Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 3

The Chinese remainder theorem for two comaximal two-sided ideals $I,J$ of a ring states that the natural ring homomorphism inducesIndeed, the map $r\mapsto (r+I,r+J)$ has kernel of a ring homomorphism $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving surjectivity. The first isomorphism theorem for rings gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.

Take $R=\mathbb{Z}$, $I=a\mathbb{Z}$ and $J=b\mathbb{Z}$. Since $a,b$ are coprime integers, Bézout's identity gives $ra+sb=1$, so $I+J=\mathbb{Z}$. Their intersection is $ab\mathbb{Z}$: it consists of integers divisible by both $a$ and $b$, and coprimality makes this equivalent to divisibility by $ab$. Thus the Chinese remainder theorem givesas a ring isomorphism. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.

Let $N$ count everyone present, including the host. The grouping, seating and departure conditions give the modular congruencesThe first two imply $N=39+45k$, and oddness forces $k$ even. By the Chinese remainder theorem, the complete solution isBecause not every invitee attended, at most $y-1$ guests came, so $N\le y$. For $y=120$, the only positive possibility is 39 people. It is feasible: seven groups of five and one of four; three tables of ten and one of nine; nineteen departing pairs and the host.

For $y=150$, both 39 and 129 people are possible. The latter uses twenty-five groups of five and one of four, three tables of ten and eleven of nine, and sixty-four departing pairs. Thus the larger invitation count destroys uniqueness; it does not force a different attendance.