Digit Problem from SMO, 2012 | Problem 14

Find the four digit number \(\overline{abcd}\) satisfying

\(2 \overline {abcd} + 1000 = \overline {dcba}\)

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 ...................

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.

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)

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