Difference between revisions of "2008 AMC 12A Problems/Problem 17"
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<math>\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004</math> | <math>\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004</math> | ||
− | ==Solution== | + | ==Solution 1== |
All positive integers can be expressed as <math>4n</math>, <math>4n+1</math>, <math>4n+2</math>, or <math>4n+3</math>, where <math>n</math> is a nonnegative integer. | All positive integers can be expressed as <math>4n</math>, <math>4n+1</math>, <math>4n+2</math>, or <math>4n+3</math>, where <math>n</math> is a nonnegative integer. | ||
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Since one fourth of the positive integers <math>a_1 \le 2008</math> can be expressed as <math>4n+3</math>, where <math>n</math> is a nonnegative integer, the answer is <math>\frac{1}{4}\cdot 2008 = 502 \Rightarrow D</math>. | Since one fourth of the positive integers <math>a_1 \le 2008</math> can be expressed as <math>4n+3</math>, where <math>n</math> is a nonnegative integer, the answer is <math>\frac{1}{4}\cdot 2008 = 502 \Rightarrow D</math>. | ||
− | == | + | ==Solution 2== |
After checking the first few <math>a_n</math> such as <math>1</math>, <math>2</math> through <math>7</math>, we can see that the only <math>a_1</math> that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for <math>a_n=3</math>, we notice the sequence yields <math>10</math>, <math>5</math>, and <math>16</math>, a valid sequence. | After checking the first few <math>a_n</math> such as <math>1</math>, <math>2</math> through <math>7</math>, we can see that the only <math>a_1</math> that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for <math>a_n=3</math>, we notice the sequence yields <math>10</math>, <math>5</math>, and <math>16</math>, a valid sequence. | ||
So we can set up an equation, <math>3x + 1 = 2(2k - 1)</math> where x is equal to <math>a_1</math>. Rearranging the equation yields <math>(3x + 3)/4 = k</math>. Experimenting yields that every 4th <math>x</math> after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to <math>2008/4 = 502 \Rightarrow D</math>. | So we can set up an equation, <math>3x + 1 = 2(2k - 1)</math> where x is equal to <math>a_1</math>. Rearranging the equation yields <math>(3x + 3)/4 = k</math>. Experimenting yields that every 4th <math>x</math> after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to <math>2008/4 = 502 \Rightarrow D</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}} | ||
[[Collatz Problem]] | [[Collatz Problem]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:52, 21 December 2020
Contents
Problem
Let be a sequence determined by the rule if is even and if is odd. For how many positive integers is it true that is less than each of , , and ?
Solution 1
All positive integers can be expressed as , , , or , where is a nonnegative integer.
- If , then .
- If , then , , and .
- If , then .
- If , then , , and .
Since , every positive integer will satisfy .
Since one fourth of the positive integers can be expressed as , where is a nonnegative integer, the answer is .
Solution 2
After checking the first few such as , through , we can see that the only that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for , we notice the sequence yields , , and , a valid sequence.
So we can set up an equation, where x is equal to . Rearranging the equation yields . Experimenting yields that every 4th after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to .
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
Collatz Problem The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.