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Find a natural number n, in canonical form, such that:

$n/2=a^2$

$n/3=b^3$

$n/5= c^5$

for some a,b and c (natural numbers).

Gerry Myerson
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GO VEGAN
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3 Answers3

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$$n=2^{15}3^{10}5^{6}$$

(Method: assuming the number is of the form $2^x3^y5^z$, you need $x$ to be divisible by $3\cdot5=15$, but to be $1\pmod2$, $y$ to be divisible by $2\cdot5=10$, but also be $1\pmod3$, and $z$ to be divisible by $2\cdot3=6$ but also be $1\pmod5$).

Alfonso Fernandez
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Attempt an $n$ of the form $n = 2^x 3^y 5^z$.

You have $n/2 = 2^{x-1} 3^y 5^z$ as a square, so $x \equiv 1 (\bmod 2),\; y \equiv z \equiv 0 (\bmod 2).$

Analogously, $x \equiv z \equiv 0 (\bmod 3)$, $y \equiv 1 (\bmod 3)$ and

$x \equiv y \equiv 0 (\bmod 5)$, $z \equiv 1 (\bmod 5)$.

The Chinese remainder theorem guarantees that you will find such natural numbers $x,y,z$.

Cocopuffs
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  • why n has no other factors besides 2,3,5 n=2^x3^y5^z*more factors?. – GO VEGAN Mar 27 '13 at 12:29
  • @user1932595 Sure, it can have more factors. You could multiply a $7^{30}$ onto $n$ and it would still work. – Cocopuffs Mar 27 '13 at 12:35
  • ok, thank u. In The Chinese remainder theorem there is one X we look for and here there is x,y,z.... – GO VEGAN Mar 27 '13 at 12:46
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    Do it separately; here, $x \equiv 1 (\bmod 2)$, $x \equiv 0 (\bmod 3)$, $x \equiv 0 (\bmod 5)$ is solved by $x \equiv 15 (\bmod 30)$, for example. – Cocopuffs Mar 27 '13 at 12:48
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Take $n=2^{15}3^{10}5^{6}\lambda^{30},\lambda\in{N}$

Reason for taking $\lambda^{30}$ is $30=lcm(2,3,5)$

$15=2k_1+1=3k_2=5k_3$

$10=2l_1=3l_1+1=5l_3$

$6=2d_1=3d_2=5d_3+1$

Now I think the reason is clear.

Abhra Abir Kundu
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