Question

If the three-digit number 6 y 8 is divisible by 9, then the value of y is​

Answer 1

Answer:

The divisibility of Nine shows that any number can be exactly divided by Nine , if the sum of the numbers is equal to any of the multiple of Nine.

Given ,

6 y 8 is divisible by 9 .

so , y should be the number which make the whole number ..Any of the multiple of nine .

Therefore y can be ( 4 ) makes sum of the whole number 18 which is the multiple of Nine.

( y = 4 )

Answer 2

Answer:

Required value of y is 4

Explanation:

Given,the three-digit number 6 y 8 is divisible by 9.

Here we want to find value of y.

We know,if sum of all digits of any number is divisible by 9 then the number will be divisible by 9.

Here sum of all digits of 6y8 is

$(6 + y + 8) = 14 + y$

If we are putting y = 4 then we get

$(14 + 4) = 18$

Which is divisible by 9.

So, required value of y is 4.

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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