Find the smallest n

Thank you for watching.

[prateekbhurkay]

N is 72

If 2n is a perfect square, then N has an odd power of 2 as a factor.
If 3n is a perfect cube, then N has 2^(3A) as a factor and 3^(3B - 1) as a factor.

Putting A & B = 1, we get 2^3 * 3^2 = 72.

72*2=144=12^2
72*3=216=6^3

[pojuantsalo]

2n = a²
3n = b³

where a, b and n are positive integers. From here we get

6n = 3a²
6n = 2b³

This leads to the equation

3a² = 2b³ ⟹ a² = 2b³/3 ⟹ a = √(2b³/3).

Here obviously 2b³/3 is an integer meaning b has to be divisible by 3:

b = 3c ⟹ 2b³/3 = 2*3*3*c³

where c is positive integer. 2*3*3*c³ has to be a perfect square. This mean that the prime factorisation must have even copies of every prime factor. The smallest c to make that happen is

c = 2 ⟹ 2*3*3*c³ = 2*2*2*2*3*3 = 12² = a² = 2n ⟹ n = 12²/2 = 72.

[stealth]

Your solution is very elegant.
I found the answer by thinking about prime factors, in a square the prime factors appear in pairs and in a cube they appear in triplets. The value must then have some number of prime factors as 2 and some as three. We can see that there must be an even amount of 3s and an odd amount of 2s. If we look at having only a single 2 as a prime factor and two 3s then it would be a perfect square when multiplied by 2 but not a perfect cube when multiplied by 3. If we instead have three 2s and two 3s as prime factors then it fulfil both conditions.
n = 2x2x2x3x3 = 72
2n = (2x2)(2x2)(3x3) = 144
3n = (2x2x2)(3x3x3) = 216

[gedcowburn]

There are quite a lot of people who say they can't do maths. I think much of this is down to poor teaching. Your teaching is brilliant. The world needs more maths teachers like you. Never stop

[vloyolaa]

Excelent analysis. Simple and direct but most of all, logical. Thank and greetings from Perú

[vilmerpaulsson]

A general solution would be:
if an = perfect square,
bn = pefect cube
solution:
n=ak^2
abk^2=q^3
k=ab
n=ak^2=a*a*a*b*b=a^3*b^2
The lowest integar value of n will always be n = a^3*b^2

[juanpallautapulido]

Great explanation. the solution you provide is simple and elegant.

[paologat]

A more elegant solution is to consider the factorization of n.

1) Since 2n is a square and 3n is a cube, the minimal n can only include factors of 2 and 3. So n=2^A*3^B.

2) Since 2n is a square, A must be odd and B even.

3) Since 3n is a cube, A must be a multiple of 3 and B a multiple of three minus one.

4) A is the smallest odd multiple of 3, that is 3 itself. B is the smallest even number that is one less than a multiple of 3, that is 2.

5) So n=2^3*3^2 = 72

[NicolasMiari]

You're doing the Lord's work.

[misterj.a]

2n is a perfect square => n is even
3n is a perfect cube mean => a) n contains at least the cube of 2 which is 8
b) n contains at least the square of 3 which is 9

Smallest possible value of n is 8*9=72

[aurochrok]

For 2n to be a perfect square n has to be even - divisible by 2. For 3n to be a perfect cube n has to be divisible by 9 and, since it’s even and 3 is not, by 8 as well. The smallest natural number that is divisible by both 8 and 9 is 8x9 = 72 - and it fits .

[Mothuzad]

I solved by just looking at properties of n.

- Has only factors of 2 and 3, otherwise it would be unnecessarily bigger.
- Has a number of factors of 2 divisible by 3 (a factor of 8) and a number of factors of 3 divisible by 2 (a factor of 9).
- The count of factors of 2 is one less than a multiple of 2. The count of factors of 3 is one less than a multiple of 3. 8 and 9 satisfy this constraint already. Thus multiplying by 2 gives a perfect square, and multiplying by 3 gives a perfect cube.

Any number meeting these conditions is sufficient, and the smallest such number is 8 * 9 = 72.

[RegisMichelLeclerc]

Intuitively, you take 2 and 3 (the factors of n), you cube one and square the other, multiply one by the other: 8*9=72. It works because 2 and 3 are primes, and I'm pretty sure it works with other pairs of primes (not too sure about the powers involved, though, you probably need to jump into logs to prove it).

[OlavRH]

Peace and good health to you, brother. Another fantastic video

[tambuwalmathsclass]

Good job 👍
Permit me to solve it differently.

[AmlanSarkar-wrpr]

How it's going man.Where you live btw.I am loving maths because of you.I love your teaching style.U are a genius.

[Inspirator_AG]

*Here is a solution using exponents of primes...*





2^1 * n = 2^(1 + A) * 3^B
3^1 * n = 2^A * 3^(1 + B)
n = 2^A * 3^B

A must be threeven and odd.
B must be even and congruent to 2 mod 3.
Both must be at their minimum given the conditions above.

This means A and B are 3 and 2 respectively.

n = 2^3 * 3^2 = 72

[TonboIV]

There is a very intuitive method which requires no algebra:

Consider the characteristics of perfect squares and cubes. A perfect square has only prime factors that come in pairs (such as 2x2) while a perfect cubes has only prime factors which come in triplets (2x2x2). When we add a single 2 to the prime factorization of "n", we get a perfect square, so "n" must already have at least one 2 for that new 2 to pair with. Likewise, n must have at least two 3s. Okay, we now know that "n" has 2s and 3s in its factorization, and if we add a single 2, all the factors will be in pairs, and if we add a single 3, all the factors will be in triplets. Effectively that means that the number of 2s is a multiple of 3 which is one less that a multiple of 2, while the number of 3s is a multiple of 2 which is one less then a multiple of 3. Three 2s and two 3s meets these conditions, and if we starts removing any pairs or triplets of factors then we'll either run out of 2s or 3s, and we've already established that "n" has 2s and 3s in it, so this is also the smallest factorization which will work, and 2x2x2x3x3 is 72.

And honestly, this is all belabouring the point quite a bit. Once you start thinking about the prime factorizations and how squares and cubes work, the answer just kind of falls out.

[chaosredefined]

Consider the prime factorization of n. If n has a factor of any prime other than 2 or 3, then we could make a smaller number by taking out all factors of that prime. So, the only prime factors are 2 and 3. So, we define a and b such that n = (2^a) (3^b).

Note that 2n has to be a perfect square. So 2^(a+1) 3^b is a perfect square, and hence a+1 and b are both even. If a+1 is even, then a is odd.

Furthermore, note that 3n has to be a perfect cube. So 2^a 3^(b+1) is a perfect cube and hence a and b+1 are both multiples of 3. If b+1 is a multiple of 3, then b = 3k+2 for some integer k.

The smallest odd multiple of 3 is 3, so a is 3. The smallest even number of the form 3k+2 is 2, so b is 2. Thus, the smallest value of n is (2^3) (3^2) = (8)(9) = 72

[danteangotti]

Who else just did this through trial and error by comparing both perfect squares and perfect cubes?