Answer to Question #297635 in Combinatorics | Number Theory for Allan

Question #297635

Six professor begin courses on Monday, Tuesday, Wednesday, Thursday, friday, and Saturday, and announce their intention of lecturing at interval of 3,2,5,6,1and 4 days respectively the regulations of the university forbid Sunday a lecture(so that a Sunday a lecture must be omitted) when first will all six professors find themselves compelled to omit a lecture

1
Expert's answer
2022-02-16T11:38:28-0500

If the day in question is the xth (counting from and including the first monday), then "\\mathrm{x}=1+2 \\mathrm{k}_{1}=2+3 \\mathrm{k}_{2}=3+4 \\mathrm{k}_{3}=4+\\mathrm{k}_{4}=5+6 \\mathrm{k}_{5}=6+5 \\mathrm{k}_{6}=7 \\mathrm{k}_{7}"

where the \mathrm{k} are integers ie

 "(1) \\mathrm{x} \\equiv 1(\\bmod 2),\\\\(2) \\mathrm{x} \\equiv(\\bmod 3),\\\\(3) \\mathrm{x} \\equiv 3(\\bmod 4) \n\n\\\\ (4) \\mathrm{x} \\equiv 4(\\bmod 1),\\\\(5) \\mathrm{x} \\equiv 5(\\bmod 6),\\\\(6) \\mathrm{x} \\equiv 6(\\bmod 5) \n\n \\\\(7) \\mathrm{x} \\equiv 0(\\bmod 7)"

Of these congruences, (4) is no restriction and (1) and (2) are included in (3) and (5).

Of the two later, (3) shows that "\\mathrm{x}" is congruent to 3,7, or "11(\\bmod 12)" , and (5) that "\\mathrm{x}" is congruent

to 5 or 11 , so that (3) and (5) together are equivalent to "\\mathrm{x} \\equiv 11(\\bmod 12)" . Hence the problem is that of solving

"x \\equiv 11(\\bmod 12), x \\equiv 6(\\bmod 5), x \\equiv 0(\\bmod 7)"

"\\mathrm{x} \\equiv-1(\\bmod 12), \\mathrm{x} \\equiv 1(\\bmod 5), \\mathrm{x} \\equiv 0(\\bmod 7)"

Here "\\mathrm{m}_{1}=12, \\mathrm{~m}_{2}=5, \\mathrm{~m}_{3}=7, \\mathrm{~m}=420, \\mathrm{M}_{1}=35, \\mathrm{M}_{2}=84, \\mathrm{M}_{3}=60" .

The "\\mathrm{n}" are given

"35 \\mathrm{n}_{1} \\equiv 1(\\bmod 12), 84 \\mathrm{n}_{2} \\equiv 1(\\bmod 5), 60 \\mathrm{n} \\equiv 1(\\bmod 7)"

or

"-\\mathrm{n}_{1} \\equiv 1(\\bmod 12),-\\mathrm{n}_{2} \\equiv 1(\\bmod 5), 4 \\mathrm{n}_{3} \\equiv 1(\\bmod 7)"

and we can take "\\mathrm{n}_{1}=-1, \\mathrm{n}_{2}=-1, \\mathrm{n}_{3}=2" . Hence

 "\\mathrm{x}=(-1)(-1) 35+(-1) 1.84+2.0 .60=-49 \\equiv 371(\\bmod 420)"

The first "\\mathrm{x}" satisfying the condition is 371 .


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