Aishwarya Rai Bank Problem!!

Let the amount which should've been given by the bank be $\$\overline{x.y}$ . Instead, she is given $\$\overline{y.x}$ . Expressing the money in cents and setting up an equation using the given data, we have,

$100y+x-5=2(100x+y)\implies 199x=98y-5\implies y=\dfrac{5+199x}{98}$

Since the amount of cent must be a non-negative integer, we must have,

$98\mid (5+199x)\\ \implies 5+199x\equiv 0\pmod{98}\\ \implies 5+3x\equiv 0\pmod{98}\\ \implies 3x\equiv (-5)\equiv 93\pmod{98}\implies x\equiv 31\pmod{98}$

So, the solution for $x$ is given by $x=31+98w~,~w\in \mathbb{W}$ . We can note that $x$ is the dollar amount. We must take $x$ such that $0\leq y\leq 99~,~y\in \mathbb{W}$ .

From observation, we can conclude that $w\geq 1\implies y\gt 99$ . So, we must have $w=0$ and this gives us $x=31$ and $y=63$ .

$\therefore$ Her cheque amount $=\$\overline{x.y}=\$31.63$

et x is the dollar amount And let y is change amount Then the check value equals : =x+y/100

By applying interchange and decrease the change by 5 cents then the equation would be

2(x+y/100)=y+(x-5)/100

So x =(98y-5)/199

As x and y are integers then by apply the following c++ code

// start of program

int main()

{

int x=0;

for(int y=0;y>=100;y++) {

x=(98*y-5)%199;

//no reminder so x is integer

if(x==0) {

x=(98*y-5)/199;

cout<<"the check amount is"<<x<<"."<<y<<endl;

}

}

return 0;

}

//the output is

The check amount is 31.63

Solution: The check's value was $31.63

variables: d = check's dollar amount c = check's change (cents) amount

So, check's value is 100d + c cents.

Because of the bank's mistake, of interchanging the check's dollar's and cent's amounts the bank gives her 100c + d cents.

She spends 5 cents, leaving her 100c + d - 5 cents.

She now has exactly twice the value of the original check.

So, 100c + d - 5 = 2( 100d + c )

98c = 199d + 5 Note: We must now use the facts that, d and c are integers

c = (199d + 5) / 98

c = 2d + 3d/98 + 5/98

c = 2d + (3d + 5)/98 ... (1) Since c and 2d are integers, (3d + 5)/98 must be an integer, say k.

integer k = (3d + 5)/98 ... (1b) Note: equation (1) becomes: c = 2d + k ... (1c)

from eqn. (1b), d = (98k - 5) / 3 d = 33k - 2 - (k - 1)/3 ... (2)

Since d, 33k and 2 are integers, (k-1)/3 must be an integer, say m.

integer m = (k-1)/3 k = 3m + 1

meaning any k of the form 3m + 1 will yield an integer, d, in (2).

Substituting k in eqn (2), we get, d = 98m + 31 ... (3a)

Substituting d in eqn (1c), we get, c = 2( 98m+31) + 3m+1 c = 199m + 63 ... (3b)

Setting m=0 in eqns (3a) and (3b) yields c=63 and d=31which are legal values for the check

since c and d are non-negative integers with c<100.

The check's value was $31.63