Henyo74idn
Answered

Kumonekta sa mga eksperto at makakuha ng mga sagot sa IDNStudy.com. Ang aming mga eksperto ay nagbibigay ng mabilis at eksaktong sagot upang tulungan kang maunawaan at malutas ang anumang problema.

Find consecutive odd number whose sum is 645​

Sagot :

Answer:

To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 645

3X + 3 = 645

3X + 3 - 3 = 645 - 3

3X = 642

3X/3 = 642/3

X = 214

Which means that the first number is 214, the second number is 214 + 1 and the third number is 214 + 2. Therefore, three consecutive integers that add up to 645 are 214, 215, and 216.

214 + 215 + 216 = 645

We know our answer is correct because 214 + 215 + 216 equals 645 as displayed above.

Find consecutive odd number whose sum is 645

ㅤ

For Brainly app users, refer to attached photo for the full solution.

ㅤ

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
View image Devoured
Natutuwa kami na ikaw ay bahagi ng aming komunidad. Huwag kalimutang bumalik upang magtanong at magbahagi ng iyong karanasan. Ang iyong kaalaman ay mahalaga sa ating komunidad. May mga katanungan ka? Ang IDNStudy.com ang may sagot. Salamat sa iyong pagbisita at sa muling pagkikita.