Difference between revisions of "2018 AMC 10A Problems/Problem 22"
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− | We can say that a and b 'have' <math>2^3 * 3</math>, that b and c have 2^2 * 3^2, and that c and d have 3^3 * 2. Combining 1 and 2 yields b has (at a minimum) 2^3 * 3^2, and thus a has 2^3 * 3 (and no more powers of 3 because otherwise | + | We can say that <math>a</math> and <math>b</math> 'have' <math>2^3 * 3</math>, that <math>b</math> and <math>c</math> have <math>2^2 * 3^2</math>, and that <math>c</math> and <math>d</math> have <math>3^3 * 2</math>. Combining <math>1</math> and <math>2</math> yields <math>b</math> has (at a minimum) <math>2^3 * 3^2</math>, and thus <math>a</math> has <math>2^3 * 3</math> (and no more powers of <math>3</math> because otherwise <math>gcd(a,b)</math> would be different). In addition, <math>c</math> has <math>3^3 * 2^2</math>, and thus <math>d</math> has <math>3^3 * 2</math> (similar to <math>a</math>, we see that <math>d</math> cannot have any other powers of <math>2</math>). We now assume the simplest scenario, where <math>a = 2^3 * 3</math> and <math>d = 3^3 * 2</math>. According to this base case, we have <math>gcd(a, d) = 2 * 3 = 6</math>. We want an extra factor between the two such that this number is between <math>70</math> and <math>100</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{D}</math> |
Solution by JohnHankock | Solution by JohnHankock | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}} |
Revision as of 18:54, 8 February 2018
Let and be positive integers such that , , , and . Which of the following must be a divisor of ?
Solution
We can say that and 'have' , that and have , and that and have . Combining and yields has (at a minimum) , and thus has (and no more powers of because otherwise would be different). In addition, has , and thus has (similar to , we see that cannot have any other powers of ). We now assume the simplest scenario, where and . According to this base case, we have . We want an extra factor between the two such that this number is between and . Checking through, we see that is the only one that works. Therefore the answer is
Solution by JohnHankock
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |