 is repeating two times and no more factors which are repeating.
Writing in the exponential form, we get
52 = 2^{2} x 13
Problem 2 :
90
Solution :
The given number 90 ends with 0. Then, it must be the multiple of 2.
90 = 2 x 5 x 3 x 3
3 is repeating two times and no more factors are repeating. Writing in the exponential form, we get
= 2 x 5 x 3^{2}
Problem 3 :
72
Solution :
The given number 72 ends with 2. Then, it must be the multiple of 2.
72 = 2 x 2 x 2 x 3 x 3
The factor 2 is repeating three times and the factor 3 is repeating two times. Then, writing the repeating factors in exponential form.
= 2^{3} x 3^{2}
Problem 4 :
100
Solution :
Since 100 ends with 0, it must be multiple of 2 or 5.
100 = 2 x 2 x 5 x 5
The factor 2 is repeating two times and the factor 5 is repeating two times. Then, writing the repeating factors in exponential form.
100 = 2^{2} x 5^{2}
Problem 5 :
64
Solution :
Since 64 ends with 4, it must be multiple of 2.
64 = 2 x 2 x 2 x 2 x 2 x 2
The factor 2 is repeating six times. Then, writing the repeating factors in exponential form.
64 = 2^{6}
Problem 6 :
105
Solution :
Since 105 ends with 5, it must be divisible by 5.
105 = 5 x 3 x 7
Since there is no repeating factor, so we cannot express it in exponential form.
Problem 7 :
120
Solution :
Since 120 ends with 0, it must be divisible by 2.
120 = 2 x 2 x 2 x 3 x 5
The factor 2 is repeating three times and no more factor is repeating. Then, writing the repeating factors in exponential form.
120 = 2^{3} x 3 x 5
Problem 8 :
85
Solution :
Since 85 ends with 5, it must be divisible by 5.
85 = 5 x 17
Problem 9 :
96
Solution :
Since 96 ends with 6, it must be divisible by 2.
96 = 2 x 2 x 2 x 2 x 2 x 3
The factor 2 is repeating five times and no more factor is repeating. Then, writing the repeating factors in exponential form.
96 = 2^{5} x 3
Problem 10 :
112
Solution :
Since 112 ends with 2, it must be divisible by 2.
112 = 2 x 2 x 2 x 2 x 7
The factor 2 is repeating four times and no more factor is repeating. Then, writing the repeating factors in exponential form.
112 = 2^{4} x 7
Problem 11 :
84
Solution :
Since 84 ends with 4, it must be divisible by 2.
84 = 2 x 2 x 3 x 7
Since 2 is repeating 2 times and no more factors are repeating. We can write it in exponential form.
84 = 2^{2} x 3 x 7
Problem 12 :
108
Solution :
Since 108 ends with 8, it must be divisible by 2.
108 = 2 x 2 x 3 x 3 x 3
Since 2 is repeating 2 times and 3 is repeating three times. So, we can write in the exponential form.
108 = 2^{2} x 3^{3}
Problem 13 :
56
Solution :
Since 56 ends with 6, it must be divisible by 2.
56 = 2 x 2 x 2 x 7
Since 2 is repeating 3 times and no more factor is repeating. So, we can write in the exponential form.
56 = 2^{3} x 7
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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