July 2nd, 2015, 02:40 PM  #21 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,662 Thanks: 652 Math Focus: Wibbly wobbly timeywimey 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 
July 2nd, 2015, 10:11 PM  #22 
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus 
The answer is 6, 3*2=6.

July 3rd, 2015, 06:34 AM  #23 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
Do you care, Denis?

July 3rd, 2015, 06:52 AM  #24 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,399 Thanks: 731  
July 12th, 2015, 11:13 AM  #25 
Newbie Joined: Jul 2015 From: ludhiana,punjab Posts: 1 Thanks: 0 
Ans is 6 for sure... 8*7=56 7*6=42 6*5=30 5*4=20 4*3=12 3*2=6 
July 12th, 2015, 12:02 PM  #26 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,399 Thanks: 731 
Geezzzzzzzz......................

January 5th, 2016, 12:17 AM  #27 
Newbie Joined: Jan 2016 From: at home Posts: 2 Thanks: 1 
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(x1), 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 themexcept 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)(π6)x^4(13/60)(π6)x^3+(1/120)(251π1386)x^2+(1/60)(3138533π)x+14(π6) Finally, in general if you want the '?'=k, i.e., f(3)=k where k is the value of your choice, then (1/120)(k6)x^4(13/60)(k6)x^3+(1/120)(251k1386)x^2+(1/60)(3138533k)x+14(k6) More details here: https://www.scribd.com/doc/260182194...tarySequences . . . I know what you're thinking, "But you just generated the pattern using a fourth degree polynomialanybody can do that!" Hence, I present to you a nonpolynomial alternative . . . 
January 5th, 2016, 05:14 AM  #28 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
Yes, given a finite number of data points there exist an infinite number of functions giving those values.

January 6th, 2016, 07:20 PM  #29 
Newbie Joined: Nov 2015 From: Texas Posts: 4 Thanks: 0 Math Focus: Algebra 
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. 
January 28th, 2016, 04:06 AM  #30 
Newbie Joined: Jan 2016 From: new york Posts: 7 Thanks: 0 
if 8*7then 56 7*6=42 6*5=30 5*4=20 so 3*2 will be 6 

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