Since \(x\) and \(y\) are positive integers\(3y\) should not exceed \(763\).
Now we see that \(x\) can only be positive when \(3y\) is odd.
Also, since \(y\) is positive it can only take values up to \(\frac{762}{3}=254\) (\(3y\) will never be 763 due to the positive constraint).

Now since \(y\) has to be odd there are \(127 \) odd numbers(between 1 and 254), so no of possible positive integer solution is \(127\).

we have 2x + 3y = 763
divide throughout by 2, the smaller coefficient; thus
x + y + y/2 = 381 + 1/2
=> x + y + (y-1)/2 = 381 .................................1.)
since x and y are to be integers , therefore (y-1)/2 = integer
we multiply (y-1)/2 by 5 in order to make the coefficient of y differ by unity from a multiple of 2.

=> (5y-5)/2 = integer
=> 2y-2 + (y-1)/2 = integer
and therefore (y-1)/2 = integer =p (suppose)
therefore y-1 =2p
=> y= 2p+1..................................................................2.)
substitute this value of y in the first equation
=> x + 3p + 1 = 381
=> x =380 -3p................................................................3.)
but if p>126 , we get negative value of x
since we need only positive values of x and y, hence p can take values between 0 and 126 only

hence corresponding values of x and y we will get....
so to get positive integral values for x and y , p= 0,1,2,3, ........126
therefore we will have 127 solutions altogether

y=(763-2x)/3
Minimum value of x = 2 and maximum value of x=380 such that it makes the above equation divisible by 3.
x=2,5,8........,380. i.e. it forms an A.P.
So,
380= 2+(n-1)3
378/3=n-1
127=n.

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## Comments

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TopNewestWe can rewrite \(x=\frac{763-3y}{2}\).

Since \(x\) and \(y\) are positive integers\(3y\) should not exceed \(763\). Now we see that \(x\) can only be positive when \(3y\) is odd. Also, since \(y\) is positive it can only take values up to \(\frac{762}{3}=254\) (\(3y\) will never be 763 due to the positive constraint).

Now since \(y\) has to be odd there are \(127 \) odd numbers(between 1 and 254), so no of possible positive integer solution is \(127\).

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Good solution!

For more information on the general case, you should read Modulo Arithmetic, esp Worked Example 4.

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we have 2x + 3y = 763 divide throughout by 2, the smaller coefficient; thus x + y + y/2 = 381 + 1/2 => x + y + (y-1)/2 = 381 .................................1.) since x and y are to be integers , therefore (y-1)/2 = integer we multiply (y-1)/2 by 5 in order to make the coefficient of y differ by unity from a multiple of 2.

=> (5y-5)/2 = integer => 2y-2 + (y-1)/2 = integer and therefore (y-1)/2 = integer =p (suppose) therefore y-1 =2p => y= 2p+1..................................................................2.) substitute this value of y in the first equation => x + 3p + 1 = 381 => x =380 -3p................................................................3.) but if p>126 , we get negative value of x since we need only positive values of x and y, hence p can take values between 0 and 126 only

hence corresponding values of x and y we will get.... so to get positive integral values for x and y , p= 0,1,2,3, ........126 therefore we will have 127 solutions altogether

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y=(763-2x)/3 Minimum value of x = 2 and maximum value of x=380 such that it makes the above equation divisible by 3. x=2,5,8........,380. i.e. it forms an A.P. So, 380= 2+(n-1)3 378/3=n-1 127=n.

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