If $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube, where $$p, q, r$$ and $$s$$ are positi

Question

If $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube, where $$p, q, r$$ and $$s$$ are positive integers, then the smallest value of $$p + q + r + s$$ is :

Accepted Answer

9. It has been given that N = ( 11 p + 7 ) ( 7 q − 2 ) ( 5 r + 1 ) ( 3 s ) N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s}) N = ( 1 1 p + 7 ) ( 7 q − 2 ) ( 5 r + 1 ) ( 3 s ) is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. p , q , r p,

If $N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :

If $N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :

If N=(11p+7)(7q−2)(5r+1)(3s)N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})N=(11p+7)(7q−2)(5r+1)(3s) is a perfect cube, where p,q,rp, q, rp,q,r and sss are positive integers, then the smallest value of p+q+r+sp + q + r + sp+q+r+s is :

If N=(11p+7)(7q−2)(5r+1)(3s)N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})N=(11p+7)(7q−2)(5r+1)(3s) is a perfect cube, where p,q,rp, q, rp,q,r and sss are positive integers, then the smallest value of p+q+r+sp + q + r + sp+q+r+s is :

If $N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :

If $N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :