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A 4 digit number is equal to the digit sum of the number multiplied by 109.Find the biggest numbber satisfied the above condition

Guest Aug 31, 2017
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#1
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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 !!!!!!!

CPhill  Aug 31, 2017
edited by CPhill  Sep 1, 2017
#2
+18564
+1

A 4 digit number is equal to the digit sum of the number multiplied by 109.

Find the biggest numbber satisfied the above condition

$$\begin{array}{|rccccl|} \hline 1. & 1 & 0 & 9 & 0 \\ 2. & 1 & 3 & 0 & 8 \\ 3. & 1 & 4 & 1 & 7 \\ 4. & 1 & 5 & 2 & 6 \\ 5. & 1 & 6 & 3 & 5 \\ 6. & 1 & 7 & 4 & 4 \\ 7. & 1 & 8 & 5 & 3 \\ 8. & 1 & 9 & 6 & 2 \\ 9. & 2 & 2 & 8 & 9 \\ 10. & \mathbf{2} & \mathbf{3} & \mathbf{9} & \mathbf{8} & \text{max} \\ \hline \end{array}$$

heureka  Sep 1, 2017

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