2016 Digital Root
For each number n , we define the Digital Root of n in the following manner: Take the sum of the digits to obtain a new number. Repeat this process until the result is a single digit.
Find the digital root of 2 0 1 6 times 2 0 1 6 … 2 0 1 6 .
As an explicit example, for n = 3 4 8 7 , 3 4 8 7 → 3 + 4 + 8 + 7 = 2 2 → 2 + 2 = 4 . Therefore, the digital root of 3487 is 4.
Clarification : In the number above, 2016 appears 2016 times in its decimal representation.
The answer is 9.
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3 solutions
Firstly, note that the digital root of 2 0 1 6 is 2 + 0 + 1 + 6 = 9 . As 2 0 1 6 appears 2 0 1 6 times in the number we are asking for, then its digital root is equal to 2 0 1 6 ⋅ 9 = 1 8 1 4 4 → 1 + 8 + 1 + 4 + 4 = 1 8 → 1 + 8 = 9 .
Coincidentally, is the same digital root of 2 0 1 6 .
Note: The term in the question is actually digitial root . The digital sum is just the sum of the digits. I've edited the problem/solution accordingly.
It is not a co-incidence.
Check this out
inspired by your problem.
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Hi Vighnesh. Can you post the demonstration of why it is not a coincidence?
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Sum of digits is 9. If it occurs 'n' times then it is 9 n . For a number divisible by 9 , the sum of digits is also a multiple of 9. So , repeating this process till the end we get 9.
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@A Former Brilliant Member – Mmm clear. I thought that you had demonstrated when does the digital sum of n is equal to the digital sum of n t i m e s n . . . n . What should I check out?
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@Mateo Matijasevick – I guess that would be a more complicated problem.
2016=2+0+1+6=9 Number of occurance is 2016 which is independent to the answer
∴ answer is 9
y o u c a n a l s o c h e c k
2016*9=18144=1+8+1+4+4=18=1+8=9
Since the digital sum of 2016 = 9 So no matter how many time 2016 is there, the digital sum will always be 9.
Because every multiple of 9 has digital sum of 9...that's why that number is divisible by 9.