ABCD multiplied by 2 equals four natural numbers of DCBA

ABCD multiplied by 2 equals four natural numbers of DCBA

Let's analyze it first
DCBA is twice of ABCD, DCBA must be even, so a must be even: 2,4,6,8
If you look at the highest position, D is either double a or carry 100 when multiplied by 2, then D is double a plus 1
If a is 2 and D is 4 or 5, but the bit of the double of 4 or 5 is not a, it does not match
If a is 4 and D is 8 or 9, the bits of double of 8 and 9 are not 4 and do not match
If a is larger, two times of ABCD is not a four digit number
So. No solution
I changed a small program to check, indeed no solution. Please carefully look at the problem is not input error

Which three continuous natural numbers are 99

It doesn't exist
Because 99 = 3 × 3 × 11
One of the three natural numbers must be 11, then the minimum product of the other two numbers is 9 × 10 = 90, and the product of the three natural numbers is at least 990
So the product of three consecutive natural numbers is 99
Hope to help you

In four digit ABCD, a + D = 10, and D is prime. If you remove a and D, the remaining two digit BC is also a prime. It is also known that four digit ABCD can be divided by 72
How much is ABCD?

If a + D = 10, and D is prime and ABCD can be divisible by 72, then d must be the only even prime 2. If a = 8.8, BC2 can be divisible by 9 and 8, then 8 + 2 + B + C can be divisible by 9, and B + C = 8 and 17bc2 can be divisible by 8