Equation problem

Find the number of integral solutions of 2x + 3y = 763 such that x and y are both positive....

#Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

We 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 $127$ .

Aditya Parson - 8 years, 1 month ago

Good solution!

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

Calvin Lin Staff - 8 years, 1 month ago

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

Mishti Angel - 8 years, 1 month ago

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.

Vinayak Agarwal - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$