Given a paper on which the numbers $1,2,3\dots ,14,15$ are written. Andy and Bobby are bored and perform the following operations, Andy chooses any two numbers (say $x$ and $y$) on the paper, erases them, and writes the sum of the numbers on the initial paper. Meanwhile, Bobby writes the value of $xy(x+y)$ in his book. They were so bored that they both performed the operation until only $1$ number remained. Then Bobby adds up all the numbers he wrote in his book, let’s call $k$ as the sum. $a$. Prove that $k$ is constant which means it does not matter how they perform the operation, $b$. Find the value of $k$.
Problem
Source: 2nd Final Mathematical Cup Senior Division P3 (2020)
Tags: combinatorics, invariant
01.10.2020 14:11
I suppose that the problem is $1, 2, \dots, 15$. Notice that \[ 3xy(x + y) = (x + y)^3 - x^3 - y^3 \]Let $z$ be the number remaining on the paper. Then, the sum of the number from his book must be \[ \frac{z^3 - 1^3 - 2^3 - \dots - 15^3}{3} \]But we know that $z$ is simply the sum of all the numbers (as the sum is invariant in the paper), which implies $z = \frac{1}{2} \cdot 15 \cdot 16 = 120$. Thus, we have \[ k = \frac{1}{3} (120^3 - (1^3 + 2^3 + \dots + 15^3)) = 571200\]which is a constant. Remark. Same exact idea as Belarus TST 2019
01.10.2020 15:59
Induction also works here.
01.10.2020 16:01
l agree with gghx
08.07.2023 02:51
Miku3D wrote: Given a paper on which the numbers $1,2,3\dots ,14,15$ are written. Andy and Bobby are bored and perform the following operations, Andy chooses any two numbers (say $x$ and $y$) on the paper, erases them, and writes the sum of the numbers on the initial paper. Meanwhile, Bobby writes the value of $xy(x+y)$ in his book. They were so bored that they both performed the operation until only $1$ number remained. Then Bobby adds up all the numbers he wrote in his book, let’s call $k$ as the sum. $a$. Prove that $k$ is constant which means it does not matter how they perform the operation, $b$. Find the value of $k$. 120 is the sum of the numbers from 1 to 15. Therefore,k has a number that is equal to or greater than 120. 1+2=3,3+4=7,5+6=11,7+8=15,9+10=19,11+12=23,13+14=27,15 2(3)=6,12(7)=84,30(11)=330,56(15)=840,90(19)=1710,132(23)=2836,182(27)=4914 Therefore,k=6+84+330+840+1710+2836+4914 and k is constant.