Question

19
Two number's both greater then 29 have HCF
29 and LCM 4147 The sum of the numbers
is :​

Answer 1

SOLUTION:

Ans :- 696

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know that

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCF

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143Or, xy = 11 × 13, or 13 × 11, . . . . . . ( the only prime factors of 143 )

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143Or, xy = 11 × 13, or 13 × 11, . . . . . . ( the only prime factors of 143 )So, x = 11 or 13 and y = 11 or 13

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143Or, xy = 11 × 13, or 13 × 11, . . . . . . ( the only prime factors of 143 )So, x = 11 or 13 and y = 11 or 13And in either of the condition (x + y) = 24

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143Or, xy = 11 × 13, or 13 × 11, . . . . . . ( the only prime factors of 143 )So, x = 11 or 13 and y = 11 or 13And in either of the condition (x + y) = 24Therefore,

Ans :- 696As the HCF of the two numbers is 29, they are the multiple of 29.Let the numbers are 29x and 29y .We know thatProduct of numbers = Product of LCM and HCFOr, 29x × 29y = 4147 × 29Or, xy = 143Or, xy = 11 × 13, or 13 × 11, . . . . . . ( the only prime factors of 143 )So, x = 11 or 13 and y = 11 or 13And in either of the condition (x + y) = 24Therefore,29x + 29y = 29(x + y) = 29 × 24 = 696

Answer 2

Answer:

The answer is 696

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