Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

the solution is 10=200

Answer 5

10 = 200

Answer 6

242.  The delta increases by 4 each time

Answer 7

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

Answer 8

200 - inaczej być nie może.

Answer 9

=152

Answer 10

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.

Answer 11

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="

Answer 12

3×3=9 (9×2)=18

4×4=16 (16×2)=32

5×5=25 (25×2)= 50

6×6=36 (36×2)=72

7×7=49 (49×2)=98

Ans-10×10=100 (100×2)=200