Submitted by Lee CP

Let the 5 consecutive numbers be $n-2 , n-1, n, n+1, n+2$ and let them correspond to $p,q,r,s,t$

This implies $p + q+ r+ s + t = 5n$ and that $q+ r+ s = 3n$

So $5n=a^2$ and $3n=b^3$

Since we are dealing with integers it is easy to see that $n=5k^2$ and that $n=3^2s^3$ .

We can eventually find a value for n which satisfies both these expressions.

That is : $n=3^2*5^3$ (our values for $k$ and $s$ would be $3*5$ and $5$ respectively ) and so $a^2=3^2*5^4$ .

Since $p + q+ r+ s + t = a^2$ ,

$\sqrt{p + q+ r+ s + t} = a=5^2*3=75$

And so the answer is $\boxed{75}$

First of all, the question should have stated that "There are 5 consecutive positive integers..." because, if this condition is not imposed on the numbers, then one can easily choose the middle number $r$ as $0$ , which would mean that the required answer is $0$ . Since this isn't the answer, it means that the condition "positive integers" has to be taken into account.

Now that the conditions are clear, let the 5 consecutive positive integers be $p=\lambda-2, q=\lambda-1, r=\lambda, s=\lambda+1, t=\lambda+2$ .

$p+q+r+s+t=a^2 \Rightarrow 5\lambda=a^2$ . Therefore, $\lambda$ must be of the form

$5^{2n_1-1}\times p^{2n_2}_{2} \times p^{2n_3}_{3}.....................................$ (1)

where $p_2, p_3...$ are prime numbers so that the R.H.S can be a perfect square.

Again, since $q+r+s= b^3 \Rightarrow 3\lambda=b^3$ . Therefore, $\lambda$ must be of the form

$3^{3m_1-1}\times p^{3m_2}_{2} \times p^{3m_3}_{3}.....................................$ (2)

where $p_2, p_3...$ are prime numbers so that the R.H.S can be a perfect cube.

Combining (1) and (2), we can say that $\lambda$ should be of the form

$5^{2n_1-1}\times 3^{3m_1-1} \times p^{6k_2}_{2} \times p^{6k_3}_{3}...$ .

Now, the primes $p_2, p_3...$ can be ${2,7,11,13...}$ . But since $r <10000$ , the only possible value of the primes $p_2, p_3...$ (if at all they occur) would be $2$ .

So combining all these, $\lambda= 5^{2n-1}\times 3^{3m-1} \times 2^{6k}$ where $n=1,2,3... m=1,2,3...$ and $k=0,1$ .

Now, taking $n=2$ and $m=1$ and $k=0$ gives us $\lambda= 5^3 \times 3^2= 125 \times 9 = 1125.$

(NOTE: Taking any other value from $n=3,4,5...$ , $m=2,3,4...$ and $k=1$ would make $\lambda > 10000$ ).

Hence, $p+q+r+s+t= (\lambda-2) +(\lambda-1) +(\lambda) +(\lambda+1) + (\lambda+2) = 5\lambda= 5\times 1125= 5625$

Therefore, $\sqrt{p+q+r+s+t}= \sqrt{5625}= \boxed{75}$

p+q+r+s+t=5p+10=a2 q+r+s=3p+6=b3

3a2=5b3 (3/5)a2=b3

for (3/5)a2 be a cube of an integer a2=(5 to the power 4)(3 to the power 2) a=75= answer

Assume that,

$p = x-2$

$q = x-1$

$r = x$

$s = x+1$

$t = x+2$

Sum of $p+q+r+s+t = 5x = m^2$

Sum of $q+r+s = 3x = n^3$

$m$ and $n$ are integers.

So, $15x^2 = m^2*n^3$ .

$15x^2 = n(mn)^2$

So, $n=15$ .

$3x = n^3 = 15^3$

$x = 9*125$

$5x = 9*625 = 75^2$

So $m = \boxed{75}$ .

from given info

we have p=r-2 q=r-1.......

therefore p+q+r+s+t=5r=a^2 & q+r+s=3r=b^3 so r must be 125*9=1125

and a^(1/2)=(5r)^(1/2)=75

just like KVPY 2013

$5(p+2)=a^2,~~3(p+2)=b^3$ . So $p+2$ has to be $3^2$ multiplied by a cube that is divisible by $5$ (Hence it must be divisible by $125$ ). Simultaneously $p+2$ has to be $5$ multiplied by a perfect square divisible by $3$ (Hence it must be divisible by $9$ ). Hence $p+2=125\times9$ , and the rest follows.

P,Q,R,S,T ARE CONSECUTIVE NUMBERS. THEREFORE, Q=P+1,R=P+2,S=P+3,T=P+4. THEREFORE P+Q+R+S+T=5P+10=A^2 5P+10=A^2. Q+R+S=3P+6=B^3 3P+6=B^3. WE HAVE TWO EQUATIONS:-5P+10=A^2 AND 3P+6=B^3. SOLVING THESE, WE GET 3 (A^2)=5 (B^3). THEREFORE WE GET A=75 AND B=15. THEREFORE PUTTING A=75 IN ONE OF THE EQUATION, WE GET P=1123. THEREFORE P+Q+R+S+T=1123+1124+1125+1126+1127=5625. THEREFORE SQUARE ROOT OF P+Q+R+S+T=SQUARE ROOT OF 5625=75. THEREFORE ANSWER IS 75.

Note that $p$ , $q$ , $r$ , $s$ , and $t$ are all consecutive integers. Therefore, we can rewrite the sets like this. $\{p, q, r, s, t\}=\{r-2,r-1,r,r+1,r+2\}$ The sum of the elements of this set is equal to $5r$ , and the sum of the middle $3$ elements is equal to $3r$ . Knowing this, we can set up this system of Diophantine equations. $5r=a^2$ $3r=b^3$ We can solve this using some simple modular arithmetic.

The first thing that should be noticed is that $r$ can be factored as $3^{3k-1}5^{2n-1}$ . We know this because multiplying this by $3$ or $5$ makes the $3^x$ and $5^x$ factors cubes or squares, respectively. We need to find values of $k$ and $n$ that, with the necessary multiplications, would give integral $a$ and $b$ . So let's set up what we need to find. $3k-1\equiv0\text{mod}2$ $2n-1\equiv0\text{mod}3$ Solving this gives this. $k\equiv1\text{mod}2$ $n\equiv2\text{mod}3$ The smallest solutions that exist for this are $\{k,n\}=\{1,2\}$ . Plugging into $r$ gives $r=3^2\times5^3$ (which obviously fits the criteria that $r≤9997$ ). The question is asking for $\sqrt{5r}=\sqrt{3^2\times5^4}$ , so the answer is $3\times5^2=\boxed{75}$

The last thing that needs to be done is to confirm that there is no other solution for $r$ . $k$ and $n$ cannot be negative or they would give a non-integer $r$ , so $k,n≥0$ . The next-smallest $r$ that can be produced occurs when $\{k,n\}=\{3,2\}$ , but this would produce an $r$ much bigger than $9997$ , so $\{1,2\}$ is the only possible solution.

Since $p,\ldots,t$ are consecutive integers, the first requirement can be written as $p+\ldots+t=5\cdot p+0+1+2+3+4=5\cdot p+10=5\cdot(p+2)=a^2$ and similarly the second statement becomes $3(p+2)=b^3$ . Since each prime factor has to occur an even number of times in a square, the first requirement thus tells us that $p+2$ is of the form $5\cdot(\text{some square})$ . Every prime factor in a cube has to occur a multiple of 3 times. Hence, the second requirement tells us that $p+2$ is of the form $3^2\cdot(\text{some cube})$ .

Combining these facts that $p+2$ has to be both of the form $3^2\cdot\text{cube}$ and $5\cdot\text{square}$ , it follows that it has to be of the form $3^2\cdot5^3\cdot c^6$ for some natural number $c$ .

We have been told that all of $p,\ldots,t$ are smaller than $10000$ , so it follows that $c=1, p=3^2\cdot 5^3-2$ and $\sqrt{p+q+r+s+t}=\sqrt{(3^2\cdot 5^3-2)+\ldots+(3^2\cdot 5^3+2)}=\sqrt{5\cdot3^2\cdot 5^3}=3\cdot5^2=75$