Difference between revisions of "1993 AIME Problems/Problem 4"
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Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are <math>405 + 465 = \boxed{870}</math> four-tuples. | Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are <math>405 + 465 = \boxed{870}</math> four-tuples. | ||
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+ | ==Solution 6== | ||
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+ | Assume <math>d = x+m, a = x-m, c = x+n, and b = x-n</math>. This clearly satisfies the condition that <math>a+d = b+c (2x = 2x)</math>. Now plug this into <math>bc-ad = 93</math>. You get <math>(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93</math> | ||
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+ | Since <math>m>n</math> (as given by the condition that <math>a>b>c>d</math>), <math>m+n>m-n</math> and <math>m</math> and <math>n</math> are integers, there are two cases we have to consider since <math>93 = 3\cdot 31</math>. We first have to consider <math>m-n = 1, m+n = 93</math>, and then consider <math>m-n=3, m+n = 31</math>. | ||
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+ | In the first case, we get <math>m=47, n=46</math> and in the second case we get <math>m=17, n=14</math>. Now plug these values (in separate cases) back into <math>a,b,c,d</math>. Since the only restriction is that all numbers have to be greater than <math>0</math> or less than <math>500</math>, we can write two inequalities. Either <math>x+47 < 500, x-47 > 0</math>, or <math>x+17 < 500, x-17 > 0</math> (using the inequalities given by <math>d</math> and <math>a</math>, and since <math>b</math> and <math>c</math> are squeezed in between <math>d</math> and <math>a</math>, we only have to consider these two inequalities). | ||
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+ | This gives us either <math>47 < x < 453</math> or <math>17 < x < 483</math>, and using simple counting, there are <math>405</math> values for <math>x</math> in the first case and <math>465</math> values for <math>x</math> in the second case, and hence our answer is <math>405+465 = \boxed{870}</math> | ||
== See also == | == See also == |
Revision as of 23:11, 19 July 2019
Problem
How many ordered four-tuples of integers with satisfy and ?
Contents
Solution
Solution 1
Let so . It follows that . Hence .
Solve them in tems of to get . The last two solutions don't follow , so we only need to consider the first two solutions.
The first solution gives us and , and the second one gives us .
So the total number of such four-tuples is .
Solution 2
Let and . From , .
Substituting , , and into , Hence, or .
For , we know that , so there are four-tuples. For , , and there are four-tuples. In total, we have four-tuples.
Solution 3
Square both sides of the first equation in order to get and terms, which we can plug in for.
We can plug in for to get on the left side, and also observe that after rearranging the first equation. Plug in for .
Now observe the possible factors of , which are . and must be factors of , and must be greater than .
work, and yields possible solutions. does not work, because if , then must differ by 2 as well, but an odd number can only result from two numbers of different parity. will be even, and will be even, so must be even. works, and yields possible solutions, while fails for the same reasoning above.
Thus, the answer is
Solution 4
Add the two conditions together to get . Rearranging and factorising with SFFT, . This implies that for every quadruple , we can replace , , etc. and this will still produce a valid quadruple. This means, that we can fix , and then just repeatedly add to get the other quadruples.
Now, our conditions are and . Replacing in the first equation, we get . Factorising again with SFFT gives . Since , we have two possible cases to consider.
Case 1: , . This produces the quadruple , which indeed works.
Case 2: , . This produces the quadruple , which indeed works.
Now, for case 1, we can add to each term exactly times (until we get the quadruple ), until we violate . This gives quadruples for case 1.
For case 2, we can add to each term exactly times (until we get the quadruple ). this gives quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence possible quadruples.
Solution 5
Let . From the equation , we have so and . We then have Since , , or . Since the prime factorization of 93 is , we must either have and , or and . We consider these cases separately.
If , then , , and . Thus can be any integer between 95 and 499, inclusive, and our choice of determines the four-tuple . We therefore have possibilities in this case.
If , then , , and . Thus can be any integer between 35 and 499, inclusive, and our choice of determines the four-tuple , as before. We therefore have possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are four-tuples.
Solution 6
Assume . This clearly satisfies the condition that . Now plug this into . You get
Since (as given by the condition that ), and and are integers, there are two cases we have to consider since . We first have to consider , and then consider .
In the first case, we get and in the second case we get . Now plug these values (in separate cases) back into . Since the only restriction is that all numbers have to be greater than or less than , we can write two inequalities. Either , or (using the inequalities given by and , and since and are squeezed in between and , we only have to consider these two inequalities).
This gives us either or , and using simple counting, there are values for in the first case and values for in the second case, and hence our answer is
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.