If 3 = 18, 4 = 32, 5 = 50, 6 = 72, 7 = 98, Then 10 = ?
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11 Answers

RONG...3 NOT = 18..............
by

Here’s a simple way to look at it.

3x6=18

4x8=24

5x10=50

6x12=72

7x14=98

The pattern is plus 2 each increase.

8x16=128

9x18=162

10x20=200

FYI,

2=8    2x4=8

1=2    1x2=2

by
3 = 18

+1   +14

4 = 32

+1  +18

5 = 50

+1 +22

6 = 72

+1 +26

7= 98

+1 +30

8= 128

+1 + 34

9= 162

+1 +38

10= 200
by
the solution is 10=200
by
10 = 200
by
242.  The delta increases by 4 each time
by
The questions says "3=18".  It does not say "3 is to 18".  Its a code, not a math question.

It has two factual components, first the first digit of the answer is an odd number increasing at each odd number interval.  The second series goes 8,2,0,2,8,...

Based on this information, the answer is: 152

3 = 18

4 = 32

5 = 50

6 = 72

7 = 98

8 = 112

9 = 130

10= 152
by
200 - inaczej być nie może.
by
=152
by
I will be assuming they are using the relation operator for assignment rather than equality, as the former has more interesting consequences.

With a little effort we can come up with say,
(-(1/30))n^5+(5/6)n^4-(49/6)n^3+(247/6)n^2-(459/5)n+84 which results in,
18, 32, 50, 72, 98, 124, 138, 116, . . . .
-------
But,
-------
2n^2 gives
18, 32, 50, 72, 98, 128, 162, 200, . . . .
(The popular choice)
------
However,
-----
n^5-25n^4+245n^3-1173n^2+2754n-2520 gives
18, 32, 50, 72, 98, 248, 882, 2720, . . . .
-----
But then there is,
-----
(8/5)n^5-40n^4+392n^3-1878n^2+(22032/5)n-4032 that gives
18, 32, 50, 72, 98, 320, 1314, 4232, . . . .
-----
And then there is ....
... I can keep going all day, but I'll just stop there.
-----
All these work, take your pick :)
-----
Finally, the point also is that given any finite set of numbers, the set does not necessarily define any single sequence.
by

I will be assuming they are using the relation operator for assignment rather than equality, as the former has more interesting consequences.

With a little effort we can come up with say,
(-(1/30))n^5+(5/6)n^4-(49/6)n^3+(247/6)n^2-(459/5)n+84 which results in,
18, 32, 50, 72, 98, 124, 138, 116, . . . .
——-
But,
——-
2n^2 gives
18, 32, 50, 72, 98, 128, 162, 200, . . . .
(The popular choice)
——
However,
—–
n^5-25n^4+245n^3-1173n^2+2754n-2520 gives
18, 32, 50, 72, 98, 248, 882, 2720, . . . .
—–
But then there is,
—–
(8/5)n^5-40n^4+392n^3-1878n^2+(22032/5)n-4032 that gives
18, 32, 50, 72, 98, 320, 1314, 4232, . . . .
—–
And then there is ….
… I can keep going all day, but I’ll just stop there.
—–
All these work, take your pick <img alt="

by

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