In the sequence of positive integers, 2022 is, of course, the 2022nd number. If we delete all squares and cubes from the sequence, in which position does 2022 now sit?
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As $44^{2}=1936<2022<45^{2}=2025$ there are 44 squares less than or equal to 2022. As $12^{3}=1728<2022<13^{3}=2197$ there are 12 cubes less than or equal to 2022. However, the sixth powers $1^{6}=1^{3}=1^{2}=1,2^{6}=4^{3}=8^{2}=64$ and $3^{6}=9^{3}=27^{2}=729$ have been double counted, hence the number of numbers remaining is $2022-44-12+3=1969$.
Problem 2
Let $a, b, c$ be integers. Prove that
\[abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\]
is always divisible by 7.
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If any of $a, b, c$ are $0 \bmod 7$ or equal mod 7 the problem is done. So they must be $1,2,3$, 4,5 , or $6 \bmod 7$, and unequal. Note that $a^{3} \bmod 7$ is congruent to 1 if $a=1,2,4$, and 6 if $a=3,5,6$. Therefore by the Pigeonhole Principle at least two of $a^{3}, b^{3}$ and $c^{3}$ must be congruent mod 7. So $a b c\left(a^{3}-b^{3}\right)\left(b^{3}-c^{3}\right)\left(c^{3}-a^{3}\right) \equiv 0 \bmod 7$ for all choices of $a, b, c$.
Credits:
Problem 1: This problem is a modified version of Question 1 from The University of Melbourne Mathematics Competition, 2016 Senior Division
Problem 2: This problem is a modified version of Question 6 from The University of Melbourne Mathematics Competition, 2016 Senior Division