Money for Nothing

Good reference to the song ! :D

Let $s$ be Satvik's money and $a$ be Agnishom's money, where $a,s\in \mathbb{N}$ . From the things that Agnishom says we can set up these two equations:

$s-x=a+x\\ 2(a-x)=s+x$

Deleting $x$ we have $7a=5s$ .

Clearly $a$ is a multiple of 5. Now, from the things that Satvik says we can set up these two equations:

$s=p\cdot 10^d+q\\ a=10q+p$

Where $p$ is a digit from 1 to 9 and $q>0$ . Since $a$ is multiple of 5, it can only end with 0 or 5, and since its last digit is $p$ , then we must have $p=5$ . Then:

$s=5\cdot 10^d+q\\ a=10q+5$

Substituting on the previous relation we have:

$7(10q+5)=5(5\cdot 10^d+q)$

Simplify:

$13q=5\cdot 10^d-7$

Testing with small values of $d$ we find that $d=5$ and $q=38461$ is the first and only integer solution with $a,b<10^6$ .

Finally $s=538461$ and $\boxed{a=384615}$ .

Let $S$ and $A$ be the amounts of money and $x$ the hypothetical amount of money involved in the transactions. Then $A + x = S - x\ \ \therefore\ S - A = 2x \\ 2(A-x) = S + x\ \ \therefore\ 2A - S = 3x,$ and addition of these equations tells me that $A = 5x$ , from which it follows that $S = 7x$ .

Let $1 \leq b \leq 9$ be the last digit of $A$ , and $n$ the number of digits of $A$ , then we have $\frac{A-b}{10} + 10^{n-1}b = S \\ A + (10^n - 1)b = 10 S \\ 5x + (10^n - 1)b = 70 x \\ 65 x = (10^n -1)b.$ Since $10^n-1$ cannot be a multiple of 5, we must have $b = 5$ . Thus $13x = 10^n -1.$ This, combined with $n \leq 6$ , is only possible if $n = 6$ . (I happen to know that $13 | 1001$ so that $13 | 999\:999$ is clearly a solution.) Finally, then, $x = \frac{999\:999}{13} = 76\:923; \\ A = 5x = \boxed{384\:615}; \\ S = 7x = 538\:461.$

Here's a method using only the very basics of standard VIII.

Let Agnishom have 'x' & satvik have 'y' rupees & they exchange with each other rupees 'p' according to given statements respectively.

Two set of equations are as follows :

$x+p=y-p$ .......................... (1) $2(x-p)=y+p$ ........................... (2)

Eliminating x we get $7x=5y$ ............. (3) .

Since $x,y < 10^6$ so they are atmost 6 digit numbers.

$5|x$ so last digit of x is 5 & thus first digit of y.

Let $y = 10^5.5 + 10^4.a+10^3.b+10^2.c+10d+e$

$x= 10^5.a + 10^4.b+10^3.c+10^2.d+10e+5$

putting values of x & y successively in (3) and shifting terms and rearranging we get,

$a.10^4.65+b.10^3.65+c.10^2.65+d.10.65=25.10^5-35$

$65.y=10^5(25+65.5)-35$

$y=538461$ & thus $\boxed{x=384615}$