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Question Number 150346 by mathdanisur last updated on 11/Aug/21

If  x;y and z are positive integers, then  determine the smallest positive integer  N=x+y+z+xy+yz+zx, which is  bigger than 2022.

$$\mathrm{If}\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\:\mathrm{and}\:\boldsymbol{\mathrm{z}}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers},\:\mathrm{then} \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive}\:\mathrm{integer} \\ $$$$\boldsymbol{\mathrm{N}}=\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx},\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{bigger}\:\mathrm{than}\:\mathrm{2022}. \\ $$

Commented by Rasheed.Sindhi last updated on 12/Aug/21

N=2023 for   {x,y,z}={2,4,287}

$$\mathrm{N}=\mathrm{2023}\:\mathrm{for}\: \\ $$$$\left\{\mathrm{x},\mathrm{y},\mathrm{z}\right\}=\left\{\mathrm{2},\mathrm{4},\mathrm{287}\right\} \\ $$

Commented by mathdanisur last updated on 12/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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