Question

18.The sum of two numbers is 496 and their H.C.F. is 31. The number of pairs of numbers satisfying the above conditions is:​

Answer 1

Step-by-step explanation:

8 is answer please say wrong or right

Answer 2

Answer:

The number of pairs is 4

Step-by-step explanation:

Given sum of two number is 496

the Highest Common Factor of two number = 31

let the 2 numbers are 31a  and 31b and a and b are co-prime numbers  

from above data we can write 31$a$ +31$b$ = 496

                                           ⇒ 31( a+b) =496

                                           ⇒ a + b =  496 /31

                                           ⇒ a + b =16

∴ the two numbers which satisfies the condition  a+b =16  are the required numbers

the pairs will satisfy the above conditions (1, 15 ) (3, 13) ( 5, 11) ( 7, 9)  

number of pairs  4