The sum of the first \(n\) positive
integers is \(1+2+3 + \cdots +
n\).
We define \(a_n\) to be the ones digit
of the sum of the first \(n\) positive
integers.
For example,
\[\begin{aligned} 1=1 \ \ &\text{and} \ \ a_1 = 1,\\ 1+2=3 \ \ &\text{and} \ \ a_2 = 3,\\ 1 + 2 + 3 = 6 \ \ &\text{and}\ \ a_3 = 6,\\ 1 + 2 + 3 + 4 = 10 \ \ &\text{and}\ \ a_4 = 0,\\ 1 + 2 + 3 + 4 + 5 = 15 \ \ &\text{and}\ \ a_5 = 5. \end{aligned}\] Thus, \(a_1 + a_2 + a_3 + a_4 + a_5 = 1 + 3 + 6 + 0 + 5 = 15\).
Determine the smallest value of \(n\) such that \(a_1 + a_2 + a_3 + \cdots + a_n \geq 2024\).
Let’s start by examining the values of \(a_n\) until we start to see a pattern.
We know \(a_1=1\), \(a_2=3\), \(a_3 = 6\), \(a_4 = 0\), and \(a_5=5\).
Unfortunately, we do not have a pattern yet. We need to keep calculating values of \(a_n\). Since \(15 + 6 = 21\), \(a_6 = 1\).
Notice that we can determine the ones digit of the sum of the first \(n\) integers from the ones digit from the sum of the first \(n-1\) integers and the ones digit of \(n\). For example, to calculate \(a_7\), we simply need to know that \(a_6 = 1\) and the sum \(1+7 = 8\) has ones digit \(8\). So \(a_7=8\). Thus, continuing on, we know
\(a_8 = 6\), since \(a_7+8 = 16\)
\(a_9 = 5\), since \(a_8+9 = 15\)
\(a_{10} = 5\), since \(a_9 + 0 = 5\)
\(a_{11} = 6\), since \(a_{10} + 1 = 6\)
\(a_{12} = 8\), since \(a_{11} + 2 = 8\)
\(a_{13} = 1\), since \(a_{12} + 3 = 11\)
\(a_{14} = 5\), since \(a_{13} + 4 = 5\)
\(a_{15} = 0\), since \(a_{14} + 5 = 10\)
\(a_{16} = 6\), since \(a_{15} + 6 = 6\)
\(a_{17} = 3\), since \(a_{16} + 7 = 13\)
\(a_{18} = 1\), since \(a_{17} + 8 = 11\)
\(a_{19} = 0\), since \(a_{18} + 9 = 10\)
\(a_{20} = 0\), since \(a_{19} + 0 = 0\)
\(a_{21} = 1\), since \(a_{20} + 1 = 1\)
The values of \(a_n\) should repeat
now. Can you see why?
Since \(a_{21} = a_1\) and the ones
digit of \(22\) equals the ones digit
of \(2\), \(a_{22} = a_2\).
Similarly, since \(a_{22} = a_2\) and
the ones digit of \(23\) equals the
ones digit of 3, \(a_{23} =
a_3\).
We will also have \(a_{24} = a_4\), and
so on.
Therefore, the values of \(a_n\) will repeat every \(20\) values of \(n\).
We can calculate \[\begin{aligned} & a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} \\ & ~~~+ a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19} + a_{20}\\ = & ~ 1 + 3 + 6 + 0 + 5 + 1 + 8 + 6 + 5 + 5 \\ & ~~~+ 6 + 8 + 1 + 5 + 0 + 6+ 3 + 1 + 0 + 0 \\ = & ~ 70 \end{aligned}\] Since the values of \(a_n\) repeat every \(20\) values of \(n\), it is also true that \(a_{21} + a_{22} + a_{23} + \cdots + a_{39} + a_{40} = 70\), and \(a_{41} + a_{42} + a_{43} + \cdots + a_{59} + a_{60} = 70\), and so on.
Since \(\dfrac{2024}{70} = 28\dfrac{32}{35}\), there are \(28\) complete cycles of the \(20\) repeating values of \(a_n\).
Therefore, the sum of the first \(28\times 20 = 560\) values of \(a_n\) sum to \(28\times 70 = 1960\).
In other words, \(a_1 + a_2 + a_3 +\cdots + a_{559} + a_{560} = 1960\).
Let’s keep adding values of \(a_n\) until we reach 2024. \[\begin{aligned} a_{561} + a_{562} &+ a_{563} + a_{564} + a_{565} + a_{566} + a_{567} + a_{568} + a_{569} + a_{570}\\ &= a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} + a_{8} + a_{9} + a_{10}\\ & = 1 + 3 + 6 + 0 + 5 + 1 + 8 + 6 + 5 + 5\\ & = 40 \end{aligned}\]
Therefore, \(a_1 + a_2 + a_3 +\cdots +a_{569} + a_{570} = 1960 + 40 = 2000\).
We also know that \(a_{571} = a_{11} = 6\) , \(a_{572} = a_{12} = 8\), \(a_{573} = a_{13} = 1\) , \(a_{574} = a_{14} = 5\), and \(a_{575} = a_{15} = 0\).
Thus, \[\begin{aligned} a_1 + a_2 + a_3+ \cdots +a_{569} + a_{570} + a_{571}+ a_{572} + a_{573} + a_{574} + a_{575} &= 2000 + 6 + 8 + 1 + 5 + 0\\ &= 2020 \leq 2024 \end{aligned}\] and \[\begin{aligned} a_1 + a_2 + a_3+ \cdots +a_{569} + a_{570} + a_{571}+ a_{572} + a_{573} + a_{574} + a_{575} + a_{576} &= 2020 + 6 \\ &= 2026 \geq 2024 \end{aligned}\]
Therefore, the smallest value of \(n\) such that \(a_1 + a_2 + a_3 + \cdots + a_n \geq 2024\) is \(n = 576\).