Formula to calculate how many times a number occurs between two numbers

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Homework Help Overview

The discussion revolves around determining how many times a specific digit, in this case, the number 7, occurs within a specified range of numbers, both inclusively and exclusively. The subject area includes number theory and combinatorial counting methods.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods for counting occurrences of the digit 7, including direct counting, division, and programming approaches. Questions arise regarding the interpretation of the problem and the counting of repeated occurrences.

Discussion Status

The discussion is active, with participants sharing different approaches and calculations. Some guidance has been offered regarding counting methods, and there is acknowledgment of the complexity involved in counting repeated occurrences of the digit.

Contextual Notes

Participants note the challenge of counting occurrences accurately, especially with numbers that contain multiple instances of the digit 7. There is also mention of the range being considered as either 1 to 1,000,000 or 0 to 999,999, which affects the counting strategy.

Brandon_R
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Hello guys, i was just wondering what is the formula to calculate how many times a number occurs between two numbers, both inclusively and exclusively.

Thanks.
 
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Can you please clarify the question? As far as I know, every (real) number occurs precisely once on the number line.
 
For example:

How many times does the number 7 occur from 1 to 1,000,000 (one to one million) inclusive.
 
Brandon_R said:
For example:

How many times does the number 7 occur from 1 to 1,000,000 (one to one million) inclusive.

What have you tried so far Brandon?
 
First i counted how many times the digit 7 occurred between 1-10 which is one. Then i divided 1,000,000 by 10 and multiplied it by 1. But that's wrong. Is there a formula to solve a question like this? This doesn't account for numbers like 77 or 777.
 
Hint: It is easier to count the number of numbers that do not contain a 7.
 
Thanks.
 
I have still no idea what the question is :bugeye:

I guess you are looking for numbers that contain digit seven, or something like that?
 
Borek said:
I have still no idea what the question is :bugeye:

I guess you are looking for numbers that contain digit seven, or something like that?
Something like that, Borek. How many numbers between 1 and 1,000,000 contain the digit 7 one or more times.
 
  • #10
0-99: 1 recurrence of digit 7 (single digit) + 9 (from "ones" in double digit) + 10 (from "tens" in double digit). (1+9+10)=20

100-999: 10 (from "ones" in triple digit numbers) + 90 (from "tens" in triple digit numbers) + 100 (from "hundreds" in triple digit numbers) = 10*(1+9+10)=200

1,000-9,999: again we see: 100*(1+9+10)=2,000

10,000-99,999: again we see: 1,000*(1+9+10)=20,000

100,000-999,999: again we see: 10,000*(1+9+10)=200,000

as a result there are 20*(1+10+100+1,000+10,000) = 222,222 recurrences of the digit 7 in one million numbers.any close to the answer ? :)
 
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  • #11
You're off by quite a lot. You aren't counting the numbers with repeated occurrences of the digit. For example, you are counting numbers such as 7 and 70, but aren't also counting numbers such as 77.
Looking at one of your ranges of numbers, you have 200 numbers in the range 100 through 999 that have 7 digits. I get 252 of them. As stated earlier in this thread it's easier to count how many numbers don't have a 7 digit.
 
  • #12
Im getting 600,000 occurrences of the digit 7 here.
 
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  • #13
I think your value is high by about 70,000. How did you get 600,000?
 
  • #14
Found it using a not so mathematical approach. I wrote a PHP script to concatenate the numbers 1 to 1 million into a variable ie 1234567891011121314151617...1000000 and then searched for how many times the number 7 occurred.

I just wanted to see how far away i was from the answer.
 
  • #15
Mark44 its the second time your saying that "it's easier to count how many numbers don't have a 7 digit" but its basically the same, when you count one you basically count the other and i don't really see why its easier. just a thought.

also the number 7 and 70 count as two occurrences because they are two different numbers and each has 1 occurrence of the digit 7.

But you right I counted some numbers twice!
and you're right on the money with 252 :)

Example for 100-999:

(107,117,127,...,197) + (207,...,297) + (307,...,397) + (...) + (907,...,997) = 10*9 = 90
(170,171,..179) + (270,..179) + (...) + (970,..979) = 10*9 -9 (No. 177,277,...,977) = 81
(700,701,...,799) = 100 - 10 (No. 707,717,...,797) - 9 (No. 770-779 without 777) = 81

=> grand total = 90+81+81= 252.

now we're mostly sure how many 7's are in 100-999 :)
 
  • #16
As far as this problem is concerned, the range 1 - 1,000,000 is the same as the range 0 - 999,999.

It's easy enough to count how many times 7 doesn't appear as a digit. In each of the 6 places there are 9 digits that can appear and one (7) that can't appear. So there are 96 numbers that don't have a 7 digit in any of the 6 places. All the rest do have a 7 digit in one or more positions.
 

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