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| | ==AOPS Contributions== | | ==AOPS Contributions== |
| − | Observe that if <math>1 \le n \le 60</math> such that n is a solution to the desired equation, so is <math>n + 60\cdot m</math>, where m is an integer, <math>0 \le m \le 9</math>.
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| − | \\So we only need to consider n from 1 to 60.
| + | [[https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8#Solution_4|AIME 2022 II Problem 8 Solution 4]] |
| − | As shown in Solution 2, there are 4 cases which we will split into 2 main cases:
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| − | \\Case 1: <math>4 \mid n</math> or <math>6 \mid n</math>, <math>5 \mid (n+1)</math>
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| − | \\Case 2: <math>4 \mid (n+1)</math> or <math>6 \mid (n+1)</math>, <math>5 \mid n</math>
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| − | \\There are 4 values of n where <math>1 \le n \le 12</math> satisfying <math>4 \mid n</math> or <math>6 \mid n</math>.
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| − | \\I claim that there are 4 values of <math>n \le 60</math> satisfying Case 1. Suppose x is one value of n satisfying <math>4 \mid n</math> or <math>6 \mid n</math>, and <math>n \le 12</math>.
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| − | \\Hence the solutions satisfying <math>4 \mid n</math> or <math>6 \mid n</math>, <math>n \le 60</math> are of the form <math>x + 12m</math>, so the values of <math>n + 1</math> are <math>x + 12m + 1 \equiv x + 2m + 1 \equiv 0</math> (mod 5), so <math>2m \equiv 4 + 4x</math> (mod 5) and hence the value of m is unique since <math>0 \le m \le 4</math> to satisfy <math>1 \le n \le 60</math> and 2 and 5 are relatively prime.
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| − | \\A similar approach can be used to show the same for Case 2, that there are 4 values of <math>n \le 60</math>.
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| − | \\Hence our answer is <math>(4+4)*10 = \fbox{080}</math>.
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Revision as of 01:32, 23 February 2022
