Question

\[4 \times \left( {ABCD} \right) = DCBA\], what is \[ABCD\]?

Answer 1

Hint: To find \[ABCD\], we will check all the possibilities one by one . As we know, any number \[abcd\] can be written as \[1000a + 100b + 10c + d\]. Using this we will first write \[ABCD\] as \[1000A + 100B + 10C + D\] and \[DCBA\] as \[1000D + 100C + 10B + A\]. Then using the facts, we will eliminate the inappropriate possibilities to find \[ABCD\].

Complete step by step answer:
As we know that any number \[ABCD\] can be written as
\[ \Rightarrow ABCD = 1000A + 100B + 10C + D - - - (1)\]
In a similar way, any number \[DCBA\] can be written as
\[ \Rightarrow DCBA = 1000D + 100C + 10B + A - - - (2)\]
It is given in the question that \[4 \times \left( {ABCD} \right) = DCBA\].
Here, we can see that \[DCBA\] is a four-digit number, so \[ABCD\] must be less than \[2500\] because any number greater than \[2500\] on multiplication with \[4\] will give a five digit number which is not required.
So, \[A\] will be either \[1\] or \[2\]. But as we know a multiple of \[4\] will always be an even number. Therefore, from \[DCBA\] we can say \[A\] is \[2\].
As we get \[A = 2\], and also \[DCBA\] is a four-digit number. So, four times \[2\] will give \[D = 8\].
Now, using \[4 \times \left( {ABCD} \right) = DCBA\], \[\left( 1 \right)\] and \[\left( 2 \right)\], we can write
\[ \Rightarrow 4 \times \left( {1000A + 100B + 10C + D} \right) = 1000D + 100C + 10B + A\]
Putting values of \[A = 2\] and \[D = 8\], we get
\[ \Rightarrow 4 \times \left( {1000 \times 2 + 100B + 10C + 8} \right) = 1000 \times 8 + 100C + 10B + 2\]
On simplification we get
\[ \Rightarrow 8000 + 400B + 40C + 32 = 8000 + 100C + 10B + 2\]
On further simplification we get
\[ \Rightarrow 390B - 60C + 30 = 0\]
Dividing both the sides by \[30\], we get
\[ \Rightarrow 13B - 2C + 1 = 0\]
On rewriting we get,
\[ \Rightarrow 2C = 13B + 1 - - - (3)\]
The only possible value of \[B\] and \[C\] which satisfies equation \[(3)\] is \[B = 1\] and \[C = 7\].
So, we get \[A = 2\], \[B = 1\], \[C = 7\] and \[D = 8\].
Therefore, \[ABCD\] is \[2178\].

Note:
Here, we can check that the answer obtained is correct or not by multiplying \[ABCD\] i.e., \[2178\] with \[4\], on multiplying we get \[8712\] i.e., \[DCBA\]. Therefore, it satisfies the condition given in the question i.e., \[4 \times \left( {ABCD} \right) = DCBA\]. Therefore, the obtained value of \[ABCD\] is correct.