Math, asked by 8468, 9 months ago

The sum of two numbers is 156 and their HCF is 13. The numbers
of such number pairs is

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Answers

Answered by parmarthsharma
0

The required pairs are: i) 13 and 143 or ii) 65 and 91

Step-by-step explanation:

Consider the provided information.

HCF of the two numbers are 13.

Lets say two numbers are: 13A and 13B

Where A and B are co-prime numbers

Therefore, HCF of A and B = 1

The sum of two numbers is 156

13A + 13B = 156

13(A + B) = 156

A + B = 12

Thus, the possible case are:

1 + 11 = 12

2 + 10 = 12 (2 and 10 are not co-prime)

3 + 9 = 12  (3 and 9 are not co-prime)

4 + 8 = 12  (4 and 8 are not co-prime)

5 + 7 = 12

6 + 6 = 12 (6 and 6 are not co-prime)

For A=1 and B=11

13×1=13 and 13×11=143

For A=5 and B=7

13×5=65 and 13×7=91

Hence, the required pairs are: i) 13 and 143 or ii) 65 and 91

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Answered by shashankpaib
0

Answer:

The required pairs are: i) 13 and 143 or ii) 65 and 9

Step-by-step explanation:

Consider the provided information.

HCF of the two numbers are 13.

Lets say two numbers are: 13A and 13B

Where A and B are co-prime numbers

Therefore, HCF of A and B = 1

The sum of two numbers is 156

13A + 13B = 156

13(A + B) = 156

A + B = 12

Thus, the possible case are:

1 + 11 = 12

2 + 10 = 12 (2 and 10 are not co-prime)

3 + 9 = 12  (3 and 9 are not co-prime)

4 + 8 = 12  (4 and 8 are not co-prime)

5 + 7 = 12

6 + 6 = 12 (6 and 6 are not co-prime)

For A=1 and B=11

13×1=13 and 13×11=143

For A=5 and B=7

13×5=65 and 13×7=91

Hence, the required pairs are: i) 13 and 143 or ii) 65 and 91

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