Equation problem

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

Note by Kislay Raj
5 years, 1 month ago

No vote yet
3 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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

Sort by:

Top Newest

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 - 5 years, 1 month ago

Log in to reply

Good solution!

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

Calvin Lin Staff - 5 years, 1 month ago

Log in to reply

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 - 5 years, 1 month ago

Log in to reply

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 - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...