Postby **Numeroprime** » Wed Jul 21, 2010 8:26 pm

One of my classmates got the same answer as presented by Dr. Andrew.... using the pattern method: how many 6's from 1-100, how many from 1000-1999 etc... and making use of the special case where the first number is 2.

When I tried Dr. Andrews method for the text question: How many zeros from 1 to 1,000,000. I obtained the wrong answer. The text says 488895.

Here is what I said (probably some of my brain fog here):

i) there are 1M numbers from 1 to 1,000,000 (rows of numbers in this case if we write one on each line.

ii) For the numbers from 1 to 999,999 there are 6 columns of numerals.

iii) Since each numeral from 0-9 has an equal chance of appearing, then the number of 0's would be: 1/10 x 6,000,000 numerals (1m numbers of 6 numerals each)

iv) This would give us: 600,000 zeros: BUT there are 6 zeros not added in from the last number 1,000,000, AND the leading zeros would not be counted for example: 000,001 and 000,030, etc... so we would need to subtract leading zeros.

space intended

v) Now to subtract leading zeros we should? have:

- for numbers 0-9 we have 5 leading zeros so (- 5 x 9 ) = -45

- for numbers 10 -99 we have 4 leading zeros so ( -4 x 90) = - 360

- for numbers 100 - 999 we have 3 leading zeros so ( -3 x 900) = - 2700

- for numbers 1000-9999 we have 2 leading zeros so (-2 x 9000) = -18,000

- for numbers 10,000 - 99,999 we have 1 leading zero so (-1 x 90,000) = -90,000

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Total zeros to subtract = -111105 plus we need to add the 6 zeros from 1,000,000

The total zeros to subtract now becomes -111,111

Our original total zeros = 600,00 - 111,111 to give us = -488,889 so while close to the text answer of 488,895 I am short 4 zeros from somewhere?

Thanks for any help....I feel good that I am close....but not quite the cigar!