Today's Maths-Type puzzle

Find a ten digit number, such that the first digit in that number is the number of ZEROs in the entire number, the second digit in that number is the number of ONEs in the entire number, the thrid digit is the number of TWOs in the entire number, and so on, until the tenth digit, which is the number of NINEs in the entire number.

Don't google it! Do it on your own, you'll like it. This is quite an interesting one if you haven't done it before.

One hint I can give you is, that if you take any ten-digit number and start applying the above 'rule' to it, you'll eventually arrive at the correct answer.

Of course, it all depends upon which number you choose to start from. ;)

It is 6210001000

Solved was a 2 minute problem but a good one.

But I couldn't think of the proof that it is the unique number satisfying this property. Please give some insight in this matter.

Quote:

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Originally posted by fundu
It is 6210001000

Solved was a 2 minute problem but a good one.

But I couldn't think of the proof that it is the unique number satisfying this property. Please give some insight in this matter.

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Method?

Well, you can work backwards:

1) it is a 10-digit number
=> the sum of the digits will = 10
(6+2+1+1=....10!!!)

2) There has to be at least one 0, one other number (to make the sum of 10), and a third number to describe this number.
=> THe sum is already at least 3 (the two non-zero numbers are different)

3) => The highest number you can have is 7. However, now there must be more zeros (cos 8 and 9 are 0's), so now the count is up to 3 zeros, and two other numbers.

etc... just keep going until you realise it is always going to be of the form
X21____1___, in which case X = 10-4 = 6!

Way to go and confuse his small frog brain sql. :lol: