Solve this if you are a genius? This quiz showed up on my Facebook news feed.

8 = 56

7 = 42

6 = 30

5 = 20

3 = ?

The pattern looks pretty easy, right?

8^{2}— 8 = 56

7^{2}— 7 = 42

6^{2}— 6 = 30

5^{2}— 5 = 20

So what is the correct answer?

Here’s mine:

3 = 3 — even if it’s preceded by a bunch of false statements.

Do you agree?

**Update**. This is an *old* post and I thought it had seen all the life it was going to see. Then Angel comes and leaves a comment:

It turns out that the “?” can be whatever we want it to be!

Angel backs up his claim with math. I think we have the definitive answer.

## 106 responses to “Solve this if you are a genius”

Hmmm… Like I said. I don’t do math. 15/3 is not 3. D’oh!

normdeploom (love that name!), I think this is the same pattern many have posted, but you’ve rearranged it slightly. Instead of saying 8 × 7 = 56 you say 56/8 = 7.

What

isdifferent is how you interpret this pattern. For you, the important point seems to be that the number on the right decreases by one for each row. Thus the final row should be 9/3 = 3, with ? = 3 being your final (corrected) answer. Right?If so, you got the same answer as Jon, and for pretty much the same reason.

Thanks for dropping by!

Finally !!!!! I’ve been answering this question one hundred times, and it’s the first time I see someone who’s finally smart enough…. Yes, 3 = 3 …. other statements are false !!!!!

God, it took me more than a year to find you !!!! lol

Patac, I’m glad you found me, and I’m glad I made your day. Fight the good fight! :-D

The answer is 14

56 – 7 – 7 = 42

42 – 6 – 6 = 30

30 – 5 – 5 = 20

20 – 3 – 3 = 14

Yet another pattern. I’m not thrilled how each row is not self-contained, but instead relies on the next. That makes the 8 in the first row maybe arbitrary, and definitely useless. It does sidestep (or maybe just ignore) the skipped 4. That might be a strength instead of a fault.

Thanks for dropping by!

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 + 42and evaluate f(8), f(7), f(6), f(5), and 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 + 42also 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 + 84And here is one where f(3) = ?

f(x) = (1/120)(? – 6)x

^{4}– (13/60)(? – 6)x^{3}+ (1/120)(251? – 1386)x^{2}+ (1/60)(3138 – 533?)x + 14(? – 6)Finally, in general if you want the ‘?’ = k, i.e., f(3) = k where k is the value of your choice, then

f(x) = (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.

Angel, wow! That is

reallycool. And to think I had thought the discussion on this problem had been completely played out.I love how for the k = 6 case, the function simplifies to f(n) = n(n-1).

And btw, I appreciate your using the relation operator in your linked paper. ;-)

Thanks for dropping by and commenting. :-)

Poor horse, but I wanted to add I also calculated a non-polynomial rule.

See here: http://i.imgur.com/BHkg0Ad.png

Brent Logan! You truly do have a great mind. I was almost convinced that 3 does not equal 3! My 6 year old dog was right all along. Thanks for enlightening me.

The amusement continues….

Haha! Your dog sounds pawsome! :-D

Well I didn’t use anyone else’s logic, but I came to the conclusion that the answer is 3. Not even bringing 4 × 3 into the equation but my reasoning is 3 could only equal itself or 0. I’m not great at maths but that’s my answer.

gtt, you had me until you said that 3 could equal 0. But then you recovered by saying you’re not great at maths, so I think (once again) we agree. ;-)

Thanks for dropping by. :-)

8 = 56

7 = 42; 56 – 42 = 14; 14/2 = 7

6 = 30; 42 – 30 = 12; 12/2 = 6

5 = 20; 30 – 20 = 10; 10/2 = 5

3 = 14; 20 – 14 = 6; 6/2 = 3

Nathan, you’re not the first to propose 14, but I think you are the first to use this reasoning. I’ll have to go back and review and see whether or not yours is mathematically equivalent to another’s.

Does it bother you that 4 is skipped? It seems like you could generate the missing row for 4 and then would end up with a different answer for 3.

8 = 56

7 = 42; (56 – (7 × 2))

6 = 30; (42 – (6 × 2))

5 = 20; (30 – (5 × 2))

3 = 14; (20 – (3 × 2))

14

I may also elaborate on the “?” Being whatever we want it to be claim. Originally I figured the solution to be 6, holding no bearing on the equivalence statements. First of all, we are assuming that these numbers lie within the set of Real Numbers in order to apply various properties. This is valid since the integers do indeed lie within. I do want to point out that considering mathematics as a language and taking a crack at this problem it must retain logic. So, by saying that 3=3 is the answer is indeed false under the presumption that this so-called equivalence relation is false which means that the last statement must retain a false statement, so it could be any number not equal to three. There is really no reason to consider otherwise as it would conjure up logic that really isn’t there. In order to consider a sequence, it has to be a function and still abides by the equivalence statement. So, saying it can be anything is indeed illogical. However, assuming there is an underlying approach to be considered, we would have to define what “=” really is as a relation in this case. Consider the problem given:

8 = 56

7 = 42

6 = 30

5 = 20

3 = ?

We know that anything divided by 1 is itself due to the multiplicative identity axiom which means that the remainder of this quotient would be zero.

So, I can arbitrarily define the relation to be “=mod(1)”

So that 0=0 for each case.

Hence, ? Would indeed be any number. One could check to see if any other relation would work, but this one is guaranteed. However, like I said before this method does not retain the logic that should be taken into consideration when taking an unbiased glance at the problem.

8 = 8

7 = 7

6 = 6… Etc

So, 3 = 3

It’s that simple… If 0 = 0, then why 3 can’t equal 3?

You are correct in that every number can only equal itself. However, logic should flow as follows:

“If not B ⇒ not A” then

“A ⇒ B” and this type of logic can work for “if and only if statements”.

So, if we are considering the statements given it can be read as follows:

“If false and false and false and false” then what is the last statement?”

It must be false.

Because, the negated statement should be a true implying the rest being true. This is only logical when “3 = anything but three.”

That way when you negate that statement you have:

“If true, then true and true and true and true”

So that “if 3 = 3, then 5 = 5, 6 = 6, 7 = 7, and 8 = 8”

This is not a math problem. It is a purely a question of logic. Establish the basic pattern working from bottom up. Double the number on the left side of the equal sign and add it to the number in the right side, the sum of which equals the number above it. Repeat working upwards. Works going down too but subtracting obviously. And the creators left the 4 out for a reason, you can’t just add it in inorder to make your pattern work. ? = 14.

However, your solution does not utilize logic. Also, math is built off of logic, so they go hand-in-hand. You are assuming that there is a pattern which assumes the false equalities are indeed true which is contradictory. This is what I was attempting to explain previously. The most logical approach is to assume that in order to obtain a string of true statements one has to start with the false statement since that is what is implying the following statements.

It is also a simplification to state that the 4 is left out for a reason. Take for instance the sequence give, as previously mentioned.

f(n) = n(n-1) ⇒ f(8) = 56, f(7) = 42, f(6) = 30, f(5) = 20, f(4) = 12, f(3) = 6.

So, if one wants to illogically create a false equality between n and n(n-1) then you have a valid pattern. In general, most patterns do not include every term up to the term that is desired. Take for instance a series problem stating that the first term of the series is, say, 1 and that each consecutive term is increase by 8. Let n be contained within the natural numbers union with the singleton set 0. This allows n = 0, 1, 2, …

Now, find the 25th term of the series.

Writing out this series we have that the nth term can be found as:

A0 = A0, A1 = A0 + 8, A2 = A1 + 8 = A0 + 2(8), A3 = A2+8 = A0 + 2(8) + 8 = A0 + 3(8), ……. An = A0 +n(8)

The first term is A0 = 1 which is given for n = 0. Hence, the 25th term is determined for n = 24. This implies that A24 = 1 + 24(8).

So, we could still determine this pattern if we were given every term up to n – 1, because we have generalized the series. We are also able to find every term in between some kth term and the nth term. This is how patterns work.

Yea, pretty sure it comes down to intent, or what the intent of the creator was originally.

8 = 56

7 = 42

6 = 30

5 = 20

3 =

We can assume the author wanted us to multiply starting with 3 similar to the following:

8 × 7 = 56

7 × 6 = 42

6 × 5 = 30

5 × 4 = 20

3 × 3 = 9

But I do not believe that would be intuitive or logical to assume that would and could be the only right answer. Please let me explain.

8 × 7 = 56

7 × 6 = 42

6 × 5 = 30

5 × 4 = 20

4 × 3 = 12

3 × 2 = 6

As you can see, the pattern remains the same. 3 × 4 = 12; 4 × 5 = 20; 5 × 6 = 30, and so on. I say it comes down to the intent of the author.

I agree that there can be various interpretations, but I was trying to establish the irrefutably most logical solution. Irrespective of the intent of the author. The reason being that conjuring up any particular pattern is in contradiction to the logic set in place.

U tell brent wht wl b d ans???

Haha! I was going to tell you to read through all of the comments to find the answer, but I reread the post first. I think you’ll find my answers (both of them) near the end.

Thanks for dropping by and commenting. :-)