Question

Find the four-digit number ABCD such that ABCD x 4 = DCBA

Answer 1

Any number multiplied by 4 will give us an even number Hence the digit D when multiplied by 4 will give us a even number
Since
A is the unit digit of the product it is even Hence A = 2, 4, 6 or 8
(It cannot be 0) A is also the first digit of the multiplicand and if A =
4, 6 or 8 the product ABCD x 4 will become a 5 digit number Hence A = 2
Writing the value of A we get 2BCD x 4 = DCB2
Now for the value of D
looking at the first and last digits of the multiplicand we can see
that 4 x D gives the unit digit of 2 and 4 x 2 gives the first digit of D
Yes you got it right D = 8 Writing the multiplication again with the
value of D we get 2BC8 x 4 = 8CB2
Now for the value of B A number is
divisible by 4 if the number formed by the last two digits is divisible
by 4 Since the number 8CB2 is a multiple of 4 the number B2 should be
divisible by 4 or the number B2 = 12, 32, 52, 72 or 92 Hence the
original number ABCD is 21C8, 23C8, 25C8, 27C8 or 29C8 But the last 4
numbers when multiplied by 4 will not give you the first digit of 8 in
the product Therefore B = 1 and the original number is 21C8 We write the
multiplication again 21C8 x 4 = 8C12
Now for the value of C notice
that when you multiply 8 the unit digit of 21C8 by 4 you write 2 in the
unit digit of the product and carry 3 The tenth digit of the product is 1
Therefore 4 x C + 3 (carry over) gives a unit digit of 1 Hence C is 2
or 7 You can easily check by the hundreds digit in the product (which is
C again) that C = 7 Therefore our answer is 2178 x 4 = 8712