
1. Find HCF of $ 378,180 $ and $ 420 $ by prime factorization method. Is HCF $ \times $ LCM of three numbers equal to the product of the three numbers?
2. Find the largest positive integer that will divide $ 122,150 $ and $ 115 $ leaving remainder $ 5,7,11 $ respectively.
Answer
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Hint: Find the prime factors of all the numbers separately. After that find the HCF of the numbers then the LCM of the product of the numbers. Then equate the two answers so that you will get whether they are equal or not in this case they aren’t equal.
For the second question remove the remainder from the respective numbers. After that find the HCF of that number. Then you will get the desired answer.
Complete step by step solution:
I. Let us find factors of following numbers:
Prime factors of $ 378 = 2 \times 3 \times 3 \times 3 \times 7 $
Prime factors of $ 180 = 2 \times 2 \times 3 \times 3 \times 5 $
Prime factors of $ 420 = 2 \times 2 \times 3 \times 5 \times 7 $
So, HCF of 378, 180 and 420 is 2 × 3 = 6
And LCM of $ 378,180 $ and $ 420 $ is $ 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 7 = 3780 $
So, let us check whether LCM ( $ 378,180 $ and $ 420 $ ) $ \times $ HCF ( $ 378,180 $ and $ 420 $ ) is equal to the product of the three numbers
$
3780 \times 6 = 378\; \times 180 \times 420 \\
22680 \ne 28576800 \;
$
Hence LCM $ \times $ HCF is not equal to the product of the three numbers.
II.Find the largest positive integer that will divide $ 122,150 $ and $ 115 $ leaving remainder $ 5,7,11 $ respectively.
As the largest positive integer that will divide $ 122,150 $ and $ 115 $ leaving remainder 5, 7 and 11 respectively.
Therefore,
$
122 - 5 = 117 \\
150 - 7 = 143 \\
115 - 11 = 104 \;
$
∵ The HCF of $ 104,117 $ and $ 143 $ is $ 13 $
Hence, the largest number which divides $ 122,150 $ and $ 115 $ leaving remainder 5, 7 and 11 as remainders respectively is 13.
Note: When we come across problems like these, we have to first go with the brief reading of the problem and then the understanding of the question then thinking of the correct solution for the given question. The prime factorization plays an important role in getting the factors solved. Without them solving the problem won’t be easy. In this problem we have to be careful while simplifying the HCF and LCM.
For the second question remove the remainder from the respective numbers. After that find the HCF of that number. Then you will get the desired answer.
Complete step by step solution:
I. Let us find factors of following numbers:
Prime factors of $ 378 = 2 \times 3 \times 3 \times 3 \times 7 $
Prime factors of $ 180 = 2 \times 2 \times 3 \times 3 \times 5 $
Prime factors of $ 420 = 2 \times 2 \times 3 \times 5 \times 7 $
So, HCF of 378, 180 and 420 is 2 × 3 = 6
And LCM of $ 378,180 $ and $ 420 $ is $ 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 7 = 3780 $
So, let us check whether LCM ( $ 378,180 $ and $ 420 $ ) $ \times $ HCF ( $ 378,180 $ and $ 420 $ ) is equal to the product of the three numbers
$
3780 \times 6 = 378\; \times 180 \times 420 \\
22680 \ne 28576800 \;
$
Hence LCM $ \times $ HCF is not equal to the product of the three numbers.
II.Find the largest positive integer that will divide $ 122,150 $ and $ 115 $ leaving remainder $ 5,7,11 $ respectively.
As the largest positive integer that will divide $ 122,150 $ and $ 115 $ leaving remainder 5, 7 and 11 respectively.
Therefore,
$
122 - 5 = 117 \\
150 - 7 = 143 \\
115 - 11 = 104 \;
$
∵ The HCF of $ 104,117 $ and $ 143 $ is $ 13 $
Hence, the largest number which divides $ 122,150 $ and $ 115 $ leaving remainder 5, 7 and 11 as remainders respectively is 13.
Note: When we come across problems like these, we have to first go with the brief reading of the problem and then the understanding of the question then thinking of the correct solution for the given question. The prime factorization plays an important role in getting the factors solved. Without them solving the problem won’t be easy. In this problem we have to be careful while simplifying the HCF and LCM.
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