Answer:
git no. and how to prove it?
Lets say the number is written as ABCD (1000a+100b+10c+d).
Multiplying it by 4, gives you DCBA (1000d+100c+10b+a)
By divisibility of 4, BA is a two-digit number divisible by 4.
Now, the largest possible value of ABCD is 2499, because anything more will result in a five - digit number when multiplied by 4.
Out of the two possibilites 1 and 2 for A, [3] follows that A is even, so 2.
From [5], D is either 8 or 9. However, since D times 4 would end in 2, D is 8.
Using [2], 4000a+400b+40c+4d = 1000d+100c+10b+a, which would simplify to 2c-13b = 1 after substituting A = 2 and D = 8.
The only possible value of (b,c) where they would both be single - digit integers is (1,7). Also to note from [3], B is odd. And from [4], B is either 1 or 3.
The answer you're looking for is 2178, multiplied by 4 gives 8712.
Step-by-step explanation:
