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# A new Maths Problem

infinisa
Hello Folks

Here is a maths problem I dreamt up this morning whilst in the hills of northern Portugal.

Fill in the missing number in this sequence:

1, 4, __ , 256

Your solution must explain WHY you have chosen the number!
boinsterman

1*1 = 1.

2*2 = 4.

4*4 = 16.

!6*!6 = 256.
I say the answer is 27.

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256
infinisa
 boinsterman wrote: The answer is 16. 1*1 = 1. 2*2 = 4. 4*4 = 16. !6*!6 = 256.

Hello boinsterman

This is fine except for the first term. How does 1*1 = 1 fit in with the other numbers?

Good try...
infinisa
 Peterssidan wrote: I say the answer is 27. 1^1 = 1 2^2 = 4 3^3 = 27 4^4 = 256

Well done Peterssidan, that is spot on!

Really simple, isn't it?

I started with the number 27, immediately thought of 3^3, and then built the sequence.

Then I noticed that when I removed the 27 (leaving the blank), what's left is extremely misleading, as it appears that the sequence is about powers of 2, when in fact it has nothing to do with this!

As I just made up the problem, I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?
lucascoelho
Ok, it was a nice warm up, but if you guys really like solving math problems you should join the Project Euler. That's the real stuff; endless fun for math-addicted!
kelseymh
infinisa wrote:
 Peterssidan wrote: I say the answer is 27. 1^1 = 1 2^2 = 4 3^3 = 27 4^4 = 256

Well done Peterssidan, that is spot on!

Really simple, isn't it?

I started with the number 27, immediately thought of 3^3, and then built the sequence.

Then I noticed that when I removed the 27 (leaving the blank), what's left is extremely misleading, as it appears that the sequence is about powers of 2, when in fact it has nothing to do with this!

As I just made up the problem, I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?

16 almost works. 2^0 = 1, 2^2 = 4, 2^4 =16, 2^8 = 256. It's a nice pattern (the exponents are powers of two), except for the first term, which should be 2^1 = 2 (or you should have 1, 2, 4, 16, 256, ... as your sequence).
 infinisa wrote: I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?
There are infinitely many solutions just that some are a bit more complicated than others. Here is something I came up with that gives the sequence 1, 4, 16, 256

2^(2^0)-0^0 = 1
2^(2^1)-0^1 = 4
2^(2^2)-0^2 = 16
2^(2^3)-0^3 = 256
asnani04
It's 3*3 = 27. Quite a nice problem, and as you said, if you remove the 27, it's quite misleading too. I liked this. Such kind of questions appear regularly in National and International level exams in Mental Ability, and so I have some experience of solving them and I like them. Keep posting problems like this, it would be a good mental exercise for all of us here.
cybersa
Nice Math problem.
I'm going to tell this to my friends.
therimalaya
Nice Problem, for only these four numbers, multiple solution exists and the multiple solution will eventually decrease to unity when the sequence move forward to infinity.
zhybsc
i think it is 5.
beacuse 5 is after 4. i do not care 256 where it is ....
zhybsc
ok go to bed
good night !
_AVG_
 infinisa wrote: I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?
There are infinitely many solutions just that some are a bit more complicated than others. Here is something I came up with that gives the sequence 1, 4, 16, 256

2^(2^0)-0^0 = 1
2^(2^1)-0^1 = 4
2^(2^2)-0^2 = 16
2^(2^3)-0^3 = 256

This is a nice observation, but I don't understand the need for the powers of zero (or did you just put them in for fun? )
You could have just written:
2^(2^0)
2^(2^1)
2^(2^2)
2^(2^3)

The sequence you have described is interesting for the following reason. An arithmetic series goes as:
a_n = a_1 + n*x [this involves addition and multiplication]
A geometric series:
a_n = a_1 * (x^n) [this involves multiplication and exponentiation]
a_n = a_1^(x^n) [this involves exponentiation and tetration i.e. the next hierarchical operation]

Interesting ...
kelseymh
[quote="_AVG_"]
 infinisa wrote: I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?
There are infinitely many solutions just that some are a bit more complicated than others. Here is something I came up with that gives the sequence 1, 4, 16, 256

2^(2^0)-0^0 = 1
2^(2^1)-0^1 = 4
2^(2^2)-0^2 = 16
2^(2^3)-0^3 = 256

This is a nice observation, but I don't understand the need for the powers of zero (or did you just put them in for fun? )
You could have just written:
2^(2^0)
2^(2^1)
2^(2^2)
2^(2^3)
[quote]

In fact, those are not the same. 2^0 = 1, so 2^(2^0) = 2. However, 0^0 = 1 (take the limit x^x as x->0, in the complex plane; it converges to unity from both directions, and the phase goes to zero). Consequently, the first term in the posted series is 1, not 2. All of the finite integer powers of zero are zero.
_AVG_
[quote="kelseymh"][quote="_AVG_"]
 infinisa wrote: I couldn't be sure that this was the only solution. In fact, a friend of mine phoned me this evening with a perfectly good justification for the missing number to be 16. Of course, his pattern was not as simple as mine. Can you find it?
There are infinitely many solutions just that some are a bit more complicated than others. Here is something I came up with that gives the sequence 1, 4, 16, 256

2^(2^0)-0^0 = 1
2^(2^1)-0^1 = 4
2^(2^2)-0^2 = 16
2^(2^3)-0^3 = 256

This is a nice observation, but I don't understand the need for the powers of zero (or did you just put them in for fun? )
You could have just written:
2^(2^0)
2^(2^1)
2^(2^2)
2^(2^3)
 Quote: In fact, those are not the same. 2^0 = 1, so 2^(2^0) = 2. However, 0^0 = 1 (take the limit x^x as x->0, in the complex plane; it converges to unity from both directions, and the phase goes to zero). Consequently, the first term in the posted series is 1, not 2. All of the finite integer powers of zero are zero.

Oh i get it now. Thanks
I should have noticed that! Anyway, 0^0 is a whole other debate ....
asnani04
nice problem liked solving it.
evablast
we've got some maths gurus here
Well, I would try and figure it out and give an answer, but I see that's already been done. Good job. I did guess 16 first, but can see the error in my calculation (after continuing reading the following comments).
zimmer
so what is the final answer?

Note: for those who give final answer please put reference on how you derived from it? i guess we need to know the real and reliable answer you got.
kelseymh
 zimmer wrote: so what is the final answer? Note: for those who give final answer please put reference on how you derived from it? i guess we need to know the real and reliable answer you got.

Did you read the postings? Peterssidan posted the correct answer on 5 Feb 2012, confirmed by the original writer.
Bikerman