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How many combinations of two or more consecutive positive intergers have the sum of 45?
Hello, Ashley!
\(\displaystyle \;\;\)We have: \(\displaystyle n\) and \(\displaystyle n+1\)
\(\displaystyle \;\;\)The equation is: \(\displaystyle \.n\,+\,(n+1)\:=\:45\;\;\Rightarrow\;\;n\,=\,22\)
The numbers are: \(\displaystyle \,\{22,\,23\}\)
Three cons.pos.integers:
\(\displaystyle \;\;\)We have: \(\displaystyle n,\,n+1,\,n+2\)
\(\displaystyle \;\;\)The equataion is: \(\displaystyle \,n\,+\,(n+1)\,+\,(n+2)\:=\:45\;\;\Rightarrow\;\;n\,=\,14\)
The numbers are: \(\displaystyle \,\{14,\,15,\,16\}\)
Four cons.pos.integers:
\(\displaystyle \;\;\)We have: \(\displaystyle n,\,n+1,\,n+2,\,n+3\)
\(\displaystyle \;\;\) The equation is: \(\displaystyle \,n\,+\,(n+1)\,+\,(n+2)\,+\,(n+3)\:=\:45\;\;\Rightarrow\;\; n\,=\,\frac{39}{4}\,??\)
Four consecutive positive integers can <u>not</u> have a sum of 45.
Keep going . . .
\(\displaystyle \;\;\)[You'll find that \(\displaystyle n\,=\,9\) is a solution.]
Two consecutive positive integers:
\(\displaystyle \;\;\)We have: \(\displaystyle n\) and \(\displaystyle n+1\)
\(\displaystyle \;\;\)The equation is: \(\displaystyle \.n\,+\,(n+1)\:=\:45\;\;\Rightarrow\;\;n\,=\,22\)
The numbers are: \(\displaystyle \,\{22,\,23\}\)
Three cons.pos.integers:
\(\displaystyle \;\;\)We have: \(\displaystyle n,\,n+1,\,n+2\)
\(\displaystyle \;\;\)The equataion is: \(\displaystyle \,n\,+\,(n+1)\,+\,(n+2)\:=\:45\;\;\Rightarrow\;\;n\,=\,14\)
The numbers are: \(\displaystyle \,\{14,\,15,\,16\}\)
Four cons.pos.integers:
\(\displaystyle \;\;\)We have: \(\displaystyle n,\,n+1,\,n+2,\,n+3\)
\(\displaystyle \;\;\) The equation is: \(\displaystyle \,n\,+\,(n+1)\,+\,(n+2)\,+\,(n+3)\:=\:45\;\;\Rightarrow\;\; n\,=\,\frac{39}{4}\,??\)
Four consecutive positive integers can <u>not</u> have a sum of 45.
Keep going . . .
\(\displaystyle \;\;\)[You'll find that \(\displaystyle n\,=\,9\) is a solution.]