8=56; What does 3=? Answer.
There is a puzzle plaguing the very fabric of sanity on Facebook and the rest of social media. The debate over the answer rages on. For those here only looking for the answer, scroll down below the quote box. If you want to try to solve it first, stop reading now.
Solve if you are Genius!
8=56
7=42
6=30
5=20
3= ?
The answer is 3. If your timelines or feeds are like mine you have no doubt seen this puzzle. Another form of this quiz adds “9=72” in the list of statements and replaces the cartoon Einstein with Calvin from the Calvin and Hobbes comic strip. Responses range from potential answers of 3, 6, 9, 14 and 42 (referring to The Hitchhiker’s Guide To The Galaxy) to long debates and explanations of how they arrived at their answer. Most people agree on 6. An Internet search has yet to turn up the source of the problem. Apparently one iteration of this puzzle image has over 25,000 retweets on Twitter and over 16 million Twitpic views. It is a defining question of our generation!
As I stated above, the answer is 3. The reason is because the preceding statements are false statements to make you think there is a pattern to mislead you. 9 never equals 72, 8 never equals 56, and so on. 3 never equals anything other than 3. The problem doesn’t set up a premise by saying “If these statements are true then what does 3 equal?” It just presents you with incorrect answers and in an effort to solve the problem we look for a pattern. It’s similar to the kid’s word joke that leads you to the wrong answer so everyone can laugh at you.
Spell mop……
Spell hop…….
Spell pop…….
What do you do at a green light……..
Raise your hand if reading that you thought “stop” when of course the answer is “go”. It’s the same with the math problem, you are expecting to follow the pattern instead of paying attention to what you are being asked. It’s misdirection, magicians rely on it. Neil Strauss, author of The Game, demonstrates this idea using 3 coins.
Looking over the discussions, responses, and reasoning going on in these threads to try to make the answer be anything other than 3 and defending their own rationalizations, I realized it was sort of a mirror to the way we think. Even when presented with the logical answer we many times will ignore the logic, facts, and truths and argue the incorrect point we believe. We see this all the time. It reminds me of pretty much every political argument that has ever taken place. Even when presented with facts that counter their reality people will emotionally cling to the false belief or point and argue it until they are hoarse. It is something we all have done at one time or another.
Mitt Romney’s recent presidential campaign was famously guilty of this.
Mitt Romney’s campaign said on Tuesday that its ads attacking President Obama’s waiver policy on welfare have been its most effective to date. And while the spots have been roundly criticized as lacking any factual basis, the campaign said it didn’t really care.
“We’re not going to let our campaign be dictated by fact-checkers,” Romney pollster Neil Newhouse said at a panel organized by ABC News.
This is a different standard than the one Romney himself has held up for the election-season ad wars. Reacting to attacks by a pro-Obama super PAC, Romney recently told a radio station that “in the past, when people pointed out that something was inaccurate, why, campaigns pulled the ad.” huffingtonpost.com Mitt Romney Campaign: We Will Not ‘Be Dictated By Fact-Checkers’
The point is, sometimes we need to take a step back, assess the situation and question what it is we are arguing and be open to the possibility that might be incorrect.
For example, this whole post could be me doing exactly what I observed others doing in defense of their answers. I could be incorrect about the answer being 3 but I’m not.
Because I’m genius.
Jaylon Carter is a blogger, social media marketing consultant, former Congressional Campaign Media & Communications Director, national labor union vice block leader, and a Hip Hop artist who performs under the stage name Timid (@timidmc).
Doesn’t everybody know that everything is always, has been always and will be always four?
5+3=28
9+1=810
8+6=214
5+4=19
Then 7+3= ??
=10. 7+3 always equals ten.
Unless the assignment operator is being used for something other than equality. Just because we were only taught one of its uses, namely equality, in elementary school, it doesn’t mean we get to tunnel our vision to match what we have become comfortable with because we were only taught one aspect of a more general operator.
If it’s not specified that it is being used for something else then it is what it is.
Correct! It is an assignment operator, thus, unless we specify we are using it for equality it is just that–an assignment operator. We have become so accustomed to the masses’ use of this assignment operator, that we immediately assume and generalize it as implying equality. It is this tunneling of our vision that limits us from accepting in many cases the facts that surround us, and that also keep us from having an open mind. Sometimes it is ok to accept we don’t know everything, and there is more out there than what we are used to, or have learned–even if we don’t like it, or understand it.
Why can’t it just as simple as it looks
8=56. The 8 is multiplied by the lower diget witch is 7
7=42 the 7 is done the same multiplied by the lower diget
6=30
5=20 now they didn’t do 4 but the answer they gave is 20 so that tells u they multiplied it but 4 the missing number so logically the answer for next one
3=[ ]. Should be 4 multiplied by 3 witch makes the answer 12
the answer is 9
Follow the PEMDAS Order of Operations and the answer comes to 14. It is very simple. I can give the flow but I need readers to use PEMDAS ORDER and come to the correct number of 14.
LOL!!! Dr. of what Hahaha . . .
. . .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
06
The answer is 11
Indian mathematics champion , Siddhant..
Unfortunately, India must not hold very high standards for their champions since 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
Yes , the answer is 6 as – 9×8=72 , 8×7=56 , 7×6=42 , 6×5=30 , 5×4=20 , 4×3=12 , so , With 3 = 3×2=6………………. Thus , 6 is the answer.
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
Well,it’s simple
8=56,(8×(8-1))
7=42,(7×(7-1))
…and so on
therefore:3=6,(3×(3-1))
because the problem didn’t forced us to follow the pattern so we confused when after 5,next number is 3,therefore think that [4×(4-1)=12] but its false,and the main idea of this problem is to us from this pattern👉[n.(n-1)=?].So the answer is:6
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
3=3, just because someone got the rest of answers wrong does not mean you have to me wrong too. Use your brain. 1+1=2, if I said 1+1=5 and then what does 1+3=?, if you put anything other than 4, then you are saying I can not add too. Stop going with the flow even if they are wrong.
James gets it.
Correct! It is an assignment operator, thus, unless we specify we are using it for equality it is just that–an assignment operator. We have become so accustomed to the masses’ use of this assignment operator, that we immediately assume and generalize it as implying equality. It is this tunneling of our vision that limits us from accepting in many cases the facts that surround us, and that also keep us from having an open mind. Sometimes it is ok to accept we don’t know everything, and there is more out there than what we are used to, or have learned–even if we don’t like it, or understand it.
Consequently:
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
8=56
7=42 (56-14=42)
6=30 (42-12=30)
5=20 (30-10=20)
4=12 (20-8=12)
3=6 (12-6=6)
6
3=15
3=12
8×7=56
7×6=42
6×5=30
5×4=20
4×3=12
8×7=56
7×6=42
6×5=30
5×4=20
3×4=12
Definately, answer is 6..
8=56 5+3
7=42 4+3
6=30 3+3
5=20 2+3
3=00 0+3
so answer is 0
This is the very CLEAR answer to this!
(72-8)-8=56
(56-7)-7=42
(42-6)-6=30
(30-5)-5=20
(20-3)-3=14
So this is the best answer….nothing to explain further! 14 Think about it 😉
Most of you are completely wrong. There is a perfectly easy problem. You just have to c the simple patterns the answer is 6
3
if the question is wrong in itself then there is absolutely no wrong answer, therefore whoever made this isnt very smart, and no matter what answer you give, you are neither wrong nor right
I say 14, this is how I got to it..
8=56
7=42 –56 -(7*2)=42
6=30 –42 -(6*2)=30
5=20 –30 -(5*2)=20
20 -(3*2)=14
3=?(14)
It’s as logical as any…
its Real.
let me try.
you first multiply 8×7=56
then you again multiply 7 by 6=42
in a descending manner.
you do not consider 4 since is not there.
then you multiply 3×2 thus gives you SIX.
ANS=6.
I think its ans is 12
8=56
7=42
6=30
5=20
4 is not given then why shld consider 4? Forget it.
3=12
Bcos..8×5=40. So add 2=42
7×4=28. Add 2 =30
6×3=18. Add 2 =20
5×2=10 add 2 =12
So 3= 12
😊😊
Perhaps your assumption that they are trying to mislead you is incorrect? Unless you made up this post, it’s really impossible to know. They give patterns like this in school all the time, just because there is an absence of the words “if” and “then” doesn’t mean that the pattern does not exist. The point of many patterns is to figure out how they got the answers. What if they took away the entire left side of the equation and only had the answers, would you see a pattern then?
I’m sure you’ll argue that too, and so will many others because apparently I’m “not smart” for seeing this pattern.
In case you were wondering, I got 6 with this non-existent pattern. Thanks for this little article on why you think I’m wrong though.
Everyone who got a different answer then 6 is wrong we were all taught in elementary school how to find the common pattern and then use what we found out to get the correct answer. Just because they omit 4 from the pattern doesn’t mean the number is gone. 8×7=56 7×6=42 6×5=30 5×4=20 so 4×3=12 and then your answer is 3×2=6 in the great words of Spock”its only logical.”
Unfortunately, you got stuck with only an elementary school education. It’s only logical to realize 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
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)=3, 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
Errata: the popular rule f(x)=x(x-1) gives f(x)=6, not f(x)=3.
It is a puzzle to figure out the pattern. The omission of 4= is to take the attention from 5 to 3. Therefore, giving the wrong answer. You do not need to be a genius to figure it out. Just a logic question to see who’s paying attention to no. 4.
Correct!
There is No Pattern… There is No Answer…
Finally someone with some common sense
Then of course if there were a 4 = 12, then you would have to assume the next
line correctly is…
3 = 6
there is a pattern
72-56=16
56-42=14
42-30=12
30-20=10
20-X=8
x=12:
56-42=14÷7=2
42-30=12÷6=2
30-20=10÷5=2
20- 14 = 6 ÷3=2
Answer is 14
6
X * (X-1) = ?
This is the pattern.
So,
3 * (3-1) = 6
8=56 8×8=64 64-8=56
9×9=81 81-9=72
7=7×7=49-7=42
6=6×6=36-6=30
5=5×5=25-5=20
similarly u will get
3×3 that is 9 and 9-3=6
then 6 is the answer
another way is
9=9×8=72
8=8×7=56
7=7×6=42
similarly u will get 3×2=6
therefore 6 is the answer and will be the answer
8 x 7 =56
7 x 6=42
6 x 5=30
5 x 4 =20
So
3 x 3 = 9 is the answer!
I agree with Brad Smith. That’s how I used to see the figures…
7 x 8 = 56
6 x 7 = 42
5 x 6 = 30
4 x 5 = 20
3 x 4 = 12
2 x 3 = 6 <—-The Answer is 6 !
9=72 (Incorrect)
8=56 (Incorrect)
7=42 (Incorrect)
6=30 (Incorrect)
5=20 (Incorrect)
3=3 (Correct)
Da answer is 6
9=72
8=56
7=42
6=30
5=20
3=6
Evry one know da mathematical formula n=n*(n-1)
so9*(9-1)=72
8*(8-1)=56
7*(7-1)=42
6*(6-1)=30
5*(5-1)=20
3*(3-1)=6
8 = 56
7 = 42 = 56 – 14 , 14 = 7+7
6 = 30 = 42 – 12, 12 = 6+6
5 = 20 = 30 – 10, 10 = 5+5
There is no 4
3 = 14 = 20 – 6, 6 = 3+3
Sometimes the relation operator ‘=’ is used for assignment rather than equality. However you are also correct. . .
https://www.scribd.com/doc/260182194/Elementary-Sequences
Of course, ‘=’ does not mean ‘equals’ – it’s just a placeholder for a number. So what numbers for ‘=’ make a sequence with explainable differences? – and also accounting for the missing ‘4’ ? ..
9472
8356 diff 1116
7242 diff 1114
6130 diff 1112
5020 diff 1110
3912 diff 1008
So the answer is 12 – and as I have not seen this explanation given anywhere else I assert this is the unique correct answer and I claim my own genius prize 🙂
*divide aha
9=72
8=56
7=42
6=30
5=20
3=?
I certainly see a pattern, whether it’s intended or not I don’t know. I think 9, 8, 7, 6, 5, 3, function as just being the 9th number, the 8th number, the 7th number and so on. I don’t think they have any value in regards to adding, multiplying or subtracting or whatever. Anyway here it goes
The amount between 72 and 56 is 16, cool. Then the amount between 56 and 42 is 14, ok. And the amount between 42 and 30 is 12, and now the pattern is forming…..
72 – 16 = 56 (the 8th number)
56 – 14 = 42 (the 7th number)
42 – 12 = 30 (the 6th number)
30 – 10 = 20 (the 5th number)
Here the 4th number is omitted, I think to throw the person, or, to devide genius from the not so genius 😉 , so if I continue with the pattern….
As we can see in the pattern with 16, 14, 12, 10, it’s fair to assume the next number must be 8 so…………
20 – 8 = 12 (the 4th number)
12 – 6 = 6 (the 3rd number)
So……
9=72
8=56
7=42
6=30
5=20
4=12 (omitted number)
3=6
To the author: thank you for this. Unfortunately, you will come to realize, as I have, that the majority of people are not very smart. I don’t mean that in a bad way, they aren’t useless or completely stupid. It is just disappointing to see the great lengths at which someone will defend a baseless opinion or assumption. Simple examples like this are one reason why I no longer believe in democracy.
“Just think of how stupid the average person is, and then realize half of them are even stupider!” – George Carlin
I’m simple I guess I think the answer is twelve because if you multiply each number by the one underneath it is how I see it
I’m simple I guess I think the answer is fifteen because if you multiply each number by the one underneath it is how I see it
今度こそ３＝７２にするための出題
9=72
8=56
7=42
6=30
5=20
3=?
３＝７２
答えは７２
にするための出題
9=72 ９×８＝７２・・・９：７２
8=56 ８×７＝５６・・・８：５６
7=42 ７×６＝４２・・・７：４２
6=30 ６×５＝３０・・・６：３０
5=20 ５×４＝２０・・・５：２０
3=? 【４×３＝１２・・３：７２・・１２＋６０＝７２】
2=66・・・【２×３＝６・・２：６６・・６＋６０＝６６】
4=ー１１８・・・【１×２＝２・・・４：？？・・２＋(ー１２０)＝ー１１８・・４；ー１１８】
ならば
答えは７２
この出題なら凡人の私も納得できる
これだけ考えてる私って・・・つくずくアホですね～＾＾
今度こそ３＝７２にするための出題
9=72
8=56
7=42
6=30
5=20
3=?
３＝７２
答えは７２
にするための出題
9=72
8=56
7=42
6=30
5=20
3=?
2=66・・・２×３＝６・・・６＋６０＝６６
4=ー１１８・・・１×２＝２・・・２＋(ー１２０)＝ー１１８
答えは７２
この出題なら凡人の私も納得できる
これだけ考えてる私って・・・アホですね～＾＾
x(x-1) applies to all lines.
9*9-9 = 72
8*8-8 = 56
7*7-7 = 42
6*6-6 = 30
5*5-5 = 20
3*3-3 = 6
8=56
7=42
6=30
5=20
4=12
3=6 is the and because 8=56-14=7=42-12=30………..so3=6
Simplest is: x=x*x-x
When you assume ‘what if 8=56’ etc.
The answer is 6 also when you visualize numbers so that the number on right is divided by the number under it on left, e.g. 56/7.
So, you get this:
8=56/7
7=42/6
6=30/5
5=20/4
4=a?/3
3=b?/2
a=12
b=6
And proof is: x=x*(x-1)/(x-1)
E.g. 6=30/5 or 6=6*5/5
Meaning x=x
😉
That’s my point Boyan. It doesn’t say “assume that” or “what if” that’s why it’s a riddle because people will automatically make the false assumption. Thanks for your comment.
3 is the most stupidest of all the possible answers…yes, most stupidest. There are multiple answers and that is the answer. It would not be logical to assume there are misstatements in a quiz. 8=56 etc cannot be false in the context of asking a question. If there is an error in the question figuring the correct answer could only be random. 3 is the childish silly answer. Any answer can be correct, even “undefined” or “wheels on a tricycle”
8=56 is false when asking a question without the conditional “if”.
8×7=56
7×6=42
6×5=30
5×4=20
4×3=12
3×2=6
“Has ‘genius’ been redefined?”
As demonstrated here and by the comments we’re left to figure out what the originator intended. Analytics might wonder where is 4=?. Perhaps what we have here really is 4*3=12 with the 4 being unstated, so 3 = 12. Kindergartners will say two wrongs, or in this case 4 wrongs, do not make a right. We all know 8 does not equal 56 and 7 does not equal 42. So of course 3 = 3. Or perhaps majority rules even to the point of determining what the rules actually are, so then the answer from the stream looks to be 6. Yet, if majority is the answer am left to ponder “Has ‘genius’ been redefined?”
I think in both the usage of “genius” in my response and the original challenge is tongue-in-cheek. I even left that door open saying that I possibly could be guilty of defending an incorrect premise but I don’t think so. The wording doesn’t say “what if” or “let”s say”. In my view, it presents several incorrect statements attempting to mislead the challenger. The “genius” will recognize this as the case and ignore them.
Thanks for the reply. Diana you are making an assumption to fit a misleading pattern. The problem sets up false statements to confuse you. It doesn’t say 8 (x what) =56 it just says 8 = 56. That is a false statement to send you down the path that you are on so that you don’t see that 3 = 3.
Oops. Top line should say
Correct answer is 9
The answer is 6
8 (x 7) =56
7 (x 6)=42
6 (x 5)=30
5 (x 4) =20
So
3 (x 3) = 9 is the answer!
Though one could argue that the answer could be 6
3 (x 2)= 6 If one were to argue that the next logical starting number after 5 should have been 4 in which case then 4 (x 3)= 12 is missing and then 3 (x 2) = 6 would follow as the next logical step in the sequence
But since no blank line is shown, I would lean more towards the correct answer being 9
🙂
Wrong. You’ll only be right if 5=15, but in fact 5=20. So the right answer is 3-6.
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
Seems like the question is more about exposing reproductive vs productive thinking.
If you follow a prayer it would be six
i dono