Difference between revisions of "Talk:2022 AIME I Problems/Problem 2"
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Which solutions are you referring to? | Which solutions are you referring to? | ||
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| + | All possible remainders modulo <math>71</math> are <math>\{0,1,2,\ldots,70\}.</math> | ||
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| + | Although <math>72</math> and <math>143</math> are not in the set, it is still true that <math>72\equiv143\pmod{71}.</math> Recall that the numbers <math>a</math> and <math>b</math> are congruent modulo <math>71</math> if and only if <math>a-b\equiv0\pmod{71}.</math> | ||
Latest revision as of 15:20, 5 March 2023
seems the most beautiful answers using modulus and inequality are flawed, and need minor revisions to fit the rigor of math
(modulus) 8c mod 71 can be 1 if c=9 instead of 8c
(inequality) 99(a-b) can be negative, and 4(2c-7b) can have a few negative values to get them equal.
Which solutions are you referring to?
All possible remainders modulo 
are 
Although 
and 
are not in the set, it is still true that 
Recall that the numbers 
and 
are congruent modulo if and only if 
