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dem82 [27]
3 years ago
5

Would any one know how to solve this arithmetic series on the number of magic squares created per order. n=order, A=integer, D=d

ifference between terms. The solution is n=1,2,... are 1, 0, 1, 880, 275305224, ...

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
7 0

Answer:

Step-by-step explanation:

Pachacha [2.7K]3 years ago
6 0

Answer

1.7761 * 10^19 (5*5 magic square estimation)

Step-by-step explanation:

Excluding rotations and reflections, there is exactly one 3×3 magic square, exactly 880 4×4 magic squares, and exactly 275,305,224 5×5 magic squares. 778-783 gives the 880 4 X 4 squares.

9

down vote

Given an n×nn×n magic square, write MnMn for its magic constant. There are nn rows, each of which has sum MnMn, so the sum of all the entries in the square is n⋅Mnn⋅Mn. Each whole number between 11 and n2n2 appears once, so

nMn=1+2+…+n2=n2(n2+1)2,Mn=n(n2+1)2.

nMn=1+2+…+n2=n2(n2+1)2,Mn=n(n2+1)2.

In particular, every n×nn×n magic square has the same constant.

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A wire that is 22 feet long connects the top of a pole to the ground. The wire is attached to the ground at a point that is 10 f
krok68 [10]

✰ <u>Concept</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>

⠀

In this question, we can clearly observer that the diagram shows a right angled triangle. And, we have been provided with the value of base, and the value of hypotenuse, using the pythagoras theorem, now we can easily find out the value of the perpendicular i.e. the value of the side h. According to the pythagoras theorem, square of hypotenuse is equal to the sum of square of perpendicular and square of side respectively. Therefore, square of side is equal to the difference of square of hypotenuse and square of perpendicular.

⠀

✰ <u>Given</u><u> </u><u>Information</u><u> </u>:-

⠀

  • Hypotenuse = 22 ft.
  • Base = 10 ft.

⠀

✰ <u>To Find</u><u> </u><u>:</u><u>-</u>

⠀

  • The value of side or the perpendicular

⠀

✰ <u>Formula</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>

⠀

\star \:  \underline{ \boxed{ \purple { \sf  {Side}^{2}  =  {Hypotenuse}^{2}  -  {Base}^{2}  }}} \:  \star

⠀

✰ <u>Solution</u><u> </u><u>:</u><u>-</u>

⠀

\sf \longrightarrow  {Side}^{2} =   {(22 \: ft)}^{2}  -  {(10 \: ft)}^{2}  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{2} =   {484 \: ft}^{2}  -  {100 \: ft}^{2}  \:  \:  \: \:  \:  \:   \\  \\  \\ \sf \longrightarrow  {Side}^{2} =   {384 \: ft}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{} =   \sqrt{ {384 \: ft}^{2} }  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{} =   \underline{ \boxed{ \frak{ \green{19.60 \: ft}}}} \:  \star \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\

Thus, option B. 19.60 ft. is the correct option.

⠀

\underline{\rule{230pt}{2pt}} \\  \\

4 0
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