Difference between revisions of "Chittur Gopalakrishnavishwanathasrinivasaiyer Lemma"
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− | We know that an <math>x</math> exists that equal to <math>42\, \cdot</math> <math>\text{mod} \sqrt{4761}.</math> This <math>x</math> is very powerful in competition math problems. Usually coming up on JMO and AMO geo problems. The Euler Line intersects the radical axis at <math>(x^n, n^x)</math> where <math>n</math> is the number of composite factors the radius has. This theorem is also used in Newton's Sums, as the <math>n</math>th root unity is the same thing as <math>x^n</math> <math>\text{mod}</math> <math>(42*10\cdot(70-1)^n).</math> Finally, you'll se it in combo! The number ways you can shuffle <math>n</math> things into <math>n^2 + nk + 1</math> items where <math>k</math> is the number of partitions in an item is the <math>x^{23\cdot3}.</math> My | + | We know that an <math>x</math> exists that equal to <math>42\, \cdot</math> <math>\text{mod} \sqrt{4761}.</math> This <math>x</math> is very powerful in competition math problems. Usually coming up on JMO and AMO geo problems. The Euler Line intersects the radical axis at <math>(x^n, n^x)</math> where <math>n</math> is the number of composite factors the radius has. This theorem is also used in Newton's Sums, as the <math>n</math>th root unity is the same thing as <math>x^n</math> <math>\text{mod}</math> <math>(42*10\cdot(70-1)^n).</math> Finally, you'll se it in combo! The number ways you can shuffle <math>n</math> things into <math>n^2 + nk + 1</math> items where <math>k</math> is the number of partitions in an item is the <math>x^{23\cdot3}.</math> My coaches Iyer Sir and Barnes approved this nice lemma. |
Latest revision as of 20:59, 20 September 2021
We know that anexists that equal to
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This
is very powerful in competition math problems. Usually coming up on JMO and AMO geo problems. The Euler Line intersects the radical axis at
where
is the number of composite factors the radius has. This theorem is also used in Newton's Sums, as the
th root unity is the same thing as
![]()
![]()
Finally, you'll se it in combo! The number ways you can shuffle
things into
items where
is the number of partitions in an item is the
My coaches Iyer Sir and Barnes approved this nice lemma.