Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 1"
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From <math>4c = \frac{d}4</math> we have that 16 [[divisor | divides]] <math>d</math>. From <math>a + 4 = \frac d4</math> we have <math>d \geq 20</math>. Minimizing <math>d</math> minimizes <math>a, b</math> and <math>c</math> and consequently minimizes our dragon. The smallest possible choice is <math>d = 32</math>, from which <math>a = 4, b = 12</math> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>. | From <math>4c = \frac{d}4</math> we have that 16 [[divisor | divides]] <math>d</math>. From <math>a + 4 = \frac d4</math> we have <math>d \geq 20</math>. Minimizing <math>d</math> minimizes <math>a, b</math> and <math>c</math> and consequently minimizes our dragon. The smallest possible choice is <math>d = 32</math>, from which <math>a = 4, b = 12</math> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>. | ||
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+ | ==Solution2== | ||
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+ | Note: a+b+c+d=a+a+8+(a+4)/4+4a+16 | ||
+ | =(25a+100)/4 | ||
+ | = 25(a+4)/4 . Therefore, we know that a+4/4 has to be an integer. since a>0, the smaller integer we can have is 2, so a=4. | ||
+ | Therefore, the smallest value is 50. | ||
+ | |||
+ | ~Please convert to latex | ||
==See Also== | ==See Also== |
Latest revision as of 00:27, 11 April 2024
Contents
Problem
A positive integer is called a dragon if it can be written as the sum of four positive integers and such that Find the smallest dragon.
Solution
From we have that 16 divides . From we have . Minimizing minimizes and and consequently minimizes our dragon. The smallest possible choice is , from which and so our desired number is .
Solution2
Note: a+b+c+d=a+a+8+(a+4)/4+4a+16 =(25a+100)/4 = 25(a+4)/4 . Therefore, we know that a+4/4 has to be an integer. since a>0, the smaller integer we can have is 2, so a=4. Therefore, the smallest value is 50.
~Please convert to latex
See Also
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |