If $a$ , $b$ are integers and $s=a^3+b^3-60ab(a+b)\geq 2012$ , find the least possible value of $s$.
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crastybow
08.02.2015 03:42
In mod 7, the LHS rewrites itself as $(a+b)^3 \equiv 0, 1, -1$. Since $7 \mid 2009 < 2012$, the minimal possible value of s is 2015. Factoring s normally yields $(a+b)(a^2-61ab+b^2)$. Letting $a+b = 5$ and $a^2-61ab+b^2 = 403$ yields $a=6, b=-1$, hence 2015 is obtainable, and we're done.
raghoodah1m
09.01.2020 09:22
crastybow wrote:
In mod 7, the LHS rewrites itself as $(a+b)^3 \equiv 0, 1, -1$. Since $7 \mid 2009 < 2012$, the minimal possible value of s is 2015. Factoring s normally yields $(a+b)(a^2-61ab+b^2)$. Letting $a+b = 5$ and $a^2-61ab+b^2 = 403$ yields $a=6, b=-1$, hence 2015 is obtainable, and we're done.
Sorry I did not understand how did you know that the min value is 2015 can you explain please?
DannyBeast
04.03.2020 17:32
raghoodah1m wrote: crastybow wrote:
In mod 7, the LHS rewrites itself as $(a+b)^3 \equiv 0, 1, -1$. Since $7 \mid 2009 < 2012$, the minimal possible value of s is 2015. Factoring s normally yields $(a+b)(a^2-61ab+b^2)$. Letting $a+b = 5$ and $a^2-61ab+b^2 = 403$ yields $a=6, b=-1$, hence 2015 is obtainable, and we're done.
Sorry I did not understand how did you know that the min value is 2015 can you explain please? Look, 2012 = 3 (mod 7), which is not equal to 0, 1 or - 1 mod 7. The same goes to 2013, 2014, but 2015 = -1 (mod 7), Hence it's the answer
sttsmet
08.05.2021 13:18
Nice! In the bigining this exercise seemed really difficult to me.... but now its clear