Question

Find the value of A,B,C,D. If ABCD × 4 = DCBA

Answer 1

Obviously ‘A’ can not be greater than 2 since 3BCD*4>12000 which is 5-digit number.

Also, as DCBA is even (multiple of 4), 'A' must be even. So, A=2

Since 8000<2BCD*4=DCB2<10000, 'D' can be either '8' or '9'.

But 9*4=36 whose last digit is not equal to A i.e. 2. So, D=8

Now,

ABCD*4=DCBA implies

=>(1000A+100B+10C+D)*4=1000D+100C+10B+A (where, A=2 & D=8)

=>8000+400B+40C+32=8000+100C+10B+2

=>60C=390B+30

=>C=(13B+1)/2<20/2

B must be odd (for 13B+1 to be even) such that (13B+1)<20 which is possible only when B=1. So, C=7

Hence,

A=2

B=1

C=7

D=8

i.e. 2178*4=8712