# Find the four digit numbers abcd that satisfy, 2(abcd)+1000=dcba?

Oct 4, 2016

See below.

#### Explanation:

Number $a b c d$ can be represented as

${n}_{1} = a {x}^{3} + b {x}^{2} + c x + d$ also

${n}_{2} = d {x}^{3} + c {x}^{2} + b x + a$

and

${n}_{0} = 1000 = 1 {x}^{3} + 0 {x}^{2} + 0 x + 0$ so

$2 {n}_{1} + {n}_{0} = {n}_{2}$ requires

$\left\{\begin{matrix}2 d - a = 0 \\ 2 c - b = 0 \\ 2 b - c = 0 \\ 1 + 2 a - d = 0\end{matrix}\right.$

and this cannot be accomplished with $a , b , c , d$ integers.

Oct 4, 2016

$a b c d = 2996$

#### Explanation:

write:
$\left[\begin{matrix}a & b & c & d \\ a & b & c & d \\ 1 & 0 & 0 & 0 \\ \text{-" & "-" & "-" & "-} \\ d & c & b & a\end{matrix}\right]$
this implies that: $2 a + 1 = d$
We also know that:
$2 a + 1 \le d \le 9$ why? d us a digit number...
also notice from the last expression $2 d = a$ that is $a$ is even.
This limits a to be $\left\{2 , 4\right\}$
So now let's try if $a$ can be 2 or 4:
Pick $a = 4$ then
$d \le 2 a + 1 = 9$ this means the last digit of $2 d = 8$ a contradiction.
If we pick $a = 2$ then $d \le 2 a + 1 = 5$
The last digit of $2 d \implies 2$ thus $d = 6$
$d + d = 6 + 6$ which will write 2 carry $1$. Now with $a = 2 \mathmr{and} b = 6$ the carry over the remaining equation is:
$\left[\begin{matrix}\text{ " & b & c \\ " " & b & c \\ " " & " " & 1 \\ "-" & "-" & "-} \\ 1 & c & b\end{matrix}\right]$

We have 2 scenarios:

Scenario :
$2 c + 1 = b \mathmr{and} 2 b = 10 + c$, unfortunately no integer solution

Scenario:
$2 c + 1 = 10 + b \mathmr{and} 2 b + 1 = 10 + c$, this yields
$b = c = 9$.

Therefore the answer is: $a b c d = 2996$