CEMC Fermat

We consider first the integers that can be expressed as the sum of exactly 4 consecutive positive integers. The smallest such integer is $1+2+3+4=10$. The next smallest such integer is $2+3+4+5=14$. We note that when we move from $k+(k+1)+(k+2)+(k+3)$ to $(k+1)+(k+2)+(k+3)+(k+4)$, we add 4 to the total (this equals the difference between $k+4$ and $k$ since the other three terms do not change). Therefore, the positive integers that can be expressed as the sum of exactly 4 consecutive positive integers are those integers in the arithmetic sequence with first term 10 and common difference 4. Since $n \leq 100$, these integers are $10,14,18,22,26,30,34,38,42,46,50,54,58,62,66,70,74,78,82,86,90,94,98$. There are 23 such integers. Next, we consider the positive integers $n \leq 100$ that can be expressed as the sum of exactly 5 consecutive positive integers. The smallest such integer is $1+2+3+4+5=15$ and the next is $2+3+4+5+6=20$. Using an argument similar to that from above, these integers form an arithmetic sequence with first term 15 and common difference 5. Since $n \leq 100$, these integers are $15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100$. When we exclude the integers already listed above (30, 50, 70, 90), we obtain $15,20,25,35,40,45,55,60,65,75,80,85,95,100$. There are 14 such integers. Next, we consider the positive integers $n \leq 100$ that can be expressed as the sum of exactly 6 consecutive positive integers. These integers form an arithmetic sequence with first term 21 and common difference 6. Since $n \leq 100$, these integers are $21,27,33,39,45,51,57,63,69,75,81,87,93,99$. When we exclude the integers already listed above $(45,75)$, we obtain $21,27,33,39,51,57,63,69,81,87,93,99$. There are 12 such integers. Since $1+2+3+4+5+6+7+8+9+10+11+12+13+14=105$ and this is the smallest integer that can be expressed as the sum of 14 consecutive positive integers, then no $n \leq 100$ is the sum of 14 or more consecutive positive integers. (Any sum of 15 or more consecutive positive integers will be larger than 105.) Therefore, if an integer $n \leq 100$ can be expressed as the sum of $s \geq 4$ consecutive integers, then $s \leq 13$. We make a table to enumerate the $n \leq 100$ that come from values of $s$ with $7 \leq s \leq 13$ that we have not yet counted: $s$ & Smallest $n$ & Possible $n \leq 100$ & New $n$ \\ 7 & 28 & $28,35,42,49,56,63,70,77,84,91,98$ & $28,49,56,77,84,91$ \\ 8 & 36 & $36,44,52,60,68,76,84,92,100$ & $36,44,52,68,76,92$ \\ 9 & 45 & $45,54,63,72,81,90,99$ & 72 \\ 10 & 55 & $55,65,75,85,95$ & None \\ 11 & 66 & $66,77,88,99$ & 88 \\ 12 & 78 & 78,90 & None \\ 13 & 91 & 91 & None. In total, there are $23+14+12+6+6+1+1=63$ such $n$.