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.