2009 AMC 8 Problems/Problem 17: Difference between revisions
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From the question's requirements, we can figure out <math>x</math> is 10. Then we can use the answer choices to find what <math>y</math> is. | From the question's requirements, we can figure out <math>x</math> is 10. Then we can use the answer choices to find what <math>y</math> is. | ||
Let's start with <math>A</math>. If <math>A</math> was right, then <math>y=70</math>. We can multiply <math>70</math> by <math>360</math> and get <math>25200</math>, which isn't a perfect cube. Then we move to <math>B</math>. <math>85-10=75</math>, so <math>y=75</math> if <math>B</math> is right. Then we multiply <math>75</math> by <math>360</math> to get <math>27000</math>, which is <math>30^3</math>. Therefore, our answer is <math>\boxed{\textbf{(B)}\ 85}</math> because <math>y=75</math> and <math>75+10=85</math>. | Let's start with <math>A</math>. If <math>A</math> was right, then <math>y=70</math>. We can multiply <math>70</math> by <math>360</math> and get <math>25200</math>, which isn't a perfect cube. Then we move to <math>B</math>. <math>85-10=75</math>, so <math>y=75</math> if <math>B</math> is right. Then we multiply <math>75</math> by <math>360</math> to get <math>27000</math>, which is <math>30^3</math>. Therefore, our answer is <math>\boxed{\textbf{(B)}\ 85}</math> because <math>y=75</math> and <math>75+10=85</math>. | ||
~Trex226 | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=16|num-a=18}} | {{AMC8 box|year=2009|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:04, 10 January 2023
Problem
The positive integers 
and 
are the two smallest positive integers for which the product of 
and is a square and the product of and is a cube. What is the sum of and ?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=2768
Video Solution
https://www.youtube.com/watch?v=ZuSJdf1zWYw
Solution
The prime factorization of 
. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of x. Similarly, y can be found by making all the exponents divisible by 3, so the minimum possible value of is 
. Thus, our answer is 
.
Solution 2 (Using Answer Choices)
From the question's requirements, we can figure out is 10. Then we can use the answer choices to find what is.
Let's start with 
. If was right, then 
. We can multiply 
by and get 
, which isn't a perfect cube. Then we move to 
. 
, so 
if is right. Then we multiply 
by to get 
, which is 
. Therefore, our answer is 
because and 
.
~Trex226
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
