Value of a four digit number

Question

Let a four digit number be $abcd$ ie. $$abcd =1000a+100b+10c+d.$$ When the digits are reversed ie. $dcba$, the new number is $9$ times $abcd$. What is the number? I've tried making an equation using the given relation and equating different coefficients of powers of $10$ but couldn't reach anywhere. Any help is appreciated.

Answer 1

From $9\times dcba<10000$ it follows immediately that $d\in\{0,1\}$.

If $d=1$ then it follows that $a=9$ and we have: $$9\times(1000+100c+10b+9)=9000+100b+10c+1$$ leading to $89c=b-8$ and consequently $c=0$ and $b=8$.

So $abcd=9801$ is a solution.

It remains to check the case $d=0$.

Try it out yourself.

Answer 2

just do it.

$dcba = 1000d + 100c + 10b + a = 9(1000a+100b+10c+d)=9000a + 900b + 90c + 9d$

If $a \ge 2$ then $9*abcd > 18,000$ which is more than four digits.

So $a \le 1$. If $a = 0$ then $abcd$ is not a four digit number so $a = 1$

$1000d + 100c + 10b + 1 = 9000 + 900b + 90c + 9d$.

$100c + 10b + 1 \le 991$ so $1000d > 8009$ so $d > 8$ so $d = 9$.

$9000 + 100c + 10b + 1 = 9000 + 900b + 90c + 81$

$100c + 10b = 900b + 90c + 80$

$10c + b = 90b +9c + 8$

$c = 89b + 8$.

Well... $c < 10$ so $b = 0$.

And $c = 8$.

So $abcd = 1089$ and $9801 = 9*1089$

Answer 3

To avoid "overflow", we must have $a=1$, which implies $d=9$.

Now,

$$1000\cdot9+100\cdot c+10\cdot b+1\cdot1=9000\cdot1+900\cdot b+90\cdot c+9\cdot9$$

or after simplification

$$c=89\cdot b+8$$

Obviously $b=0$ and

$$\color{green}{1089}.$$

Answer 4

Hint: Note that There is two other conditions:

1 : 3 must divide the sum of digits of required number. So if $a=1$ and $ d=9$ then $3|10+b+c$.

2 : $b+c<10$

It is not hard to find these couples and see that only $b=0$ and $c=8$ satisfies all conditions, i.e $abcd=1089$ and $dcba=9801$