Two mathematicians are planning their wedding celebration and are trying to arrange the seating plan for the guests. The only restriction is that all tables must seat the same number of guests and each table must have more than one guest. There are fewer than 350 guests, but they have forgotten the exact number. However they remember that when they try to seat them with two at each table there is one guest left over. The same happens with tables of 3, 4, 5 and 6 guests. When there were 7 guests per table there were none left over. Find the number of guests.

let x be the number of guests

\(x \equiv 1(\bmod 2)\)

\(x \equiv 1(\bmod 3)\)

\(x \equiv 1(\bmod 4)\)

\(x \equiv 1(\bmod 5)\)

\(x \equiv 1(\bmod 6)\)

\(x \equiv 0(\bmod 7)\) congruence (i)     (M1)(A2)

the equivalent of the first five lines is

\(x \equiv 1\left( {\bmod ({\text{lcm of 2, 3, 4, 5, 6}})} \right) \equiv 1(\bmod 60)\)     A1

\( \Rightarrow x = 60t + 1\)

from congruence (i) \(60t + 1 \equiv 0(\bmod 7)\)     M1A1

\(60t \equiv  - 1(\bmod 7)\)

\(60t \equiv 6(\bmod 7)\)

\(4t \equiv 6(\bmod 7)\)

\(2t \equiv 3(\bmod 7)\)     A1

\( \Rightarrow t = 7u + 5\,\,\,\,\,{\text{(or equivalent)}}\)     A1

hence \(x = 420u + 300 + 1\)     A1

\( \Rightarrow x = 420u + 301\)

smallest number of guests is 301     A1     N6

Note: Accept alternative correct solutions including exhaustion or formula from Chinese remainder theorem.

[10 marks]

There were a number of totally correct solutions to this question, but many students were unable to fully justify the result. Some candidates had learnt a formula to apply to the Chinese remainder theorem, but could not apply it well in this situation. Many worked with the conditions for divisibility but did not make much progress with the justification.