IT’S DAY 4 of Maths Week, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!

TODAY, WE ASK some simple sounding questions about a hypothetical place where there are two types of coin – and no bank notes.

Imagine if there were only €3 and €5 coins available: it turns out that any whole number of euros from €8 onwards (say €17 or €91) could be made up using only these coins, though many coins might be involved. [Some lower totals are impossible to achieve, namely, €1, €2, €4 and €7.]

The same cannot be said if only €3 and €6 coins are available; can you see why?

Let’s transport ourselves to the Kingdom of Kerry, and pretend that the only currency there is the mighty dingle.

We ask some questions that can be answered with a little effort, and end with a similar question whose answer is surprisingly elusive.

First, supposed that there are only 5 and 7 dingles coins, nothing else.

We can’t pay exactly for items costing under 5 dingles, or for things costing 6 (= 3 + 3), 8 (= 3 + 5), 9 (= 3 + 3 +3) or 10 (= 5 + 5) dingles in total. Items costing 11 (or 13) dingles also cannot be paid for exactly. Can you see why?

It turns out that things costing 100 dingles or more can always be paid for exactly.

So the crossover from “may not be possible” to “is always possible” is somewhere between 13 and 100. Can you find it?

What is the smallest positive whole number so that for that amount or any larger whole number amount, it can be made up using 5 and 7 coins?

Let’s call this the tipping number for 5 and 7. Would something like this have worked if we had used only 2 and 6 dingle coins? Or 6 and 15 dingle coins?

Now repeat using several other possible pairs of values, such as 4 and 9, or 6 and 7. Namely, find the tipping numbers for those pairs of values.

Do you see a pattern? Is there a general formula that gives the tipping number no matter what two coin values we start with?

If there were, then you could speedily find the tipping number for 8 and 11 without doing any hard work!

Finally, try the same kind of question assuming there are three coin values available, say, 5, 7 and 11 dingle coins. Repeat using three other coin values? Do you notice a pattern here?

Is there a formula that gives the tipping number for any three coin values?

Come back tomorrow for the answers to today’s puzzle.

Related Reads

Maths Week: Your Monday puzzle

Maths Week: Sunday puzzle

Maths Week: Your Saturday puzzle

Monday’s puzzle: the answer

A. Sinead should give Imelda €85.

Sinead pays for the train (€86) and the concert tickets  (€120), totalling €206. Imelda pays €330 for accommodation and €46 for drinks. A total of €376.

Imelda paid €170 more than Sinead. If Sinead pays half of this to Imelda, they will both have contributed €291.

B. We agreed that Imelda spends more than Sinéad, so divide that excess expense into two equal amounts. If Sinéad pays one of those (a half of the discrepancy) to Imelda, all is well. Now each person has spent the same amount.

C. Suppose for the sake of argument that Tom, Dick and Harry respectively spend €320, €300, and €250. Using the method from before, if Dick first gives €10 to Tom, then Tom, Dick and Harry respectively are out of pocket €310, €310, and €250. So Harry is contributing €60 less than either Tom or Dick. If he gives €20 (a third of the discrepancy) to each of his two pals, the overall contributions of the three lads will finally be equal, namely, €290 each. There are other ways to arrive at the same result, but the upshot is the same.

D. It should be clear how the ideas above can be extended to four (or more) people faced with the same challenge of evening up expenses. It is helpful to list the people in order from biggest to smallest spender before doing the calculations.