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Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on digit.

## Digit Problem – Cubic Equation – SMO Test

Find the four digit number $\overline{abcd}$ satisfying

$2 \overline {abcd} + 1000 = \overline {dcba}$

• 2996
• 2775
• 1611
• 2001

### Key Concepts

Algebra

Digit Problem

Cubic Equation

But try the problem first…

Source

Challenges and Thrills – Pre – college Mathematics

## Try with Hints

First hint

If you got stuck with this sum lets try to start from here :

Let us consider the left hand side of the given equation –

$2 \overline {abcd} + 1000$ . Here if we express it in equation

suppose $n_{1} = ax^3 +bx^2+cx+d = \overline {abcd}$

$n_{0} = 1x^3 +0x^2+0x+0 = 1000$

$n_{0} = dx^3 +cx^2+bx+a = \overline {dcba}$ = which is the right hand side of the given equation.

So the equation becomes = > $2 n_{1} + 1000 = n_{2}$

Now compare the coefficients and try to do the problem ……………….

Second Hint

Lets continue after hint 1 we get :

2a + 1 = d

2b = c

2c = b

2d = a

from this we can understand that $2a +1 \leq d \leq 9$ as d is th digit number.

Now as 2d = a so a should be even number. Then the limits will be {2,4}.

Now implement the values of a and d and try to find the answer.

Final Step

Here is the rest of the problem:

If we consider a = 4

Then $d \geq 2a+1 = 9$ ; thus d = 9. But $2d = a \neq 8$.Its a contradiction.

So again , a = 2

$d \leq 2a+1 =5$ ; thus 2d = a = 2

d = 6; So d + d = 6+6 which can be written as 2 carry 1.

So, a = 2 and b = 6. Now the carry over remaining equation is

b c

b c

0 1

is equal to 1 c b

so either 2c +1 = b and 2b = 10 + c

r,

2c +1 = 10 + b and 2 b + 1 = 10 + c so b = c = 9

$\overline {dcba} = 2996$(Answer)