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# Eight Candle-based Maths Puzzles

Around this time of year, people light a lot of candles in celebration of the season. Here is a selection of candle-based maths puzzles you can share with friends and family! We’ll share the solutions in a later post.

• You go to the shop to buy a box of candles and a box of matches. The total cost is €1.10, and the shopkeeper says that the candles cost €1 more than the matches. How much did the matches cost?
• 200 people are gathered in a room for a party, and everyone has an unlit candle. One person lights their candle, and then walks to another person in the room whose candle is not yet lit, and uses it to light that person’s candle. Then both people whose candles are burning go and find someone else’s candle to light, and so on – at each stage, anyone whose candle is already lit finds one other person and lights their candle. If it takes 30 seconds for each person (including the first) to find a friend and light a candle, how long does it take from when the first candle is lit to when all the candles in the room are burning?
• You have two candles which each take 1 hour to fully burn, and they burn at a constant rate along their length. You also have a lighter. You’re given the task of measuring exactly 90 minutes using the objects you have (and nothing else – no clock, weighing scale or cutting equipment). How can you do this?
• You have nine identical candles, but one is made from a different type of wax which means it is heavier than the others. You have only a balance scale to compare the weights of objects. What’s the fewest times you need to use the balance to work out which candle is the heavy one?
• The end pieces of wax left over after burning ten candles will yield one extra candle if you melt them all together. If you burned 100 candles, how many extra candles could you make (assuming you have as much wick as you need)?
• Assuming you follow the standard pattern of blowing out the same number of candles on your cake as your age each birthday, how many birthday candles will you blow out in your lifetime, if you live to be N years old?
• It’s your seventh birthday, and in front of you is a round cake with seven candles arranged in a circle around the edge. Your goal is to blow out all of the seven candles – but you’re not very good at directing your breath (you are seven, after all – you haven’t had much practice at blowing out candles). If you blow out one candle, the two candles either side are also blown out. But your friends have set up the cake with trick candles! When they are blown on, if they’re lit they go out, but if they’re currently unlit, they light up again. Whichever candle you blow on, that one and the two candles either side change from lit to unlit depending on what they previously were. How can you blow out all the candles using the fewest breaths?
• I like using number-shaped candles on birthday cakes, but I only have five 0-shaped candles and five 1-shaped candles. If I represent numbers in binary, what’s the smallest age I won’t be able to represent on a cake?

### Posted by Katie Steckles

is a mathematician based in Manchester, who gives talks and workshops on different areas of maths. She finished her PhD in 2011, and since then has talked about maths in schools, at science festivals, on BBC radio, at music festivals, as part of theatre shows and on the internet. Katie writes blog posts and editorials for The Aperiodical, a semi-regular maths news site.

### 1 comment

1. Katie Steckles wrote (10. Dec 2019):
> […] in front of you is a round cake with seven candles arranged in a circle around the edge. Your goal is to blow out all of the seven candles
>
Whichever candle you blow on, that one and the two candles either side change […] if they’re lit they go out, but if they’re currently unlit, they light up again.
> How can you blow out all the candles using the fewest breaths?

 ******* ooo.... ---**** .ooo... -**-*** ..ooo.. -*-*-** ...ooo. -*--*-* ....ooo -*---*- o....oo **----* oo....o ------- 

Chrys schrieb (09.12.2019, 00:18 Uhr):
> Die interessante Frage ist dabei doch, welchen Wert t […]

… jemand der Anzeige B_Q
(d.h. der Anzeige der Bahnsteig-Ausfahrts-Kante B beim Treffen/Passieren des vorderen Zugendes “Lok” Q)
als Ablesewert zuordnen würde ?

Wer weiß ?? …
Ganz egal ob und zu welchem t-Wert sich wer-auch-immer entschließen mag:

Nach Versuchsanordnung war und bleibt die Anzeige A_P
(d.h. die Anzeige der Bahnsteig-Einfahrts-Kante A beim Treffen/Passieren des hinteren Zugendes “Caboose” P)
gleichzeitig zu Anzeige B_Q.

p.s.
> Für β ≠ 0 sind der geometr. modellierte Zug und sein “Lorentz-projizierter Schatten” auf der x-Achse zwei verschiedene Objekte.

»sein Lorentz-projizierter Schatten auf der x-Achse«

»der Bahnsteig, dessen Bestandteile alle gleichzeitig das Passieren des Zuges anzeigten«

?
.

> Zu sagen, der bewegte Zug sei kontrahiert, mag für Lorentz gerechtfertigt sein

Dadurch würde H. A. Lorentz zu Unrecht unterstellt, er habe überhaupt einen nachvollziehbaren Begriff von “Länge (und wie man sie vergleicht)” gehabt — und Einsteins Leistung würde entsprechend geschmälert.

p.p.s.
Frank Wappler schrieb (03.12.2019, 12:20 Uhr):
> Raumzeit-Intervalle, s^2 […] Nach Überprüfung halte ich die folgenden Verhältniswerte für vorzeigbar: […]

Bzgl. der gerade ein weiteres Mal erwähnten Versuchsanordnung, schematisch (also nicht unbedingt maßstäblich): ::::::::::::: A ============ M ============ B ::::::::: F :::::               ε ~~>                         ε ::: K ::::::: P ::::::::::::::::::::::::::: Q :::::::::::: 

gefolgt von ::::::: A ::::::::::::: M ::::::::::::: B ::::::::: F :::::         ε                                       ~~> ε ::::::: K ======= P ========= N =================== Q :::::::: 

erweisen sich die folgenden Verhältniswerte der betreffenden Raum-Zeit-Intervalle als … noch vorzeigbarer:

(s^2[ ε_AP, ε_BQ ] / s^2[ ε_AK, ε_FQ ]) = (1 – β) / (1 + β),

(s^2[ ε_AP, ε_FQ ] / s^2[ ε_AK, ε_FQ ]) = 0,

(s^2[ ε_BQ, ε_FQ ] / s^2[ ε_AK, ε_FQ ]) = -1,

(s^2[ ε_AP, ε_AK ] / s^2[ ε_AK, ε_FQ ]) = (β – 1) / (1 + β),

(s^2[ ε_BQ, ε_AK ] / s^2[ ε_AK, ε_FQ ]) = 0.