Difference between revisions of "2018 AMC 10A Problems/Problem 22"
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| − | 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>, and this new factor cannot be divisible by <math>2</math> or <math>3</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{\textbf{D) } 13}</math> | + | 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>, and this new factor cannot be divisible by <math>2</math> or <math>3</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{\textbf{(D) } 13}</math> |
Solution by JohnHankock | Solution by JohnHankock | ||
Revision as of 19:44, 9 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 
, and this new factor cannot be divisible by or . 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 | ||
