Fun with Perfect Powers

If $\sqrt{2k}$ , $\sqrt[3]{3k}$ and $\sqrt[5]{5k}$ is all integer, then so as the product of the three. Combining gives $(\sqrt{2k})(\sqrt[3]{3k})(\sqrt[5]{5k}) = (\sqrt{2})(\sqrt[3]{3})(\sqrt[5]{5}) (k^{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}})$ This form may be reduced to $(\sqrt{2})(\sqrt[3]{3})(\sqrt[5]{5})(k^{\frac{31}{30}}) = k\sqrt[30]{k(2^{15})(3^{10})(5^6)}$ Here we can conclude that the least $k$ satisfying all condition shall be $k = (2^{30-15})(3^{30-10})(5^{30-6}) = (2^{15})(3^{20})(5^{24})$ The question asked us to sum the exponents of all prime bases, which is $15+20+24 = 59$

Let $k=2^{a}3^{b}5^{c}$ . Then, we get: $a+1\equiv b\equiv c\equiv0\space(\text{mod}2)$ $a\equiv b+1\equiv c\equiv0\space(\text{mod}3)$ $a\equiv b\equiv c+1\equiv0\space(\text{mod}5)$ Solving this system, we are able to obtain the minimum values of $a$ , $b$ and $c$ , which are 15, 20 and 24 respectively. Thus, the answer is $15+20+24=\large 59$ .

To minimize the number, we use only the prime factors 2, 3, and 5. (All other solutions can be produced by multiplying our solution by any 30th power.) $k = 2^a \cdot 3^b \cdot 5^c.$ The conditions in the problem can be translated as follows.

• $2k$ is a perfect square: $a+1$ , $b$ and $c$ are multiples of 2.

• $3k$ is a perfect cube: $a$ , $b+1$ and $c$ are multiples of 3.

• $5k$ is a perfect fifth power: $a$ , $b$ , and $c+1$ are multiples of 5.

Thus

• $a$ is a multiple of $3\cdot 5 = 15$ , and one less than a multiple of two. This implies $a \equiv 15$ mod 30.

• $b$ is a multiple of $2\cdot 5 = 10$ , and one less than a multiple of three. This implies $b \equiv 20$ mod 30.

• $c$ is a multiple of $2\cdot 3 = 6$ , and one less than a multiple of five. This implies $c \equiv 24$ mod 30.

In conclusion, $k = 2^{15}\cdot 3^{20}\cdot 5^{24}\cdot N^{30},\ \ \ N = 1, 2, \dots$ The answer to the problem is $15 + 20 + 24 = \boxed{59}.$

I didn't get fancy on my scratch paper, but I observed that if K= $2^a 3^b 5^c$ , then

a = a multiple of 15 (3*5), that is an odd number (2k+1)

b = a multiple of 10 (2*5), that is two more than a multiple of three

c = a multiple of 6 (2*3), that is four more than a multiple of five

From here, it was easy to deduce that the smallest a is 15, the smallest B is 20 , and the smallest C is 24 .

Just how the heck is the answer 59, with 59% getting it right?

Just think about what condition is imposed by each statement:

2k is a perfect square: k contains a multiple of 2 of every factor, also, 2 is included 2m-1 times where m is a whole number 3k is a perfect cube: k contains 3n of every factor, other than 3 which is included 3n-1 times where n is a whole number 5k is a perfect 5th power: k contains 5p of every factor, other than 5 which is included 5p-1 times where p is a whole number

Clearly we need a number composed of 2's, 3's and 5's as factors to minimally satisfy the condition. Introducing other factors would make the number bigger so we don't want to do that.

We can find the number of 5s since it must be a number of the form 5p-1 which is divisible by 2 and 3. Just check the multiples of 6: 6 (no), 12 (no), 18 (no), 24 (yes: 5(5)-1=24). Therefore the number of 5s in the factorisation is 24.

Repeat for 2 and 3:

The number of 3s is of the form by 3n-1 and is divided by 5 and 2. Check multiples of 10: 10 (no), 20 (yes: 3(7)-1=20). Therefore the number of 3s in the factorisation is 20.

The number of 2s is of the form by 2m-1 and is divided by 5 and 3. Check multiples of 15: 15 (yes: 2(8)-1=15). Therefore the number of 2s is 15.

Therefore our number is acutally:

$k = 2^{15} 3^{20} 5^{24}$

And the sum of exponents in the factorisation:

$15 + 20 + 24 = 59$