Problem B
Years and Years of Threepeats'
Find all years between 1000 BCE and 2020 CE, inclusive, which have exactly
three digits the same. For example, the years 1011 and 1222 each have exactly
three digits the same, but 1111 and 1123 do not.

The years are:

Explanation:

1. Years BC:

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.

2. Years CE

a) Years with three digits:

b) Years with four digits

i) Starting with 1:

ii) Starting with 2:

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.

3. Total

10 years BC and 22 years CE have exactly three digits the same: 10 + 22 = 32.

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

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step-by-step explanation: