A 4 digit number is equal to the digit sum of the number multiplied by 109.Find the biggest numbber satisfied the above condition
Here's my best attempt at this one......whether it's the correct answer, I don't know.....!!!
Note that any four digit number can be written as
1000a + 100b + 10c + d
And let the sum of its digits be a + b + c + d
And we're told that
1000a + 100b + 10c + d = 109 (a + b + c + d) distribute the 109
1000a + 100b + 10c + d = 109a + 109b + 109c + 109d subtract the right side from both sides
891a - 9b -99c - 108d = 0 divide through by 9
99a - b - 11c - 12d = 0 add 12d , b to both sides
99a - 11c = 12d + b factor the left side
11 [ 9a - c ] = `12d + b divide both sides by 11
9a - c = [ 12d + b ] / 11
Note that we need to have "a" as large as possible.......
And as large as the right side can be, 10, is when d = 9 and b = 2
And this implies that a = 2 and c =8
And when the right side is 9, d = 8 and b = 3
And this implies that a = 2 and c = 9
And when the right side is 8, d = 7 and b = 4
And this implies that a = 2, but c = 10 which is impossible
So....it appears that the largest our number can be is when a = 2, b = 3, c = 9 and d = 8
So.....the number is 2398
P.S. - any corrections / additional answers by other mathematicians are welcome !!!!!!!
A 4 digit number is equal to the digit sum of the number multiplied by 109.
Find the biggest numbber satisfied the above condition
\(\begin{array}{|rccccl|} \hline 1. & 1 & 0 & 9 & 0 \\ 2. & 1 & 3 & 0 & 8 \\ 3. & 1 & 4 & 1 & 7 \\ 4. & 1 & 5 & 2 & 6 \\ 5. & 1 & 6 & 3 & 5 \\ 6. & 1 & 7 & 4 & 4 \\ 7. & 1 & 8 & 5 & 3 \\ 8. & 1 & 9 & 6 & 2 \\ 9. & 2 & 2 & 8 & 9 \\ 10. & \mathbf{2} & \mathbf{3} & \mathbf{9} & \mathbf{8} & \text{max} \\ \hline \end{array} \)