We looked again.

The paper, as printed, is definitely 8^(1/3)

However, the line underneath the 1 and above the 3, looks longer than necessary. There is room for a minus sign in front of the 1.

He will check with his teacher today.

We looked again.

The paper, as printed, is definitely 8^(1/3)

However, the line underneath the 1 and above the 3, looks longer than necessary. There is room for a minus sign in front of the 1.

He will check with his teacher today.

I’ll post a solution later today or tomorrow.

Tried doing this in my head but had to resort to pen and paper. I think something like this:

t=0: 7, 4

t=4: 3, 0, flip 4

t=4: 3, 4

t=7: 0, 1, flip 7

t=7: 7, 1

t=8: 6, 0 flip 7

t=8: 1, 0

t=9: 0, 0

Well done Ravvie.

And very nicely explained.

However, Mrs D stumbled when she read your 6th line t=8: 6, 0 flip 7 and the 7th line t=8: 6, 0.

It took a while before she realised that she had allowed the 7-minute timer to become confused with t=8. She had started to think she had an 8-minute timer instead, which now had 6 minutes gone and would have 2 minutes to run when Flipped !!

She twigged, eventually, after an extended ‘discussion’

Looks like Valentine’s Day is going to be expensive for me this year …

A bit more spring-cleaning, with another recent teaser.

I’ll post a solution tomorrow or Sunday …

I have nothing to say on this teaser

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Nicely coded response Ravvie. Well done.

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Hi Ravvie.

Would you like to outline the solution for the benefit of others watching this thread.

Or should I

(or possibly Mike ?)

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I’m happy to let Mike have the honour of opening the casket

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Sorry, it has me stumped.

Sorry Mike. I apologise.

I’ll post my solution in a short while. Meanwhile …

The original teaser, from c.20 years ago, involved a Baron, his three sons and his barber. The solution is similar.

The key point is that the Barber, or in today’s teaser, the Duke of Salisbury, gets a * chest* of diamonds, not just a single diamond.

Its then a matter of showing that the remaining diamonds can * always* be divisible by 6 (or in the case of the previous teaser, 3) with none remaining.

The ceramic box doesn’t contain any diamonds.

Mrs R used an empirical approach, going through the possibilities of numbers of rooms, giving an answer of zero each time. She stopped at six rooms, claiming she had exhausted all the possibilities. I made the mistake of asking “what about seven rooms?”

After a Paddington stare, she then spotted the same solution as mine.

If there are r rooms, there are r^2 caskets with r^3 diamonds of which r have been given already. Hence r^3-r = r(r^2-1) diamonds to share initially amongst six sons.

As (r^2-1) = (r-1)(r+1), we have (r-1)r(r+1) diamonds.

Any sequence of three consecutive integers must be divisible by both three and two. Hence there will always be a multiple of six diamonds, so sadly the casket must be empty.

I don’t recall the earlier version of the puzzle, but the relationship (r^2-s^2)=(r-s)(r+s) is a fairly common occurrence in maths so worth remembering. In this case s=1.

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The above was posted in this Thread about 4 years ago. I’d forgotten I’d done that ! Eoink provided the solution.

It was also posted about 20, possibly 24 years ago in the previous Forum and is probably still there in some archive !!

I first saw it in The Sunday Times, whilst sunbathing on a beach in Dubai or Abu Dhabi back in 1969

Nicely done Ravvie.

At school we always used (a^2 - b^2) = (a + b)(a - b)

(rather than r and s).

And we were made to remember it !!!

A bit like Pythag and the various tan/sin/cos rules etc etc etc …

Nicely set out explanation Ravvie.

I trust Mrs R awarded you 10/10 … despite your mistake of asking about seven rooms .

**Flippin’ heck !**

Mrs D has a couple of old two-penny piece coins.

She was pretty sure that with age, the coins had become ‘biased’.

I reconned they were probably ‘fair’

I picked one of the coins and flipped it 250 times. The results were Heads 128 times and Tails 122 times.

I considered this showed this coin was ‘fair’

Mrs D wasn’t convinced so flipped the coin an extra 50 times and firmly asserted that the coin was ‘biased’

After all 300 flips, the relative frequency of Heads was exactly 0.49

For the extra 50 flips that Mrs D made, how many were Heads and how many were Tails ? (1 Mark for a correct answer)

Do you think Mrs D’s assertion that the coin is ‘biased’ is correct ? (9 Marks for a correct answer)

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Hi Don,

good to see that you are still providing us with new teasers. I think it’s fair to say that it’s relatively easier to collect the nine marks than the one

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Mrs D needs to work on her flipping technique. Of her 50 flips, 19 were heads and 31 were tails. As the relative frequency of heads was 0.49 after 300 slips, this would suggest that the coins are not biased. Simply, the relatively frequency will tend towards 0.50 as the number of flips (or sample size) increases. There is probably some complicated graph that shows relative frequency trending towards 0.50 over an infinite sample size, but I’m not going there.

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There were more Heads in the first sample of 250 flips and more Tails in the second sample of 50 flips. As both samples have a sufficient number of flips to show a bias, it would have shown in both samples.

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Hi Mulberry, Hi Mike,

Well done ! You have both provided the mathematically correct answers, and Mike, your figures of 19 and 31 are spot on. I hope you both enjoyed this one.

Clearly, the coin is ‘fair’.

The 9 marks are intended as a bit of fun given that trying to explain to my dearly beloved that she (might, possibly) be wrong is going to be difficult … scary … I might know she is mathematically incorrect … but wrong !!! Never

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