Question

5 digit number that when multiplied by 4 gives the number in reverse order

Answer 1

Let $\smalln=10,000a+1,000b+100c+10d+e=abcde$ be our 5 digit number.

Our number when multiplied by 4 gives its reverse, i.e., $\small4n=edcba=10,000e+1,000d+100c+10b+a.$

[N.B.: Here $\smalla,\ b,\ c,\ d$ and $\smalle$ are digits of the numbers and they're not multiplied to each other.]

So $\small4n$ is a 5 digit number which is less than 1 lakh, i.e.,

$\small\longrightarrow 4n<1,00,000$

$\small\longrightarrow n<25,000$

Since $\smalln$ is a 5 digit number,

$\small\longrightarrow n>10,000$

(10,000 is not an answer.)

Then,

$\small\longrightarrow 10,000<n<25,000$

This means ten-thousands digit of $\smalln$ should be either 1 or 2.

$\small\Longrightarrow a\in\{1,\ 2\}$

This $\smalla$ is also ones digit of $\small4n$ which is even. So should $\smalla$ be.

$\small\Longrightarrow\underline{a=2}$

So ones digit of $\small4n$ is 2, then ones digit of $\smalln$ must be either 3 or 8.

$\small\Longrightarrow e\in\{3,\ 8\}$

But since ten-thousands digit of $n$ is 2,

$\small\longrightarrow n>20,000$

$\small\longrightarrow 4n>80,000$

Then ten-thousands digit of $\small4n$ should be either 8 or 9.

$\small\Longrightarrow\underline{e=8}$

Now $\smalln=2bcd8$ and $\small4n=8dcb2.$

$\small4n$ is a multiple of 4, so should the 2 digit number formed by joining its ones and tens digits be, i.e., $\smallb2$ should be a multiple of 4.

Then $\smallb$ should be odd. (12, 32, 52, 72 and 92 are multiples of 4.)

$\small\longrightarrow b\in\{1,\ 3,\ 5,\ 7,\ 9\}$

But since ten-thousands digit of $\small4n$ is 8,

$\small\longrightarrow80,000<4n<90,000$

$\small\longrightarrow20,000<n<22,500$

Then thousands digit of $\smalln$ should be either 0, 1 or 2.

$\small\Longrightarrow\underline{b=1}$

Now $\smalln=21cd8$ and $\small4n=8dc12.$

$\small4n$ ends in 12 so $\smalln$ should end in any of these four 2 digit numbers:- 03, 28, 53, 78.

(because for every non - negative integer $\smallk,$ one-fourth of $\small100k+12$ is $\small25k+3$ which ends in any of the above 4 numbers.)

Since ones digit of $\smalln$ is 8, $\smalln$ should end in either 28 or 78.

Assume it's 28. Then $\small4n$ starts with 82.

But the thousands digit of $\smalln$ is 1 which when multiplied by 4 gives 4. So thousands digit of $\small4n$ should be $\smalld=4+m$ where $\smallm$ is the remainder carried out on multiplying hundreds digit of $\smalln,$ i.e., $\smallc,$ by 4.

Then $\small4+m=2$ followed by $\smallm=-2$ which contradicts our assumption since possible values of $\smallm$ are 0, 1, 2 and 3 because of the multiplier 4.

$\small\Longrightarrow\underline{d=7}$

Now, the following happens when we multiply $\smalln$ by 4.

→  e = 8 multiplied by 4 gives a = 2 with remainder 3 (8 × 4 = 32).

→  d = 7 multiplied by 4 gives b = 1 with remainder 3 (7 × 4 + 3 = 31).

→  c multiplied by 4 makes one digit of 4c + 3 equal to c, with remainder 3.

→  b = 1 multiplied by 4 gives d = 7 with remainder 0 (1 × 4 + 3 = 7).

→  a = 2 multiplied by 4 gives e = 8 (2 × 4 + 0 = 8).

As in 3rd point the remainder must be 3 in order to advance to 4th point.

And according to the 3rd point,

$\small\longrightarrow 4c+3=10(3)+c$

$\small\longrightarrow 4c+3=30+c$

$\small\longrightarrow\underline{c=9}$

Hence our number is 21978.

Answer 2

Answer:

21978 is the answer.

21978 x 4 = 87912

I just used a simple loop from 10000 to 99999 and multiplied each number by 4, converted to a String, and checked to see if the String reversed was equal to the original number as a String.