How many 6-digit numbers have at least one 7 in them, that is, in numbers 100,000 to 999,999?. I thank you for any help.

Guest Apr 12, 2019

#1**+2 **

\(\text{We're talking a 6 digit number that is made up of the digits 0-6, 8-9}\\ \text{For the first digit we have 8 choices}\\ \text{For the remaining 5 digits we have 9 choices each}\\ 8 \cdot 9^5 = 472392\)

.Rom Apr 12, 2019

#2**+2 **

Integers that contain no 7s =

(8 choices) for 1st digit [ not 0 or 7 ]

(9 choices) for each of the next 5 digits [ no 7s ] = (9)^5

So....

8 * 9^5 = 472392

And we have 999,999 - 100,000 + 1 = 900,000 possible 6 digit integers

So.......the number that contain at least one 7 =

900,000 - 472392 =

427608

CPhill Apr 12, 2019

#3**+2 **

CPhill: You have a reversal typo in your last answer. I wrote a short computer code to count them and it agrees with yours(corrected).

Rom: Yours is to be subtracted from 900,000 to get the right number.

def getCount(x) : count = 0; for i in range(100000, 999999) : count = count + has7(i); return count; print("Total = ", getCount(x)): Total =**427,608 numbers that have at least one 7 in them.**

Guest Apr 12, 2019