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July 2nd, 2015, 01:40 PM | #21 |
Math Team Joined: May 2013 From: The Astral plane Posts: 1,764 Thanks: 707 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 |
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus |
The answer is 6, 3*2=6.
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July 3rd, 2015, 05:34 AM | #23 |
Math Team Joined: Jan 2015 From: Alabama Posts: 3,101 Thanks: 850 |
Do you care, Denis?
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July 3rd, 2015, 05:52 AM | #24 |
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,117 Thanks: 800 | |
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July 12th, 2015, 10:13 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 |
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July 12th, 2015, 11:02 AM | #26 |
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,117 Thanks: 800 |
Geezzzzzzzz......................
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January 4th, 2016, 11:17 PM | #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(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, 04:14 AM | #28 |
Math Team Joined: Jan 2015 From: Alabama Posts: 3,101 Thanks: 850 |
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, 06:20 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 06:22 PM. |
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January 28th, 2016, 03:06 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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can you solve this,can u solve this,solve this,can you solve,can u solve,CAN YOU SOLVE THIS?,you can solve this,solve this picture,Solve this..,7x8=56 6x7=42 5x6=30 3=?,Are you able to solve this question,can u solve this,5x6, 6x7 7x8 is what kinds of maths,can you solve this math??,do u solve it
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