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Find the tactic and solve the puzzle, simple. Solve if you are Genius, check the puzzle image and find which number equals to three. Share your answer through comments.

#### Puzzle Question:

⇒ 8 = 56

⇒ 7 = 42

⇒ 6 = 30

⇒ 5 = 20

⇒ 3 = ??

**Options:** A) 15, B) 12, C) 06

## Math Puzzles Image – Genius Puzzles

Got it? share your answer via comments. Let see what you got! Keep scrolling for the answer. You can also check the video with solution.

**Puzzle Video:**

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⇒ 8 = 56 ⇒ 8 × 7 = 56

⇒ 7 = 42 ⇒ 7 × 6 = 42

⇒ 6 = 30 ⇒ 6 × 5 = 30

⇒ 5 = 20 ⇒ 5 × 4 = 20

⇒ 3 = ?? ⇒ 3 × 2 = 6

**Answer =** **6**

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It turns out the ‘?’ can be whatever we want it to be!

Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

f(8)=56,

f(7)=42,

f(6)=30,

f(5)=20,

f(3)=9.

Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them–except with a different value of f(3)!

Here’s another one that also works but gives f(3)=12:

f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

And here is one where f(3)=π

f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

Finally, in general if you want the ‘?’=k, i.e., f(3)=k where k is the value of your choice, then

(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

More details here: https://www.scribd.com/doc/260182194/Elementary-Sequences

For a non-polynomial rule see here: http://i.imgur.com/BHkg0Ad.png