Update README.md

README.md

@@ -112,12 +112,52 @@ In decimal arithmetic, adding the numbers 1 and 1 results in the number 2.
 
 **User:**   
 ```
-Define\n\\[p = \\sum_{k = 1}^\\infty \\frac{1}{k^2} \\quad \\text{and} \\quad q = \\sum_{k = 1}^\\infty \\frac{1}{k^3}.\\] Find a way to write\n\\[\\sum_{j = 1}^\\infty \\sum_{k = 1}^\\infty \\frac{1}{(j + k)^3}\\] in terms of $p$ and $q.
+What is the smallest positive perfect cube that can be written as the sum of three consecutive integers?
 ```
 
 **Thinkless-1.5B:**  
 ```
+<think>
+Okay, let's try to figure out the smallest positive perfect cube that can be written as the sum of three consecutive integers. Hmm, let's start by understanding the problem step by step.
 
+First, a perfect cube is a number raised to the power of 3. So, we need to find the smallest positive integer \( n \) such that \( n = a + (a+1) + (a+2) \), where \( a \) is an integer. Then, \( n \) should also be a perfect cube. Let me write that down:
+
+\( n = a + (a+1) + (a+2) \)
+
+...
+
+Next, 27:
+
+Is 27 a perfect cube? Yes, \( 3^3 = 27 \). Let's check:
+
+...
+**Final Answer**
+The smallest positive perfect cube that can be written as the sum of three consecutive integers is \boxed{27}.
+</think>
+
+To find the smallest positive perfect cube that can be written as the sum of three consecutive integers, we start by expressing the sum of three consecutive integers. Let the three consecutive integers be \( a \), \( a+1 \), and \( a+2 \). Their sum is:
+
+\[
+a + (a+1) + (a+2) = 3a + 3
+\]
+
+This sum must be a perfect cube. Let \( n \) be the perfect cube, so we have:
+
+\[
+n = 3(a + 1)
+\]
+
+Thus, \( n \) must be a multiple of 3. The smallest perfect cubes are 1, 8, 27, 64, 125, etc. We check each of these to see if they can be written as \( 3(a + 1) \):
+
+- For \( n = 1 \): \( 3(a + 1) = 1 \) gives \( a = -\frac{2}{3} \), not an integer.
+- For \( n = 8 \): \( 3(a + 1) = 8 \) gives \( a = \frac{5}{3} \), not an integer.
+- For \( n = 27 \): \( 3(a + 1) = 27 \) gives \( a = 8 \), which is an integer. The three consecutive integers are 8, 9, and 10, and their sum is \( 8 + 9 + 10 = 27 \).
+
+Checking the next perfect cubes (64, 125, etc.) confirms they do not yield integer values for \( a \). Therefore, the smallest positive perfect cube that can be written as the sum of three consecutive integers is:
+
+\[
+\boxed{27}
+\]
 ```