https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Lab&feedformat=atomAoPS Wiki - User contributions [en]2022-09-26T09:04:10ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=379182011 AIME II Problems/Problem 62011-03-31T14:40:39Z<p>Lab: </p>
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<div>Problem:<br />
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Define an ordered quadruple (a, b, c, d) as interesting if <math>1 \le a<b<c<d \le 10</math>, and a+d>b+c. How many ordered quadruples are there?<br />
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Solution:<br />
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There is probably some really complicated formula for this, but as I didnt know it and had 3 hours to "do my best", I listed all possible combinations out. The answer is 80.<br />
You can do casework with the value of a. If a = 1, we can do casework on b. If b = 2, we can do casework on c. if c = 3, d = 5,6,7,8,9, or 10. That's 6. If c = 4, d can be 6,7,8,9,10. That's 5. Continuing this pattern, we get 21 options is b = 2. if b=3 and c = 4, d = 7, 8, 9, or 10, c= 5 gets d = 8, 9, 10 and by quick inspection we get 10 options with b = 3. With b = 4, c = 5 we get d = 9,10, c= 6 we get d = 10, that's 3, with b = 5 there are no solutions.<br />
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So if a = 1 there are 21+10+3 = 34 options<br />
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If a = 2, b = 3, c= 4, then d = 6,7,8,9,10 so 15 options with b = 3 (same pattern as above). if b = 4, c=5 gives d = 8, 9, 10 so 6 options. if b = 5, c=6 gives just one option, d = 10, and if b=6, there are no options.<br />
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So if a = 2 there are 15+6+1 = 22 options<br />
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If a = 3, b = 4, c = 5, then d = 7, 8, 9, 10 so 10 options (keeping with the same pattern). if b=5 and c = 6, d = 9,10 so 3 options. If b = 6 and c = 7, there are no options.<br />
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So if a = 3 there are 10+3 = 13 options<br />
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If a=4, b=5, c= 6, then d = 8, 9, 10 so 6 options. If b = 6, c= 7 there is just one option<br />
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So if a = 4 there are 6+1 = 7 options.<br />
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If a = 5, b = 6, c = 7, d can be 9, 10 so 3 options. If b = 7, c = 8 there are no options<br />
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So if a = 5 there are 3 options.<br />
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Finally, If a = 6, b = 7, c =8, there is just one option, d = 10<br />
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So if a = 6 there is just 1 option<br />
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Summing, 34 + 22 + 13 + 7 + 3 + 1 = 080, answer</div>Labhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_7&diff=379172011 AIME II Problems/Problem 72011-03-31T14:32:16Z<p>Lab: </p>
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<div>Problem:<br />
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Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal tot he number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let ''m'' be the maximum number of red marbles for which such an arrangement is possible, and let N be the number of ways he can arrange the ''m''+5 marbles to satisfy the requirement. Find the remainder when N is divided by 1000.<br />
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Solution:<br />
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We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 "not the same colors" and 0 "same colors." Now, for every red marble we add, we will add one "same color" pair and keep all 10 "not the same color" pairs. It follows that we can add 10 more red marbles for a total of 16 = ''m''. We can place those ten marbles in any of 6 "boxes": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as (n+k)Choose(k) where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each "box" must contain at least one, are not counted here) and 5 "bars," the green marbles. So the answer is 15Choose5 = 3003, take the remainder when divided by 1000 to get the answer: 003</div>Labhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_7&diff=379162011 AIME II Problems/Problem 72011-03-31T14:32:02Z<p>Lab: </p>
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<div>Problem:<br />
<br />
Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal tot he number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let ''m'' be the maximum number of red marbles for which such an arrangement is possible, and let N be the number of ways he can arrange the ''m''+5 marbles to satisfy the requirement. Find the remainder when N is divided by 1000.<br />
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Solution:<br />
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We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 "not the same colors" and 0 "same colors." Now, for every red marble we add, we will add one "same color" pair and keep all 10 "not the same color" pairs. It follows that we can add 10 more red marbles for a total of 16 = ''m''. We can place those ten marbles in any of 6 "boxes": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as (n+k)Choose(k) where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each "box" must contain at least one, are not counted here) and 5 "bars," the green marbles. So the answer is 15Choose5 = 3003, take the remainder when divided by 100 to get the answer: 003</div>Labhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_3&diff=379052011 AIME II Problems/Problem 32011-03-31T12:14:06Z<p>Lab: </p>
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<div>Problem:<br />
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The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.<br />
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Solution:<br />
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Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.<br />
You get 18''x''+153''y''=2880, because (x+0y)+(x+y)+(x+2y)+...(x+17y)=18''x''+153''y''=the total angle measures of all of the angles in an 18-gon=2880<br />
Solving the equation for integer values (or a formula that I don't know) you get ''x''=7, and ''y''=18<br />
The smallest angle is therefore 7.<br />
However, we aren't done here. The smallest possible angle for a 18-gon with an arithmetic sequence is 7 degrees, we also need ''x''+17''y''<180 because it is convex. By working down from (7,18) to (24,16) etc. we get to the final possibility (143, 2) which satisfies ALL of the requirements.<br />
The smallest angle is therefore 143</div>Labhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_3&diff=379042011 AIME II Problems/Problem 32011-03-31T12:13:18Z<p>Lab: </p>
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<div>Problem:<br />
<br />
The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.<br />
<br />
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Solution:<br />
<br />
Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.<br />
You get 18''x''+153''y''=2880, because (x+0y)+(x+y)+(x+2y)+...(x+17y)=18''x''+153''y''=the total angle measures of all of the angles in an 18-gon=2880<br />
Solving the equation for integer values (or a formula that I don't know) you get ''x''=7, and ''y''=18<br />
The smallest angle is therefore 7.<br />
However, we aren't done here. The smallest possible angle for a 18-gon with an arithmetic sequence is 7 degrees, we also need "x"+17"y"<180 because it is convex. By working down from (7,18) to (24,16) etc. we get to the final possibility (143, 2) which satisfies ALL of the requirements.<br />
The smallest angle is therefore 143</div>Lab