Number of 6's between 1 to 18000  TOPIC_SOLVED

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Number of 6's between 1 to 18000

Postby Numeroprime on Wed Jul 14, 2010 10:40 pm

Is there a smart way to solve this..?

I can see a brute force type of way.... by looking at the number of 6's from 1-100, then 101 to 200 etc... and special cases for 601-700 etc... but it seems very long, tedius and prone to error?

Thanks for any ideas.
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Postby stapel_eliz on Wed Jul 14, 2010 10:53 pm

I would suspect you would work in "chunks" of numbers, and see if you can find any patterns which might simplify things at all.

For instance, between 1 and 9 (which is the same number as between 0 and 9), there is one 6. Between 10 and 19, there is another 6, in "16". Between 20 and 29, there is another 6, in "26". And so forth, except for the sixties, where you get another ten 6's.

Then start working with the hundreds. You know how many 6's there are between 0 and 99. How many then will there be between 100 and 199? How many between 200 and 299? How many extra 6's do you get between 600 and 699? And so forth.

Then start working with the thousands. You should be able to use patterns, and values from previous "chunks" of numbers, to arrive at a final value.

Have fun! :wink:
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Re: Number of 6's between 1 to 18000

Postby Numeroprime on Wed Jul 14, 2010 11:12 pm

Thanks!

That was my guess, but I was hoping for a motrin free method.... Patterns makes sense and its a good exercise.

Mahalo,

Tory
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Re: Number of 6's between 1 to 18000  TOPIC_SOLVED

Postby Numeroprime on 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
------------------------------------------------------------------------------------------------------
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!
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Postby stapel_eliz on Wed Jul 21, 2010 8:28 pm

What is the "Doctor Andrew's method"?
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