Two merchants share a telephones package such that one keeps 2/7 of the total and the other with 5/7. The first resells them in batches of 51 and the second in batches of 31. When they finish, the first has 46 telephones left and the second 19. What is the lowest initial number of telephones that makes this possible?
I've noticed that
Let $x$ the initial number. The number of telephones that the first merchant has is $2x/7=y= 46+51\lambda\equiv 46 \mod(51)$ and the second $5x/7=z=19+31\mu\equiv19\mod(31)$.
Also $x=y+z=46+51\lambda+19+31\mu=65+51\lambda+31\mu$.
How can I continue? .. can you give any hint please?
lets make $7x$ the intial number of phones. (instead of x)
$51 \lambda + 46 = 2x\\ 31\mu + 19 = 5x$
$2x \equiv 46\pmod {51}\\ 5x \equiv 19 \pmod {31}$
The smallest $x$ that satisfies the first equation is $23$
$5(23+51\lambda) \equiv 19 \pmod {31}\\ 22 + 7\lambda \equiv 19 \pmod {31}\\ 7\lambda \equiv 28 \pmod {31}\\ \lambda \equiv 4\pmod {31}$
$x = 23 + 51\cdot 4 = 227\\ 7x = 1589$