Mary wrote a book of 1536 pages
She numbered all the pages by hand
How many times did she write the digit 6?
Answers
Answer:
Step-by-step explanation:
Logically, there will be an even distribution of digits through a rollover point (i.e. 999 to 1000)
Single digit page numbers = 9 (9-1 + 1)… The +1 comes from the fact that we are looking at all digits one through nine, inclusive. That is, if we want to count all digits from one through nine, we can do a subtraction, but need to include one more to include the full set.
Double digit page numbers = 99-10 + 1 = 90
Triple digit page numbers = 999-100 + 1 = 900
The total number of digits written by hand through 1000 = 9+(90*2)+(900*3) = 2889. Since the digits are all equal to this point, Mary wrote all digits (2889+1)/10 = 290 times to get to 1000. (note that I'm adding one to 2889 to get from 999 to 1000)
Is a little bit harder to determine how many times she wrote the number 6 between 1001 and 1537, but we can take the following into account: 500 is one half of one thousand, so we figure another 290/2 = 145.
From 1500 to 1537, she would've written the 6 with the following numbers: 1506, 1516, 1526, and 1536, so that's 4 additional times.
If my logic doesn't fail me here, we have a total of 290 + 145 + 4 = 439 total 6s
Answer:
404 times
Step-by-step explanation:
The reasoning could the following: think about each digit place (units, tens, hundreds, thousands)
Thousands: there are 0 "6s" since we only go until 1 thousand.
Hundreds: there are 100 "6s", with the numbers from 600-699, inclusive (only considering the hundreds digit place).
Tens: there are 150 "6s.". A 6 appears 10 times in a row every 100 numbers (60-69, inclusive). Dividing 1500 by 100 (not considering the final 36, since it doesn't reach the 60's) we have 15 x 10 =150.
Units: there are 154 "6s". A 6 appears every 10 numbers. Dividing 1536 by 10 we get 153,6. Since there is a six in the last 10 digits, we can round up and get 154.
0+100+150+154=404