Question

Considering all 2-digit natural numbers, how many values of "y" do not satisfy the equation |7x-5y| = 3, given that "x"and "y" are positive integers.

Answer 1

$|7x-5y|=3;7x-5y=\pm 3;$
$\frac{7x-3}{5}=y;\left(A\right)$
$\frac{7x-3}{5}=y;\left(B\right)$
Case A : $\frac{7x-3}{5}=y;\left(A\right)$
$\therefore x,y\to$ 2-digit natural numbers $\Rightarrow 10\le y\le 99$ and $10\le x\le 99$
7x = 498 $\Rightarrow$ x = 71 max, integral value
7x = 53 $\Rightarrow$ x = 8 min, integral value
Thus for $10\le y\le 99\Rightarrow x$ lies between 8 and 71.
But x should lie between 10 and 99,
For that y should be greater than 12
Hence $12\le y\le 99$.
For y = 12; x = 9 $\to$ value not possible.
Since, $7x=5\times y+3\Rightarrow x=\frac{5\times y+3}{7};$
we should check next value of y at 19. [12+7 = 19].
$\therefore$ Values of y will be 19, 26, 33, ...., 96.
$96=19+(n-1)\times 7\Rightarrow n=12\to$ number of values of y that satisfy the equation.

Case B : $\frac{7x-3}{5}=y;\left(B\right)$
$\therefore x,y\to$ 2-digit natural numbers $\Rightarrow 10\le y\le 99;10\le x\le 99$
7x = 492 $\Rightarrow$ x = 70 max, integral value
7x = 47 $\Rightarrow$ x = 7 min, integral value
Thus for $10\le y\le 99\Rightarrow x$ lies between 7 and 70.
But x should lie between 10 and 99,
For that y should be greater than 15
Hence $15\le y\le 99$.
For $y=16;x=11$
Since, $7x=5\times y-3\Rightarrow x=\frac{5\times y-3}{7};$
we should check next value of y at 23. [16+7 = 23].
$x=\frac{5\times 23-3}{7}=16$
$\therefore$ Values of y will be 16,23,30, ...., 93.
$93=16+(n-1)7\Rightarrow n=12\to$ number of values of y that satisfy the equation.
Total number of values = n(case A) + n(case B)=12+12=24
Total number of values of y between 10 and 99,
99=10+(n-1)1$\Rightarrow$ n=90
No. of values of y that satisfy equation = 24
Total no. of values = 90
Hence, no. of values that do not satisfy the equation = 99-24=66.