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Kevin will start with the integers 1, 2, 3 and 4 each used exactly once and written in a row in any order. Then he will find the sum of the adjacent pairs of integers in each row to make a new row, until one integer is left. For example, if he starts with 3, 2, 1, 4, then he takes sums to get 5, 3, 5, followed by 8, 8, and he ends with the final sum 16. Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4, how many possible final sums are there?

 Jan 19, 2020
 #1
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Kevin will start with the integers 1, 2, 3 and 4 each used exactly once
and written in a row in any order.
Then he will find the sum of the adjacent pairs of integers in each row
to make a new row, until one integer is left.
For example, if he starts with 3, 2, 1, 4, then he takes sums to get 5, 3, 5,
followed by 8, 8, and he ends with the final sum 16.
Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4,
how many possible final sums are there?


I assume  5 possible final sums \(\{16,~ 18,~ 20,~ 22,~ 24 \}\)

 

\(\begin{array}{|cccc|} \hline \text{Start} & a & & b & & c & & d \\ \text{First sum} & & a+b & & b+c & & c+d \\ \text{Second sum} & & & a+2b+c & & b+2c+d & & \\ \text{Final sum} & & & &a+b+c+d+2(b+c)& & & \\\hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{Final sum} &=& a+b+c+d+2(b+c) \quad | \quad a+b+c+d=1+2+3+4= 10\\ \mathbf{\text{Final sum}} &=& \mathbf{10+2*(b+c)} \\ \hline \end{array}\)

 

\(\mathbf{b+c=\ ?}\)

\(\begin{array}{|c|c|c|c|c|c|} \hline \mathbf{b+c} & c= & 1 & 2 & 3 & 4 \\ \hline b=& & & & & \\ \hline 1 && - &3 & 4 & 5 \\ \hline 2 && 3 &- & 5 & 6 \\ \hline 3 && 4 &5 & - & 7 \\ \hline 4 && 5 &6 & 7 & - \\ \hline \end{array} \)

 

\(\begin{array}{|c|c|c|c|} \hline & b+c & 2x & +10 \\ \hline 1 & 3 & 6 & 16 \\ 2 & 4 & 8 & 18 \\ 3 & 5 & 10 & 20 \\ 4 & 6 & 12 & 22 \\ 5 & 7 & 14 & 24 \\ \hline \end{array}\)

 

laugh

 Jan 19, 2020
edited by heureka  Jan 19, 2020
 #2
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heureka: Isn't this the way it is supposed to work?

 

1,2,3,4 =3,5,7 =8,1, 2 =9,3 =12  ??

 Jan 19, 2020

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