Question

21 1 If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are ​

Answer 1

$\huge\sf\fbox\purple{ ♡ Solution ♡ }$

★ Explanation ★

It is given that the sum of two numbers is 1215 and their H.C.F is 81.

Let the two numbers be x and y.

Now,

81x+81y=1215     ...[sum of two numbers are 1215]

=>81(x+y)=1215

=>x+y=15

So,

For, x=1,y=14,the numbers are 1×81+14×81=81+1134=1215

For, x=7,y=8,the numbers are 7×81+8×81=567+648=1215

For, x=2,y=13,the numbers are 2×81+13×81=162+1053=1215

For, x=4,y=11,the numbers are 4×81+11×81=324+891=1215

Therefore, the numbers of such pairs are 4.

Answer 2

Given: Sum of two numbers = 1215

           H.C.F of the numbers = 81

To find: Possible pairs of numbers

Let: The numbers be X and Y

Solution: According to the given question,

H.C.F of the numbers = 81

It means both the numbers must be multiples of 81

Hence, numbers must be in the form let X = 81x and Y = 81y

now, sum of  the numbers = 1215

⇒ 81x + 81y = 1215

⇒ 81(x + y) = 1215

i.e., the sum of the numbers must be a multiple of 81

On dividing 1215 by 81, we get 15 which means x + y = 15

The possible pairs for x + y = 15 are:-

1 and 14 as (1 + 14 = 15)

2 and 13 (2 + 13 = 15)

similarly,

3, 12 ; 4, 11 ; 5, 10 ; 6, 9 ; 7, 8.

Possible pairs for numbers X(81x) and Y(81y) are:-

81 x 1 = 81 and 81 x 14 = 1134

81 x 2 = 162 and 81 x 13 = 1053 and so on till x = 7 and y = 8

Hence, the possible number of pairs of such numbers are 7.