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Difference between revisions of "1985 AHSME Problems/Problem 12"

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==Problem==
 
==Problem==
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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]]?
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Let <math>p</math>, <math>q</math> and <math>r</math> be distinct prime numbers, where <math>1</math> is not considered a prime. Which of the following is the smallest positive perfect cube having <math>n = pq^2r^4</math> as a divisor?
  
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<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>
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<math> \mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) }\left(pq^2r^2\right)^3 \qquad \mathrm{(C) } \left(p^2q^2r^2\right)^3 \qquad \mathrm{(D) } \left(pqr^2\right)^3 \qquad \mathrm{(E) \  }4p^3q^3r^3 </math>
  
 
==Solution==
 
==Solution==
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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>.
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A number of the form <math>p^aq^br^c</math> will be a perfect cube precisely when <math>a</math>, <math>b</math>, and <math>c</math> are multiples of 3. (Clearly, since we are looking for the ''smallest'' possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of <math>pq^2r^4</math>, we must have <math>a \geq 1</math>, <math>b \geq 2</math>, and <math>c \geq 4</math>. The smallest multiple of <math>3</math> that is at least <math>1</math> is <math>3</math>, the smallest that is at least <math>2</math> is again <math>3</math>, and the smallest that is at least <math>4</math> is <math>6</math>. Hence the smallest possible perfect cube with <math>n</math> as a divisor is <math>p^3q^3r^6 = \boxed{\text{(D)} \left(pqr^2\right)^3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=12|num-a=13}}
 
{{AHSME box|year=1985|num-b=12|num-a=13}}
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{{MAA Notice}}

Latest revision as of 19:48, 19 March 2024

Problem

Let $p$, $q$ and $r$ be distinct prime numbers, where $1$ is not considered a prime. Which of the following is the smallest positive perfect cube having $n = pq^2r^4$ as a divisor?

$\mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) }\left(pq^2r^2\right)^3 \qquad \mathrm{(C) } \left(p^2q^2r^2\right)^3 \qquad \mathrm{(D) } \left(pqr^2\right)^3 \qquad \mathrm{(E) \ }4p^3q^3r^3$

Solution

A number of the form $p^aq^br^c$ will be a perfect cube precisely when $a$, $b$, and $c$ are multiples of 3. (Clearly, since we are looking for the smallest possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of $pq^2r^4$, we must have $a \geq 1$, $b \geq 2$, and $c \geq 4$. The smallest multiple of $3$ that is at least $1$ is $3$, the smallest that is at least $2$ is again $3$, and the smallest that is at least $4$ is $6$. Hence the smallest possible perfect cube with $n$ as a divisor is $p^3q^3r^6 = \boxed{\text{(D)} \left(pqr^2\right)^3}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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