on another thread now
Let's say that $\displaystyle (1/2)m= a^2$ and $\displaystyle (1/3)m= b^3$.
Then $\displaystyle m= 2a^2= 3b^3$. Since m has at least one factor of 2, b must be an integer having a factor of 2 and $\displaystyle b^3$ must have a factor of 8. Since m has at least one factor of 3, a must be an integer having a factor of 3 and $\displaystyle a^2$ must have a factor of 9. The smallest possible value for m, then must be 8(9)= 72. Check: $\displaystyle 72/2= 36= 6^2$ and $\displaystyle 72/3= 24= 8(3)$ which is NOT a perfect cube, 8 is but 3 is not. In order to maintain the "1/2 of the number is a perfect square" we must multiply by squares: $\displaystyle 3^2$ comes to mind!
If m= 9(72)= 648, then m/2= 648/2= 324= $\displaystyle 18^2$.
m/3= 648/3= 216= $\displaystyle 6^3$.
I believe that $\displaystyle m=30233088000000$ is the lowest such number.Can you find a positive integer n for which 1/2n is a perfect square, 1/3n is a perfect cube and 1/5n is a perfect fifth power?
You see, $\displaystyle m=2^{15}3^{10}5^{6}=2(2^73^55^3)^2=3(2^53^35^2)^3= 5(2^33^25^1)^5$
According to HallsofIvy's logic, what we're looking for is a number $\displaystyle m=2^a3^b5^c$ satisfying:
$\displaystyle a \equiv 1 (\bmod 2) \equiv 0 (\bmod 3) \equiv 0 (\bmod 5)$
$\displaystyle b \equiv 0 (\bmod 2) \equiv 1 (\bmod 3) \equiv 0 (\bmod 5)$
$\displaystyle c \equiv 0 (\bmod 2) \equiv 0 (\bmod 3) \equiv 1 (\bmod 5)$
The triplet 15,10,6 is the lowest set of ints satisfying this modular equation. Incidentally, if you add 30 to each of these exponents, you get the next highest solution, $\displaystyle m=2^{45}3^{40}5^{36}$, which is higher than is worth thinking, being 58 digits long!
hi, thank you guys
i didnt understand the next bit where we had to show with n
when u say mod im assuming u mean modulus? as in absoulte value? then how can the modulus of 2 * 1 = 0 * modulus of 3?
also, how did you know to multiply by 3^2?
Hello, Aquafina!
Find a positive integer $\displaystyle m$ such that $\displaystyle \tfrac{1}{2}m$ is a perfect square and $\displaystyle \tfrac{1}{3}m$ is a perfect cube.
We have: .$\displaystyle \begin{array}{ccc}\frac{1}{2}m \:=\:a^2 & \Longrightarrow & m \:=\:2a^2 \\ \\[-4mm] \frac{1}{3}m \:=\:b^3 & \Longrightarrow & m \:=\:3b^3 \end{array}$
$\displaystyle m \:=\:2a^2$ must be a cube ... all exponents are multiples of 3.
. . Then: .$\displaystyle m \;=\;2(2^2x^3)^2 \;=\;2^3x^6 $ .[1]
$\displaystyle m \:=\:3b^3$ must be a square ... all exponents are even.
. . Then: .$\displaystyle m \:=\:3(3y^2)^3 \;=\;3^4y^6$ .[2]
The LCM of [1] and [2] is: .$\displaystyle m \:=\:2^3\cdot3^4\cdot k^6$
The least $\displaystyle m$ occurs for $\displaystyle k=1\!:\;\;m = 2^3\cdot3^4 \:=\:\boxed{648}$
hi thanks! why did you say that
m=2a^2 is a cube and m=3b^2 is a square
arent they the other way around because the equations they are derived from is a square and cube the other way around?
also, how do u do the next part from here? as in including the perfect 5th power?
Sorry. By "modulus" I mean modular arithmetic, a branch of math generalizing patterns that can be found by studying the remainders of division.
Notice that what you are asking for is satisfied by the aforementioned number, factored such:
$\displaystyle m=30233088000000=2^{15}3^{10}5^{6}=2(2^73^55^3)^2= 3(2^53^35^2)^3=5(2^33^25^1)^5$
Generalizing this pattern, the subsequent modular equations can be summarized in plain English by saying we are looking for a number $\displaystyle m=2^a3^b5^c$ such that (1) b and c are even but a is odd, (2) a and c are divisible by 3 but b leaves a remainder of 1, and (3) a and b are divisible by 5, but c leaves a remainder of 1.
By properties of modular equivalences, any such system of equations contains a solution. Therefore, we can go beyond your criteria of 2,3,5, and answer this riddle for any arbitrary list of prime numbers. However, your final answer m will become quite large quite quickly. Without being bogged down by the actual answer, I can say with complete mathematical certainty that there exists a number m, for example, such that m/2 is a perfect square, m/3 a perfect cube, m/5 a fifth power, m/7 a seventh power, m/11 an eleventh power, and m/13 a thirteenth power!
also, the factorisation you did for the value of m, by taking 2 out, 3 out, and 5 out seperately, how does that prove the situation that:
(1) b and c are even but a is odd, (2) a and c are divisible by 3 but b leaves a remainder of 1, and (3) a and b are divisible by 5, but c leaves a remainder of 1.
The answer to that lies in realizing that when we "factor out" a 3 from $\displaystyle 2^{15}3^{10}5^{6}$, for example, that is the same as subtracting 1 from the exponent. Also realize that by "perfect cube" we mean literally each exponent is divisible by three, i.e. $\displaystyle 2^{12}$ is a perfect cube because it can be written as $\displaystyle (2^4)^3$
Combining these two observations, in order to get a perfect cube by factoring out a 3 from $\displaystyle 2^{a}3^{b}5^{c}$, we require a and c to be perfectly divisible by 3, and b to be exactly 1 more than a multiple of 3. Applying this same logic to the other two exponents was where I got the statement,If you plug away and actually solve this riddle, the lowest set of numbers a,b,c that satisfies it is precisely 15,10,6.(1) b and c are even but a is odd, (2) a and c are divisible by 3 but b leaves a remainder of 1, and (3) a and b are divisible by 5, but c leaves a remainder of 1.
thanks thats amazing!
that was very interesting, i'd like to think of this further in special cases, though i doubt i will be able to calculate the actual value then (number being too big!)
any way i could generalise this so a number is perfect square, cube.... 1/n is a perfect nth power?
HallsofIvy showed in his/her post that finding a number fitting your criteria is equivalent to finding a number $\displaystyle m=2x^2=3y^3=5z^5$ for x,y,z integers. Extrapolating that pattern, we hypothesize that m factorizes like such: $\displaystyle m=2^{a}3^{b}5^{c}$ , where (1) b and c are even but a is odd, (2) a and c are divisible by 3 but b leaves a remainder of 1, and (3) a and b are divisible by 5, but c leaves a remainder of 1.
Solve this riddle for a,b,c and plug back in to find m.
Well, m has to be of the form $\displaystyle m=2^a3^b5^c7^d11^e13^f$... for some finite list of primes. Since factoring out 2,3,5 doesn't affect d,e,f..., they must all be divisible by 2,3, and 5. Hence, they all must be multiples of 30. So d, for example, must be 0,30,60,90,etc. Obviously, to minimize m, just assume d,e,f,... are all zero, i.e. no prime factors greater than 5.
Therefore, although larger choices of m exist, we conclude that the smallest m is probably of the form m=$\displaystyle 2^a3^b5^c$, which brings us up to speed with the other posts.