Difference between revisions of "2019 AIME I Problems/Problem 7"
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<math>10^{210}</math> can be factored into <math>2^{210} \cdot 5^{210}</math>, and <math>m+n</math> equals to the sum of the exponents of <math>2</math> and <math>5</math>, which is <math>210+210 = 420</math>. | <math>10^{210}</math> can be factored into <math>2^{210} \cdot 5^{210}</math>, and <math>m+n</math> equals to the sum of the exponents of <math>2</math> and <math>5</math>, which is <math>210+210 = 420</math>. | ||
Multiply by two to get <math>2m +2n</math>, which is <math>840</math>. | Multiply by two to get <math>2m +2n</math>, which is <math>840</math>. | ||
− | Then, use the first equation (<math>\log x + 2\log(\gcd(x,y)) = 60</math>) to show that <math>x</math> has to have lower degrees of <math>2</math> and <math>5</math> than <math>y</math> (you can also test when <math>x>y</math>, which is a contradiction to the restrains you set before). Therefore, | + | Then, use the first equation (<math>\log x + 2\log(\gcd(x,y)) = 60</math>) to show that <math>x</math> has to have lower degrees of <math>2</math> and <math>5</math> than <math>y</math> (you can also test when <math>x>y</math>, which is a contradiction to the restrains you set before). Therefore, <math>\gcd(x,y)=x</math>. Then, turn the equation into <math>3\log x = 60</math>, which yields <math>\log x = 20</math>, or <math>x = 10^{20}</math>. |
Factor this into <math>2^{20} \cdot 5^{20}</math>, and add the two 20's, resulting in <math>m</math>, which is <math>40</math>. | Factor this into <math>2^{20} \cdot 5^{20}</math>, and add the two 20's, resulting in <math>m</math>, which is <math>40</math>. | ||
Add <math>m</math> to <math>2m + 2n</math> (which is <math>840</math>) to get <math>40+840 = \boxed{880}</math>. | Add <math>m</math> to <math>2m + 2n</math> (which is <math>840</math>) to get <math>40+840 = \boxed{880}</math>. |
Revision as of 00:18, 13 July 2020
Contents
Problem 7
There are positive integers and that satisfy the system of equations Let be the number of (not necessarily distinct) prime factors in the prime factorization of , and let be the number of (not necessarily distinct) prime factors in the prime factorization of . Find .
Solution 1
Add the two equations to get that . Then, we use the theorem to get the equation, . Using the theorem that , along with the previously mentioned theorem, we can get the equation . This can easily be simplified to , or .
can be factored into , and equals to the sum of the exponents of and , which is . Multiply by two to get , which is . Then, use the first equation () to show that has to have lower degrees of and than (you can also test when , which is a contradiction to the restrains you set before). Therefore, . Then, turn the equation into , which yields , or . Factor this into , and add the two 20's, resulting in , which is . Add to (which is ) to get .
Solution 2 (Bashier Solution)
First simplifying the first and second equations, we get that
Thus, when the two equations are added, we have that
When simplified, this equals
so this means that
so
Now, the following cannot be done on a proof contest but let's (intuitively) assume that and and are both powers of . This means the first equation would simplify to and Therefore, and and if we plug these values back, it works! has total factors and has so
Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.
Solution 3 (Easy Solution)
Let and and . Then the given equations become and . Therefore, and . Our answer is .
Solution 4
We will use the notation for and as . We can start with a similar way to Solution 1. We have, by logarithm properties, or . We can do something similar to the second equation and our two equations become Adding the two equations gives us . Since we know that , , or . We can express as and as . Another way to express is now , and is now . We know that , and thus, , and . Our equations for and now become or . Doing the same for the equation, we have , and , which satisfies . Thus, . ~awsomek
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.