A number with 6 distinct digits which get multiplied by 5 if we move the last digit to front

Question

There is a number with 6 different digits, if we pick the last digit of that number and place before that number we got $5$ times our number.

How to find such a number?

Answer 1

The number can be written $10x+y$, where $10000<x<100000$ and $0<y\le 9$. The modified number is then $100000y+x$ and so the equation is $$ 100000y+x=5(10x+y) $$ that can be rewritten $$ 99995y=49x $$ Since $99995$ is a multiple of $7$, we have $$ 14285y=7x $$

Now $7$ divides $14285y$, but doesn't divide $14285$. So $7$ divides $y$ and the condition $0<y\le 9$ gives $y=7$ and $x=14285$.

Note that the information that the number has different digits is redundant.

Answer 2

If the number is ABCDEF then FABCDE = 5*ABCDEF or in other words: 5(100000A + 10000B + 1000C + 100E + 10E + F) = 100000F + 10000A + 1000B + 100C + 10D + E.

Hint: Try subtracting one from the other.

2)James age now = y;

John age now = x;

How long ago it was when James was Johns age: m = ???? something to do with x and y ?????

i) y = 2(x - m)

ii) y = 28

iii) m = ???? something to do with x and y and maybe 28 if y = 28 ?????