Question

A two digit number is obtained by either multiplving the sum of the digits by 8 and then
subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find
the number How many such numbers are there?

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Answer 1

Answer:

Solution :-

Let the two digit number be 10x + y where x is the tens digit and y is the ones digit.

Now, according to the question.

10x + y = 8(x + y) - 5

10x + y = 8x + 8y - 5

10x - 8x + y - 8y = - 5

2x - 7y = - 5  .................(1)

And,

10x + y = 16(x - y) + 3

10x + y = 16x - 16y + 3

10x - 16x + y + 16y = 3

- 6x + 17y = 3  ................(2)

Now, multiplying the equation (1) by 17 and (2) by 7, we get

34x - 119y = - 85 ...............(3)

- 42x + 119y = 21 ..............(4)

Now, adding (3) and (4), we get

  34x - 119y = - 85

- 42x + 119y =   21

_________________

 - 8x             = - 64

_________________

⇒ 8x = 64

x = 64/8

x = 8 

So, tens digit is 8.  

Substituting the value of x = 8 in (1), we get

2x - 7y = - 5

2*8 - 7y = - 5

16 - 7y = - 5

- 7y = - 5 - 16

- 7y = - 21

7y = 21

y = 21/7

y = 3

Ones digit is 3.

So, the required number is 83.

Answer.

Answer 2

Answer:

after solving both the equation we get, x equal to 8

and y equal to 3

thus the required no. is 83