Difference between revisions of "2019 AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | Since all four terms on the left are positive integers, from <math>\sqrt{\log{a}}</math>, we know that both <math>\log{a}</math> has to be perfect square and <math>a</math> has to be a power of ten. The same applies to <math>b</math> for the same reason. Setting <math>a</math> and <math>b</math> to <math>10^x</math> and <math>10^y</math>, where <math>x</math> and <math>y</math> are the perfect squares, <math>ab = 10^{x+y}</math>. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <math>y</math> of <math>200</math> from answer choice <math>E</math>), two of those perfect squares must add up to one of the possible sums of <math>x</math> and <math>y</math> given from the answer choices (<math>52</math>, <math>100</math>, <math>144</math>, <math>164</math>, or <math>200</math>). | + | Since all four terms on the left are positive integers, from <math>\sqrt{\log{a}}</math>, we know that both <math>\log{a}</math> has to be a perfect square and <math>a</math> has to be a power of ten. The same applies to <math>b</math> for the same reason. Setting <math>a</math> and <math>b</math> to <math>10^x</math> and <math>10^y</math>, where <math>x</math> and <math>y</math> are the perfect squares, <math>ab = 10^{x+y}</math>. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <math>y</math> of <math>200</math> from answer choice <math>E</math>), two of those perfect squares must add up to one of the possible sums of <math>x</math> and <math>y</math> given from the answer choices (<math>52</math>, <math>100</math>, <math>144</math>, <math>164</math>, or <math>200</math>). |
Only a couple possible sums are seen: <math>16+36=52</math>, <math>36+64=100</math>, <math>64+100=164</math>, <math>100+100=200</math>, and <math>4+196=200</math>. By testing each of these (by seeing whether <math>\sqrt{x} + \sqrt{b} + \frac{x}{2} + \frac{y}{2} = 100</math>), only the pair <math>x = 64</math> and <math>y=100</math> work. Therefore, <math>a</math> and <math>b</math> are <math>10^{64}</math> and <math>10^{100}</math>, and our answer is <math>\boxed{\textbf{(D) } 10^{164}}</math>. | Only a couple possible sums are seen: <math>16+36=52</math>, <math>36+64=100</math>, <math>64+100=164</math>, <math>100+100=200</math>, and <math>4+196=200</math>. By testing each of these (by seeing whether <math>\sqrt{x} + \sqrt{b} + \frac{x}{2} + \frac{y}{2} = 100</math>), only the pair <math>x = 64</math> and <math>y=100</math> work. Therefore, <math>a</math> and <math>b</math> are <math>10^{64}</math> and <math>10^{100}</math>, and our answer is <math>\boxed{\textbf{(D) } 10^{164}}</math>. |
Revision as of 20:20, 10 February 2019
Contents
Problem
Positive real numbers and have the property that and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is ?
Solution 1
Since all four terms on the left are positive integers, from , we know that both has to be a perfect square and has to be a power of ten. The same applies to for the same reason. Setting and to and , where and are the perfect squares, . By listing all the perfect squares up to (as is larger than the largest possible sum of and of from answer choice ), two of those perfect squares must add up to one of the possible sums of and given from the answer choices (, , , , or ).
Only a couple possible sums are seen: , , , , and . By testing each of these (by seeing whether ), only the pair and work. Therefore, and are and , and our answer is .
Solution 2
Given that and are both integers, and must be in the form and , respectively for some positive integers and . Note that . By substituting for a and b, the equation becomes . After multiplying the equation by 2 and completing the square with respect to and , the equation becomes . Testing squares of positive integers that add to , is the only option. WLOG, let and . Plugging and to solve for and gives us and . Therefore, .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.