A number theory problem by Utkarsh Dwivedi

abcd x 4 = dcba

notice that 'a' which is a product of (d*4) must be divisible by 4 , --> hence it must be these digits (2 or 4 or 6 or 8 or 0)

but since a is also the first digit in abcd, 0 does not make any sense, 4 6 8 also not possible because *4 will gives u a 5digit value, so a is 2 for sure.

Since a is 2, d*4 which yields a unit value of 2 imply that d =8 so we have a=2, d=8 now.

Now you should be able to imagine abcd= 2BC8 By now smart guess will save a lot of work,

What 21?? x 4 = 84?? ~ 87?? , I did this by trial and error actually >< haha The answer is 2178 anyway

2178 x 4 = 8712

Method:

Since DCBA is a multiple of 4, we must have A even.

Further, since ABCD x 4 is also a four-digit number, we must have 4A be a one-digit number.

Therefore:

A = 2

Since A=2 and D≥4A we must have D=8 or D=9.

Since DCBA ends in 2 and is a multiple of 10 plus 4D, we must also have 4D end in 2.

Since 8x4=32 and 9x4=36 we conclude:

D = 8

This leaves the middle two letters. We can treat this as a two-digit calculation, remembering to carry the "3" from 4D : 4(BC) + 3 = CB

That is,

40B + 4C + 3 = 10C + B

=> 39B - 6C + 3 = 0

=> 3(13B - 2C + 1) = 0

B=1, C=7 is the most obvious solution.