Answer:
The years are:
- 1000 BCE, 999 BCE, 888 BCE, 777 BCE, 666 BCE, 555BCE, 444 BCE. 333 BCE, 222 BCE, 111 BCE
- 111 CE, 222 CE, 333 CE, 444 CE, 555 CE, 666 CE, 777 CE, 888 CE, 999 CE, 1000CE, 1011 CE, 1101 CE, 1110 CE,1222CE, 1333CE, 1444CE, 1555 CE, 1666 CE, 1777CE, 1888 CE, 1999CE, and 2000 CE
Explanation:
<u>1. Years BC:</u>
a) Years with four digits:
The first number with 3 equal digits is 1000. After that the years go decreasing: 999, 998, 997, ...
b) Years with three digits:
From 999 to 111, the numbers have three digits, thus the only that are solutions ara 999, 888, 777, 666, 555, 444, 333, 222, and 111: 9 numbers
After that the years have two digits, thus no solutions, with two digits.
Hence, we count 10 different years.
<u>2. Years CE</u>
a) Years with three digits:
- 111, 222, 333, 444, 555, 666, 777, 888, 999: 9 years
b) Years with four digits
i) Starting with 1:
- With three 0: 1000: 1 year
- With three 1: 1011, 1101, 1110: 3 years
- With three digits different to 1: 1222, 1333, 1444, 1555, 1666, 1777, 1888, 1999: 8 years
ii) Starting with 2:
- With three 0: 2000: 1 year
The next one with three equal digits is 2111 and it is after 2020 CE.
Therefore, 9 + 1 + 3 + 8 + 1 = 22 years starting with 2.
<u>3. Total</u>
<u />
10 years BC and 22 years CE have exactly three digits the same: 10 + 22 = 32.
0.1347 rounded to the nearest tenth is 0.1
There is no such number.
Note that 36x = 9 results in x = 9/36 or x = 1/4.
Could you possibly share an example of this type of problem? Right now it doesn't make sense to me.
3 times the sum of b and f, can also be written as: 3(b+f)
