Near the Jolly Quiz Master is an archery club, and one evening four of the members were having a post-competition drink. We overheard their post-match analysis, as it were, and it seemed to have been a very odd result.

From what we could gather they had been shooting at a four-ringed target, with seven points for a bullseye, five points for an inner, three points for a magpie, and one point for an outer.

Very oddly, archer A had not hit the bullseye, but had hit all the other rings; archer B had not registered an inner, but had hit all the other rings; archer C had got everything but a magpie; and archer D had not hit an outer, but had registered scores in all the other rings. No one had missed with any arrow.

This seemed very odd, but stranger still they had all scored 30 points.

Intrigued by all this we spent a couple of pints trying to work out the archers' scoring, and we'll leave you with this question: how many bullseyes were scored by the archer who got three magpies?

Archer A, who got no bullseyes, must have shot at least eight arrows, and archer D, with no outers, could have fired no more than eight arrows. So we know they all shot eight arrows.

For archer A we have

i + m + o = 8

5i + 3m + o = 30

With possible scores of five inners (25), one magpie (3) and two outers (2) or four inners (20), three magpies (9) and one outer (1).

Archer B could have got three bullseyes (21), two magpies (6) and three outers (3) or two bullseyes (14), five magpies (15) and one outer (1).

And archer D could only have got one bullseye (7), one inner (5) and six magpies (18).

Archer C got no magpies so we don't need to bother about the scores.

From this, only archer A could have got three magpies, and got no bullseyes.