Find consecutive odd number whose sum is 645
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Answer:
- 213, 215, 217
- 125, 127, 129, 131, 133
- 29, 31, 33, 35, ... ,57
Step-by-step explanation:
Assume that we have [tex]k[/tex] consecutive odd numbers. Let [tex]n[/tex] be the first odd number, then [tex]n+2,\: n+4, \: . \: .\: .\: , \: n+(k-1)2[/tex] will be the next odd numbers.
The sum of these consecutive odd numbers is equal to:
[tex]\begin{gathered} n+(n+2)+(n+4)+. \: . \: .+ n+(k-1)2=645 \end{gathered}[/tex]
[tex]\begin{gathered} \dfrac{k}{2}(2n+(k-1)2)=645 \quad \textsf{(using the A.P sum formula)} \end{gathered}[/tex]
[tex]\dfrac{k}{2}[2(n+k-1)]=645[/tex]
[tex]k(n+k-1)=645[/tex]
Since [tex]n[/tex] and [tex]k[/tex] must be positive integers, [tex]k[/tex] and [tex]n+k-1[/tex] should be factors of 645. Therefore, [tex]k[/tex] and [tex]n+k-1[/tex] may be equal to the factors in pairs of 645. Note that [tex]k[/tex] could not be too small nor too large, so we neglect the factors in pairs (1, 645), (645, 1), (215, 3), (129, 5), (43, 15). The only possible solutions are:
[tex]\begin{gathered} (k,n+k-1)=(3,215),(5,129),(15,43) \end{gathered}[/tex]
Solving for [tex]k[/tex] and [tex]n:[/tex]
[tex]\begin{array}{c|c|c}(k,n+k-1)=(3,215)& (k,n+k-1)=(5,129)& (k,n+k-1)=(15,43) \\ k=3, n+k-1=215 & k=5,n+k-1=129& k=15,n+k-1=43 \\ k=3,n+3-1=215 & k=5,n+5-1=129 & k=15,n+15-1=43 \\ k=3,n+2=215 & k=5,n+4=129& k=15,n+14=43 \\ & & \\ \boxed{k=3, n=213} & \boxed{k=5,n=125} & \boxed{k=15,n=29}\end{array}[/tex]
Therefore, there are 3 possible sequence of consecutive odd numbers:
- 213, 215, 217
- 125, 127, 129, 131, 133
- 29, 31, 33, 35, ... ,57
