c4countdown

Here's a fun little not-too-difficult maths problem I encountered in my tutorial for numbers and proofs yesterday:

What is the lowest positive integer excluding zero that: when divided by 2 gives a square number, when divided by 3 gives a cube number and when divided by 5 gives a fifth power.

I've already tried and tested this one out on my guinea pigs and so far only Howard has managed to find the solution. If you think you've solved it pm me so as not to reveal the answer to Kirk.

I'll reveal a hint later for those who are struggling, plus an extension question which Howard made up.

Dinos Sfyris wrote:I'll reveal a hint later for those who are struggling, plus an extension question which Howard made up.

And for those who have worked it out, try to work out what the hint would be and Howard's extension question.

A natural extension is to ask for the smallest that works for 2,3,5 and 7. But asking for 2,3,4,5 is potentially interesting, and harder.

Edit 1: The answer to the first question is, uh, fairly large.
Edit 2: Actually the second question has no solutions, but it's quite a neat problem to show that.

I have to admit that while I solved the original puzzle yesterday, I couldn't manage to solve my extension without a hint from Dinos.

And neither of those was my extension, Charlie.

Does knowing the extension spoil the puzzle? If not it'd be cool to post it.

Dinos Sfyris wrote:... not-too-difficult maths problem ...

lol. Define "not too difficult".

Charlie Reams wrote:Edit 2: Actually the second question has no solutions, but it's quite a neat problem to show that.

I'd pretty well come to that conclusion myself, Charlie. But was reluctant to say so, cos I'd already said that my extension had no solution until Dinos's hint suggested otherwise.

But I'm glad that I'm not totally past it, since you've confirmed that your second extra cannot be solved.

In order of solvation: Charlie, Gevin, Peter Mabey, David Walliams, Conor and Jack Hurst have all solved it so far. A big high five to them

If you're struggling: Try using canonical factorisation of the form: 2^a x 3^b x 5^c to find values of a, b and c which fit the parameters.

I'll be busy tomorrow but will be back Tuesday to offer further hints, but if you're desperate in the meantime speak to Howard.

Here's the extension: Is there an integer greater than zero that fits the criteria of the original puzzle, but is also itself a power? If such an integer exists what is its lowest possible value? Charlie mysteriously knew the extension question and answered it before I posted it, so either he has magical powers or Howard told it him (or probably both)

Happy puzzling

I also have the answer to the Reamsian extension question (2,3,5,7) although TBH it's fairly trivial once you work out the basic problem.

Another useful inroad may be to try the reduced case where it only has to be twice a square and thrice a cube (i.e. ignore the 5). The method generalises easily.

Dinos Sfyris wrote:David Walliams

Damn, I've been waiting for the right time to do that joke for ages!

Dinos Sfyris wrote:Charlie mysteriously knew the extension question and answered it before I posted it, so either he has magical powers or Howard told it him (or probably both)

Definitely both. Charlie convinced me that he'd solved the original, without actually saying so, so I gave him the extension.

Howard Somerset wrote:

Dinos Sfyris wrote:Charlie mysteriously knew the extension question and answered it before I posted it, so either he has magical powers or Howard told it him (or probably both)

Definitely both. Charlie convinced me that he'd solved the original, without actually saying so, so I gave him the extension.

The man's Fermat

Gevin has also solved the extension question, and Bob de Caux has mastered the original puzzle and is close to solving the extension.

Extension hint: For the answer of the form 2^a x 3^b x 5^c to be a power then a, b and c must all share a common factor.

edit: Bob's now got the extension too

Dinos Sfyris wrote: Bob's now got the extension

That I'd like to see.

Sue Sanders wrote:

Dinos Sfyris wrote: Bob's now got the extension

That I'd like to see.

(Just out of shot).

Lesley Hines wrote:

Howard Somerset wrote:

Dinos Sfyris wrote:Charlie mysteriously knew the extension question and answered it before I posted it, so either he has magical powers or Howard told it him (or probably both)

Definitely both. Charlie convinced me that he'd solved the original, without actually saying so, so I gave him the extension.

The man's Fermat

Careful - he already struggles to fit his head through the door!

I'd completely forgotten about this thread. Here's the answer for anyone who's not worked it out yet:

i) The trick here was to make 2^a x 3^b x 5^c where using the rules of powers (2^k x 3^l x 5^m)^z = 2^kz x 3^lz x 5^mz

a is divisible by 3 and 5 and (a-1) is divisible by 2 ie a is a solution of 30n + 15 for any non-negative integer n (lowest value = 15 when n = 0)
b is divisible by 2 and 5 and (b-1) is divisible by 3 ie b is a solution of 30p + 10 for any non-negative integer p (lowest value = 10 when p = 0)
c is divisible by 2 and 3 and (c-1) is divisible by 5 ie c is a solution of 30q + 6 for any non-negative integer q (lowest value = 6 when q = 0)

Hence the solution is 2^15 x 3^10 x 5^6 or 30233088000000

Just to check:

If we divide the answer by 2 we get 2^14 x 3^10 x 5^6 = (2^7 x 3^5 x 5^3)^2
If we divide the answer by 3 we get 2^15 x 3^9 x 5^6 = (2^5 x 3^3 x 5^2)^3
If we divide the answer by 5 we get 2^15 x 3^10 x 5^5 = (2^3 x 3^2 x 5^1)^5

Extension) Here a, b and c must still satisfy the npq equations given above but in order for the answer itself to be a power a, b and c must also have a factor in common. So the lowest possible solution will use the lowest possible common factor of

30n + 15 (cannot have 2 as a factor)
30n + 10 (cannot have 3 as a factor)
30n + 6 (cannot have 5 as a factor)

If we try 7, this can be a factor of all 3 equations at a=105, b=70, c=126

so a solution is possible and the lowest it can be is 2^105 x 3^70 x 5^126 which is (2^15 x 3^10 x 5^18)^7

or written out fully: 11935975575414855963304386072568173408508300781250000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000

Phew!

Dinos Sfyris wrote:I'd completely forgotten about this thread. Here's the answer for anyone who's not worked it out yet:

i) The trick here was to make 2^a x 3^b x 5^c where using the rules of powers (2^k x 3^l x 5^m)^z = 2^kz x 3^lz x 5^mz

a is divisible by 3 and 5 and (a-1) is divisible by 2 ie a is a solution of 30n + 15 for any non-negative integer n (lowest value = 15 when n = 0)
b is divisible by 2 and 5 and (b-1) is divisible by 3 ie b is a solution of 30p + 10 for any non-negative integer p (lowest value = 10 when p = 0)
c is divisible by 2 and 3 and (c-1) is divisible by 5 ie c is a solution of 30q + 6 for any non-negative integer q (lowest value = 6 when q = 0)

Hence the solution is 2^15 x 3^10 x 5^6 or 30233088000000

Just to check:

If we divide the answer by 2 we get 2^14 x 3^10 x 5^6 = (2^7 x 3^5 x 5^3)^2
If we divide the answer by 3 we get 2^15 x 3^9 x 5^6 = (2^5 x 3^3 x 5^2)^3
If we divide the answer by 5 we get 2^15 x 3^10 x 5^5 = (2^3 x 3^2 x 5^1)^5

Extension) Here a, b and c must still satisfy the npq equations given above but in order for the answer itself to be a power a, b and c must also have a factor in common. So the lowest possible solution will use the lowest possible common factor of

30n + 15 (cannot have 2 as a factor)
30n + 10 (cannot have 3 as a factor)
30n + 6 (cannot have 5 as a factor)

If we try 7, this can be a factor of all 3 equations at a=105, b=70, c=126

so a solution is possible and the lowest it can be is 2^105 x 3^70 x 5^126 which is (2^15 x 3^10 x 5^18)^7

or written out fully: 11935975575414855963304386072568173408508300781250000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000

Phew!

Definitely not something for the back of the envelope then. Unless it's a big envelope.

For what it's worth, I presented this problem to a Year 10 (14- and 15-yr-olds) top set (only one managed to solve it), and here is a description of the solution which was aimed at them:

Because X = 2a^2 = 3b^3 = 5c^5, we know that X must be a multiple of 2 and a multiple of 3 and a multiple of 5. In other words X has factors of 2, 3 and 5.

So given that X = 2a^2, then a^2 must itself have factors of 3 and 5.
Similarly, b^3 must itself have factors of 2 and 5.
Also c^5 must itself have factors of 2 and 3.

All numbers can be expressed as products of their prime factors. So a^2 can be expressed as 3 x 5 x n(a), where n(a) is the product of all its other prime factors. But a can also be expressed this way, so a must be 3 x 5 x m(a) where m(a) is the product of all its other prime factors.

Similarly b = 2 x 5 x m(b), and c = 2 x 3 x m(c).

Given that a, b and c are all raised to powers and then multiplied by something to make X, we know that X will not only contain factors of 2, 3 and 5, but they will all have been raised by various powers. So X has factors of 2^j, 3^k, and 5^l. This means that a, b and c must also have factors which are powers of 2, 3 and 5.

So we can re-express X as follows:

X = 2a^2 = 2 x (2^A1 x 3^A2 x 5^A3 x m(a))^2
= 3b^3 = 3 x (2^B1 x 3^B2 x 5^B3 x m(b))^3
= 5c^5 = 5 x (2^C1 x 3^C2 x 5^C3 x m(c))^5

Remember the following laws of indices:
(y^z)^k = y^zk
y^z x y^k = y^(z+k)
(y x z) ^k = y^k x z^k

Using these, we can rewrite the above statements as follows:

X = 2a^2 = 2^1 x (2^A1 x 3^A2 x 5^A3 x m(a))^2 = 2^(2xA1 + 1) x 3^(2xA2) x 5^(2xA3) x m(a)^2
= 3b^3 = 3^1 x (2^B1 x 3^B2 x 5^B3 x m(b))^3 = 2^(3xB1) x 3^(3xB2 + 1) x 5^(3xB3) x m(b)^3
= 5c^5 = 5^1 x (2^C1 x 3^C2 x 5^C3 x m(c))^5 = 2^(5xC1) x 3^(5xC2) x 5^(5xC3 + 1) x m(c)^5

(NB: x is being used here to denote “multiplied by”, rather than a value called x)

If we can find powers of 2, 3 and 5 which will make this true, then we can just make m(a), m(b) and m(c) equal to 1 – because remember we are looking for the smallest possible solution.

This means that we want to find values such that:

2^(2xA1 + 1) x 3^(2xA2) x 5^(2xA3) = 2^(3xB1) x 3^(3xB2 + 1) x 5^(3xB3) = 2^(5xC1) x 3^(5xC2) x 5^(5xC3 + 1)


...which is the same as saying
2xA1 + 1 = 3xB1 = 5xC1
and 2xA2 = 3xB2 + 1 = 5xC2
and 2xA3 = 3xB3 = 5xC3 + 1

(NB: x is being used here to denote “multiplied by”, rather than a value called x)

(If you’re confused by this, skip to “solution breakdown” at the end, which might make it easier to understand.)

So our power of 2 will be a number that can be expressed as ((2 x something) + 1), or (3 x something), or (5 x something). The smallest number which satisfies this is 15
[15 = (2 x 7) + 1 = 3 x 5 = 5 x 3].

Our power of 3 will be a number that can be expressed as (2 x something), or ((3 x something) + 1), or (5 x something). The smallest number which satisfies this is 10
[10 = 2 x 5 = (3 x 3) + 1 = 5 x 2].

Our power of 5 will be a number that can be expressed as (2 x something), or (3 x something), or ((5 x something) + 1). The smallest number which satisfies this is 6
[6 = 2 x 3 = 3 x 2 = (5 x 1) + 1].

So we have a solution!

X = 2^15 x 3^10 x 5^6 = 30,233,088,000,000 = (in standard form) 3.0233088 x 10^13


SOLUTION BREAKDOWN (and sanity check)

Does it work?

2^15 x 3^10 x 5^6 = 2 x (2^7 x 3^5 x 5^3)^2 = [using the laws of indices] 2^((7x2)+1) x 3^(5x2) x 5^(3x2)
= 3 x (2^5 x 3^3 x 5^2)^3 = [using the laws of indices] 2^(5x3) x 3^((3x3)+1) x 5^(2x3)
= 5 x (2^3 x 3^2 x 5^1)^5 = [using the laws of indices] 2^(3x5) x 3^(2x5) x 5^ ((1x5)+1)

From this we can see that if X = 2a^2 = 3b^3 = 5c^5 then:

a = 2^7 x 3^5 x 5^3 = 3,888,000 = (in standard form) 3.888 x 10^6
b = 2^5 x 3^3 x 5^2 = 21600 = (in standard form) 2.16 x 10^4
c = 2^3 x 3^2 x 5^1 = 360

Hmm, that looked an awful lot better when I could use superscript and subscript!