Maths Puzzle

14/1/2014

17 Comments

Today's Maths Puzzle came from Miss McAndrew's Class on twitter. See if you are a GENIUS!

17 Comments

ashvin

24/10/2014 05:19:51 am

answer
3=9

hari

26/12/2014 05:26:55 am

no itsc 12

ashvin

24/10/2014 05:20:19 am

answer
3=9

zeyda

2/2/2015 05:28:32 pm

3=15

karthika

3/2/2015 02:28:15 am

ans =6

karthika

3/2/2015 02:30:14 am

3*3=9 ,9-3=6 so ans 6

sunil

10/8/2015 10:42:12 am

3=5

maya devi

25/9/2015 11:49:26 am

8*(8-1)=56
7*(7-1)=42

3*(3-1)=6

maria

6/10/2015 10:29:21 pm

8=56 7=42 6=30 5=20 4=12 3=6

pratik

9/10/2015 11:15:02 am

susheela

14/10/2015 01:04:26 pm

3=15

MADHURI

24/10/2015 06:50:46 am

3=4

DARSH

10/11/2015 07:58:21 am

ANS 6

jurgen colman

26/12/2015 08:33:27 pm

3=6

Angel

5/1/2016 09:04:31 am

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/Elementary-Sequences

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

tanuja

1/5/2016 08:06:13 am

Its 6

Piyush nishad

23/12/2016 06:09:04 am

3=9

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