Count the predecessors
Imagine that I have a 10-digit number.
The first digit is the number of times 0 has occurred in the number, the 2nd digit is the number of times 1 has occurred in the number....the 10th digit is the number of times 9 occurs in the number.
You have to find the number.
Source: Finally-the source of the problem is a bit ambiguous, but I believe it is from an Indonesian Math and Coding Contest.
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1 solution
The first digit is a 6 in all of the given choices, which means the number has six zeros ⇒ Choice E is eliminated.
The second digit is the total number of 1 ′ s that appear in the number. Of the remaining four choices, A thru D, only Choice D satisfies this condition.
The last digit is the total number of 9 ′ s that appear in the number. Again, Choice D satisfies this condition as well.
Hence, the number is 6 , 2 1 0 , 0 0 1 , 0 0 0 .