Leisure time always gives you a fresh way to think about problems & when I was having such, I saw the following problem in Mathematics Olympiad group,thought of an ordinary,easy to grasp way to solve it,without going too deep in Mathematics.
Prove that the expressions 2x+3y and 9x+5y are divisible by 17 for the same integer value set x,y
Let's assume that there is a set of integer x y such that 2x+3y is divisible by 17 for an integer k.We can obtain the 9x+5y in terms of 2x+3y by 9(2x+3y)-17y
Since 17| 2x+3y (17 divides 2x+3y for an integer k) and 17|17y (17 divides 17y) , 17 |(9(2x+3y)-17y) for an integer 9k-y.
9(2x+3y)-17y => 18x+10y=> 2(9x+5y).As we deduced earlier this is divisible by 17 but 2 isn't divisible by 17 hence 9x+5y is divisible by 17.
For a given x & y, which divides 2x+3y by 17 do the same for 9x+5y, BUT is it the same,other-way round? Let's see.
Let's assume that there is a set of integer x y such that 9x+5y is divisible by 17 for an integer t.We can write the 2x+3y in terms of 9x+5y from 2(9x+5y)+17y
Since 17| 9x+5y (17 divides 9x+5y for an integer t) and 17|17y (17 divides 17y) , 17 |(2(9x+5y)+17y) for an integer 2t+y.
2(9x+5y)+17y=> 18x+27y=>9(2x+3y) As we deduced earlier this is divisible by 17 but 9 isn't divisible by 17 hence 2x+3y is divisible by 17.
So the expressions 2x+3y and 9x+5y is divisible by 17 for the same integer value set x,y
Prove that the expressions 2x+3y and 9x+5y are divisible by 17 for the same integer value set x,y
Let's assume that there is a set of integer x y such that 2x+3y is divisible by 17 for an integer k.We can obtain the 9x+5y in terms of 2x+3y by 9(2x+3y)-17y
Since 17| 2x+3y (17 divides 2x+3y for an integer k) and 17|17y (17 divides 17y) , 17 |(9(2x+3y)-17y) for an integer 9k-y.
9(2x+3y)-17y => 18x+10y=> 2(9x+5y).As we deduced earlier this is divisible by 17 but 2 isn't divisible by 17 hence 9x+5y is divisible by 17.
For a given x & y, which divides 2x+3y by 17 do the same for 9x+5y, BUT is it the same,other-way round? Let's see.
Let's assume that there is a set of integer x y such that 9x+5y is divisible by 17 for an integer t.We can write the 2x+3y in terms of 9x+5y from 2(9x+5y)+17y
Since 17| 9x+5y (17 divides 9x+5y for an integer t) and 17|17y (17 divides 17y) , 17 |(2(9x+5y)+17y) for an integer 2t+y.
2(9x+5y)+17y=> 18x+27y=>9(2x+3y) As we deduced earlier this is divisible by 17 but 9 isn't divisible by 17 hence 2x+3y is divisible by 17.
So the expressions 2x+3y and 9x+5y is divisible by 17 for the same integer value set x,y
