Question Number 202037 by syamilkamil1 last updated on 19/Dec/23
$$ \\ $$A board has 2, 4, and 6 written on it. One repeatedly chooses values (not necessarily different) for x, y, and z from the board, and writes xyz + xy + yz + zx + x + y + z if and only if those numbers are not already on the board and are also less than or equals 2013. The person repeats this process until no more numbers can be written. How many numbers will be written at the end of this process?
Answered by nikif99 last updated on 20/Dec/23
$${Just}\:\mathrm{20}.\:{See}\:{image}\:{for}\:{details}. \\ $$$${Proposition}:\:{If}\:{a}\:{number}\:{n}>=\mathrm{223}\:{is}\: \\ $$$${written},\:{it}\:{ceases}\:{procedure}\:{with}\:{this} \\ $$$${n}\:{because}\:{next}\:{result}\:{exceeds}\:\mathrm{2013}. \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{2},\mathrm{2},\mathrm{223}\right)\rightarrow\mathrm{2}×\mathrm{2}×\mathrm{223}+\mathrm{2}×\mathrm{2}+ \\ $$$$\mathrm{2}×\mathrm{223}+\mathrm{2}×\mathrm{223}+\mathrm{2}+\mathrm{2}+\mathrm{223}=\mathrm{2015}. \\ $$$${In}\:{the}\:{region}\:{A}\mathrm{1}..{C}\mathrm{11}\:{are}\:\mathrm{10}\:{possible} \\ $$$${cases}\:{when}\:{only}\:\mathrm{2},\mathrm{4},\mathrm{6}\:{are}\:{written}\:\left({see}\right. \\ $$$$\left.{col}.\:{E}\:{for}\:{start}\right). \\ $$$${For}\:{each}\:{of}\:{these}\:{cases},\:{application}\:{of} \\ $$$${the}\:{given}\:{formula}\:{for}\:{each}\:{row}\:{gives} \\ $$$${as}\:{result}\:{column}\:{N},\:{so}\:{the}\:{respective} \\ $$$${number}\:{can}\:{be}\:{written}\:{on}\:{the}\:{board}. \\ $$$$\bigstar{For}\:{the}\:\mathrm{2}{nd}\:{round}\:{you}\:{apply}\:{each} \\ $$$${number}\:{from}\:{the}\:{range}\:{N}\mathrm{2}..{N}\mathrm{11}\:{to}\:{the} \\ $$$${cell}\:{N}\mathrm{1}\:\left({in}\:{the}\:{example},\:{N}\mathrm{6}=\mathrm{62}\right),\:{which} \\ $$$${copies}\:{possible}\:{triplettes}\:{to}\:{the}\:{range} \\ $$$${A}\mathrm{13}..{C}\mathrm{22}. \\ $$$${So},\:{you}\:{have}\:\mathrm{10}\:{new}\:{cases}\:{to}\:{apply}\:{the} \\ $$$${formula}\:\left({if}\:{you}\:{omit}\:\mathrm{62}\:{and}\:{use}\:\mathrm{2},\mathrm{4},\mathrm{6}\right. \\ $$$$\left.{only},\:{you}\:{are}\:{driven}\:{to}\:{a}\:{repetition}\right). \\ $$$${New}\:{triplettes}\:{of}\:{rows}\:\mathrm{13}\:{to}\:\mathrm{22}\:{lead}\:{to} \\ $$$${results}\:{N}\mathrm{13}..{N}\mathrm{22}.\:{Not}\:{all}\:{results}\:{are} \\ $$$${acceptable}\:{since}\:{some}\:{of}\:{them}\:\left({marked}\right. \\ $$$$\left.{in}\:{red}\right)\:{are}\:>\mathrm{2013}. \\ $$$${Accepted}\:{results}\:{are}\:{written}\:{vertically} \\ $$$${in}\:{cols}\:{F}\:{to}\:{M}\:\left({in}\:{our}\:{example},\:{H}\right). \\ $$$${Repeating}\:{the}\:{same}\:{procedure}\:{from}\:\bigstar \\ $$$${for}\:{all}\:{numbers}\:{in}\:{N}\mathrm{2}..{N}\mathrm{11}\:{you}\:{get} \\ $$$${results}\:{in}\:{cols}\:{F}\:{to}\:{M}. \\ $$$${It}\:{is}\:{impossible}\:{to}\:{follow}\:\mathrm{3}{rd}\:{round} \\ $$$${because}\:{all}\:{numbers}\:{in}\:{cols}\:{F}\:{to}\:{M} \\ $$$${are}\:>\mathrm{223}\:\left({see}\:{proposition}\:{above}\right). \\ $$$${Finally},\:{collecting}\:{all}\:{produced}\:{numbers} \\ $$$${you}\:{get}\:{final}\:{result}\:{in}\:{A}\mathrm{24}..{H}\mathrm{28}. \\ $$
Commented by nikif99 last updated on 20/Dec/23
