Difference between revisions of "1985 AHSME Problems/Problem 12"
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==Problem== | ==Problem== | ||
− | Let <math> p, q </math> and <math> r </math> be distinct | + | 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? |
− | <math> \mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) \ | + | <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== | ||
− | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:48, 19 March 2024
Problem
Let , and be distinct prime numbers, where is not considered a prime. Which of the following is the smallest positive perfect cube having as a divisor?
Solution
A number of the form will be a perfect cube precisely when , , and 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 , we must have , , and . The smallest multiple of that is at least is , the smallest that is at least is again , and the smallest that is at least is . Hence the smallest possible perfect cube with as a divisor is .
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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