Question

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

Answer 1

See below.

Explanation:

Number `abcd` 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

`{(2 d-a=0), (2 c-b=0), (2 b - c=0), (1 + 2 a - d=0):}`

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

Answer 2

`abcd=2996`

Explanation:

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

We have 2 scenarios:

Scenario :
`2c + 1 = b and 2b = 10 + c`, unfortunately no integer solution

Scenario:
`2c + 1 = 10 + b and 2b + 1 = 10 + c`, this yields
`b = c = 9`.

Therefore the answer is: `abcd=2996`