Question:
The number of pairs of integers \((x , y)\) satisfying \(x|\geq y\geq-20\) and \(2x+5y=99\) is
The number of pairs of integers \((x , y)\) satisfying \(x|\geq y\geq-20\) and \(2x+5y=99\) is
Updated On: Apr 29, 2024
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Approach Solution - 1
\(2x+5y=99\)
when \(y=-19,x=97\);
since \(x≥y;\) the maximum value of y is \(13\) and corresponding value of \(x\) is \(17\).
We know that the solutions of \(y\) are in arithmetic progression with common difference \(2\).
Here \(a=-19\), \(d=2,t_n=13\)
\(t_n=a+(n-1)d\)
\(-19+(n-1)(2)=13\)
\((n-1)2=32 ⇒ n=17\)
when \(y=-19,x=97\);
since \(x≥y;\) the maximum value of y is \(13\) and corresponding value of \(x\) is \(17\).
We know that the solutions of \(y\) are in arithmetic progression with common difference \(2\).
Here \(a=-19\), \(d=2,t_n=13\)
\(t_n=a+(n-1)d\)
\(-19+(n-1)(2)=13\)
\((n-1)2=32 ⇒ n=17\)
Hence number of pairs integers is \(17\)
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Approach Solution -2
\(2x+5y=99\)
For x to be an integer, u must be odd.
With every change of 2 in y, there will be a change of 5 in y.
For example
When y=11, x=22.
When y=13, x=17
So y=15 will result in lower value of x=17-5 =12
Thus, Answer is all odd values between -20 and 14.
So, \(\frac{14−(−20)}{2}=17\)
For x to be an integer, u must be odd.
With every change of 2 in y, there will be a change of 5 in y.
For example
When y=11, x=22.
When y=13, x=17
So y=15 will result in lower value of x=17-5 =12
Thus, Answer is all odd values between -20 and 14.
So, \(\frac{14−(−20)}{2}=17\)
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