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Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?
- a)2
- b)5
- c)3
- d)6
- e)4
Correct answer is option 'E'. Can you explain this answer?
Verified Answer
Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs ...
Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
Most Upvoted Answer
Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs ...
To find the number of pairs of numbers that satisfy the given conditions, we need to consider the factors of 128 and find the pairs that have a highest common factor (HCF) of 8.
Factors of 128:
The factors of 128 are 1, 2, 4, 8, 16, 32, 64, and 128.
Pairs with a HCF of 8:
To find the pairs with a HCF of 8, we need to consider the pairs of factors that have 8 as their highest common factor.
Pairs with a HCF of 8 are:
(8, 16) - HCF = 8
(8, 32) - HCF = 8
(8, 64) - HCF = 8
(16, 32) - HCF = 8
(16, 64) - HCF = 8
(32, 64) - HCF = 8
There are 6 pairs of numbers that satisfy the given conditions, so the correct answer is option E) 6.
Factors of 128:
The factors of 128 are 1, 2, 4, 8, 16, 32, 64, and 128.
Pairs with a HCF of 8:
To find the pairs with a HCF of 8, we need to consider the pairs of factors that have 8 as their highest common factor.
Pairs with a HCF of 8 are:
(8, 16) - HCF = 8
(8, 32) - HCF = 8
(8, 64) - HCF = 8
(16, 32) - HCF = 8
(16, 64) - HCF = 8
(32, 64) - HCF = 8
There are 6 pairs of numbers that satisfy the given conditions, so the correct answer is option E) 6.
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Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs ...
Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
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Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?a)2b)5c)3d)6e)4Correct answer is option 'E'. Can you explain this answer?
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