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A-rectangular-cardboard-is-8cm-long-and-6cm-wide-What-is-the-least-number-of-beads-you-can-arrange-on-the-board-such-that-there-are-at-least-two-of-the-beads-that-are-less-than-10-cm-apart-




Question Number 113355 by Aina Samuel Temidayo last updated on 12/Sep/20
A rectangular cardboard is 8cm long  and 6cm wide. What is the least  number of beads you can arrange on  the board such that there are at least  two of the beads that are less than  (√(10))cm apart.
$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{cardboard}\:\mathrm{is}\:\mathrm{8cm}\:\mathrm{long} \\ $$$$\mathrm{and}\:\mathrm{6cm}\:\mathrm{wide}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{beads}\:\mathrm{you}\:\mathrm{can}\:\mathrm{arrange}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{board}\:\mathrm{such}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{beads}\:\mathrm{that}\:\mathrm{are}\:\mathrm{less}\:\mathrm{than} \\ $$$$\sqrt{\mathrm{10}}\mathrm{cm}\:\mathrm{apart}. \\ $$
Commented by Aina Samuel Temidayo last updated on 12/Sep/20
Solution please?
$$\mathrm{Solution}\:\mathrm{please}? \\ $$
Answered by mr W last updated on 13/Sep/20
Commented by mr W last updated on 13/Sep/20
the points on the grid are (√(10)) apart  from each other.  with a rectangle of 6×8 we can cover  at most 9 points.  it means if we put 10 points or more  on the 6×8 board there are at least two  points which are less than (√(10)) apart  from each other.
$${the}\:{points}\:{on}\:{the}\:{grid}\:{are}\:\sqrt{\mathrm{10}}\:{apart} \\ $$$${from}\:{each}\:{other}. \\ $$$${with}\:{a}\:{rectangle}\:{of}\:\mathrm{6}×\mathrm{8}\:{we}\:{can}\:{cover} \\ $$$${at}\:{most}\:\mathrm{9}\:{points}. \\ $$$${it}\:{means}\:{if}\:{we}\:{put}\:\mathrm{10}\:{points}\:{or}\:{more} \\ $$$${on}\:{the}\:\mathrm{6}×\mathrm{8}\:{board}\:{there}\:{are}\:{at}\:{least}\:{two} \\ $$$${points}\:{which}\:{are}\:{less}\:{than}\:\sqrt{\mathrm{10}}\:{apart} \\ $$$${from}\:{each}\:{other}. \\ $$

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