Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on digit.
Find the four digit number \(\overline{abcd}\) satisfying
\(2 \overline {abcd} + 1000 = \overline {dcba}\)
Algebra
Digit Problem
Cubic Equation
Answer: 2996
Singapore Mathematics Olympiad
Challenges and Thrills - Pre - college Mathematics
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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