View source for 1985 AHSME Problems/Problem 12

==Problem==
Let's write <math> p, q, </math> and <math> r </math> as three distinct [[prime number]]s, where <math> 1 </math> is not a prime. Which of the following is the smallest positive [[perfect cube]] leaving <math> n=pq^2r^4 </math> as a [[divisor]]?

<math> \mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) \ }(pq^2r^2)^3 \qquad \mathrm{(C) \  } (p^2q^2r^2)^3 \qquad \mathrm{(D) \  } (pqr^2)^3 \qquad \mathrm{(E) \  }4p^3q^3r^3 </math>

==Solution==
For a number of the form <math> p^aq^br^c </math> to be a perfect cube and a multiple of <math> pq^2r^4 </math>, <math> a, b, </math> and <math> c </math> must all be multiples of <math> 3 </math>, <math> a\ge1 </math>, <math> b\ge2 </math>, and <math> c\ge4 </math>. The smallest multiple of <math> 3 </math> greater than <math> 1 </math> is <math> 3 </math>, the smallest multiple of <math> 3 </math> greater than <math> 2 </math> is <math> 3 </math>, and the smallest multiple of <math> 3 </math> greater than <math> 4 </math> is <math> 6 </math>. Therefore, the smallest such <math> p^aq^br^c </math> is <math> p^3q^3r^6=(pqr^2)^3, \boxed{\text{D}} </math>.

==See Also==
{{AHSME box|year=1985|num-b=12|num-a=13}}