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July 2nd, 2015, 01:40 PM   #21
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Math Focus: Wibbly wobbly timey-wimey stuff.
C'mon people! It's a trick question. It is intuitively obvious that f(3)= 87653 (note the pattern of 87653.) Clearly we can immediately say:
$\displaystyle f(x) = \frac{87647}{120} x^4 - \frac{1139411}{60} x^3 + \frac{21999517}{120} x^2 - \frac{46715911}{60} x + 1227058$

Thus f(3) = 87653.

-Dan
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July 2nd, 2015, 09:11 PM   #22
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Math Focus: Calculus
The answer is 6, 3*2=6.
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July 3rd, 2015, 05:34 AM   #23
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Do you care, Denis?
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July 3rd, 2015, 05:52 AM   #24
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Originally Posted by Country Boy View Post
Do you care, Denis?
Deeply...losing sleep over this...
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July 12th, 2015, 10:13 AM   #25
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Ans is 6 for sure...
8*7=56
7*6=42
6*5=30
5*4=20
4*3=12
3*2=6
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July 12th, 2015, 11:02 AM   #26
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Geezzzzzzzz......................
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January 5th, 2016, 12:17 AM   #27
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It turns out that 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...tary-Sequences

. . . I know what you're thinking, "But you just generated the pattern using a fourth degree polynomial--anybody can do that!" Hence, I present to you a non-polynomial alternative . . .

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January 5th, 2016, 05:14 AM   #28
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Yes, given a finite number of data points there exist an infinite number of functions giving those values.
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January 6th, 2016, 07:20 PM   #29
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This is the basic question in Maths. The simplest way of answer it is already posted.

Just look out the question,

8=56
7=42
6=30
5=20
3=?

If you see that the answer comes from the descending multiplication way.

8=56 (8*7=56)
7=42 (7*6=42)
6=30 (6*5=30)
5=20 (5*4=20)
4=12 ( 4 is missing here, just put 4)
3=? the answer 6 because 3*2=6

If you want to learn more tricky questions. Connect with a tutor who helps you to solve these questions and help in your exams and as well as in homework.

Last edited by greg1313; January 6th, 2016 at 07:22 PM.
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January 28th, 2016, 04:06 AM   #30
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if 8*7then 56
7*6=42
6*5=30
5*4=20
so
3*2 will be 6
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