If-A-is-a-fifty-element-subset-of-the-set-1-2-3-100-such-that-no-two-numbers-from-A-add-up-to-100-show-that-A-contains-a-square-

Question Number 22082 by Tinkutara last updated on 10/Oct/17

If A is a fifty - element subset of the set

{1, 2, 3, \dots ., 100} such that no two

numbers from A add up to 100 show

that A contains a square .

Answered by Rasheed.Sindhi last updated on 14/Oct/17

The subset of {1, 2, \dots, 100},

(i) which {doesn}^{'} t contain perfect

square

and

(ii) whose no two elements add

up to 100

has at most 49 elements and are

as follows . All the possible

sets having at most elements

are given here . (Note thatby

notation a ∣ b here is meant

^{'} one of a or b^{'}) :

{\overset{1}{99} \overset{2}{, 2 ∣ 98} \overset{3}{, 3 ∣ 97}, \overset{4}{96} \overset{5}{, 5 ∣ 95} \overset{6}{, 6 ∣ 94} \overset{7}{, 7 ∣ 93},

\overset{8}{8 ∣ 92}, \overset{9}{91}, \overset{10}{10 ∣ 90}, \overset{11}{11 ∣ 89}, \overset{12}{12 ∣ 88}, \overset{13}{13 ∣ 87}

\overset{14}{14 ∣ 8}, \overset{15}{15 ∣ 85}, \overset{16}{84}, \overset{17}{17 ∣ 83}, \overset{18}{18 ∣ 82}, \overset{19}{19},

\overset{20}{20 ∣ 80}, \overset{21}{21 ∣ 79}, \overset{22}{22 ∣ 78}, \overset{23}{23 ∣ 77}, \overset{24}{24 ∣ 76},

\overset{25}{75}, \overset{26}{26 ∣ 74}, \overset{27}{27 ∣ 73}, \overset{28}{28 ∣ 72}, \overset{29}{29 ∣ 71},

\overset{30}{30 ∣ 70}, \overset{31}{31 ∣ 69}, \overset{32}{32 ∣ 68}, \overset{33}{33 ∣ 67}, \overset{34}{34 ∣ 66},

\overset{35}{35 ∣ 65}, \overset{36}{37 ∣ 63}, \overset{37}{38 ∣ 62}, \overset{38}{39 ∣ 61}, \overset{39}{40 ∣ 60},

\overset{40}{41 ∣ 59}, \overset{41}{42 ∣ 58}, \overset{42}{43 ∣ 57}, \overset{43}{44 ∣ 56}, \overset{44}{45 ∣ 55},

\overset{45}{46 ∣ 54}, \overset{46}{47 ∣ 53}, \overset{47}{48 ∣ 52}, \overset{48}{51}, \overset{49}{50}}

∴ 50 th number must be perfect

square .

Commented by Tinkutara last updated on 14/Oct/17

Thank you very much Sir!