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# My Teachers Three Kids: Newsletter Challenge (06/05/07)

Posted June 03, 2007 5:01 PM
Pathfinder Tags: challenge questions

The question as it appears in the 06/05 edition of Specs & Techs from GlobalSpec:

Tom meets his old math teacher John.
"Hello John, how old are your three children now?"
"Well, if you multiply their ages together you get 36, and, if you add their ages you get the number on that bus."
Tom is perplexed, so after a while John adds, "Oh! And the oldest plays the piano," at which point Tom is happy.

How old are the three children? What number is on the bus?

Thanks to Randall who submitted the original question.

(Update: June 12, 8:39 AM EST) And the Answer is...

The trick is in realising that the first two bits of information are not quite enough for Tom to arrive at a unique solution. The final snippet (somewhat obliquely) resolves his dilemma.

There are only eight possible combinations of ages which when multiplied together give 36.

1, 1, 36 (sum 38)
1, 2, 18 (sum 21)
1, 3, 12 (sum 16)
1, 4, 9 (sum 14)
1, 6, 6 (sum 13)
2, 2, 9 (sum 13)
2, 3, 6 (sum 11)
3, 3, 4 (sum 10)

If the number on the bus had been anything other than 13 then Tom would have known immediately how old all the children were. Of the two possibilities which add up to 13 only the answer [2, 2, 9] has an "oldest" child, and, this is therefore the answer.

Congrats to ve9gfi for beign the first to correctly answer the puzzle.

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#1

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/03/2007 5:42 PM

It could be 2,3,6 or 1,4,9 or 2,2,9 (twins)

A child of 9 is a better candidate for playing a piano ==> it must be one of the last two.

The owner of the bus may be superstitious and my guess is a 13 will not be used.

my answer is therefore 1,4,9 and 14 on the bus.

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#2

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/03/2007 6:43 PM

If the teacher does music his 6 year old may be capable of playing.

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#97

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 8:59 AM

also 1,3,12 1,1,36 3,3,4 1,2,18

Anonymous Poster
#102

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 9:46 AM

Since the oldest playing the piano is the clue I would suggest that that means that the youngest are twins. 2,2,9 and 13 for the number on the bus. The other combination that produces 13 for the bus number has 1,6,6 ie no singular oldest.

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#3

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/03/2007 6:55 PM

ok a math teacher may not do music.

I forgot about the 1,2,18 and 21, and 1,1,36 and 38 (sporting model twins)

but presumed the math teacher did calculate the odds.

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#4

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:35 AM

When I got into bed I realized I forgot another possibility.

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#5

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 3:47 AM

Aren't you guys forgetting the youngest may be less than 1 year old !

e.g. Zero!

Ok no one says my kid is zero, they'd say 'oh he's 8months' or 12 weeks whatever..but that's still zero in years.

Or do I deserve a slap with a wet haddock? mmmmm wet haddock

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#6

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 4:10 AM

<SLAP!>

If one of them has a (nominal) age of zero, then the product of their ages will be zero.

<SLAP!>

You're enjoying that aren't you? Perhaps I'll go slap Jarad instead and make him release Kris.

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#7

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 4:18 AM

By golly I needed that...

I s'pose 0.5, 4, 18 is a bit silly ...

And I've been waiting for that 22 & 1/2 bus for ages now!

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#30

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:50 AM

Yes, but if it's a school bus, there is a chance it has a 1/2. I rode bus 16 1/2 in high school.

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#31

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:06 AM

Excellent!...

There is plenty of room for fun with this!

Did you miss it half the time?

Did you all sit at the back....didn't we all?

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#34

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:32 AM

I used to sit at the front sometimes - worried the .... out of the driver

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#113

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/07/2007 10:49 AM

I rode the same bus...The short bus!

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#8

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 4:58 AM

0 x a x b = 0

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#9

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 5:00 AM

What about 1,3,12 and 16 on the bus?

Or 3,3,4 and 10?

Or 1,6,6 and 13?

I haven't spotted the significance of the piano playing yet.

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#10

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 5:26 AM

the factors of 36 are 2,2,3,3 so it could be;

1,1,36 bus38

1,2,18 bus21

2,2,9 bus13

or

2,3,6 bus 11

but, because tom was perplexed it must be because of the bus #38 [it is rare for a sum to be greater than a product]

=-==

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#15

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 11:57 AM

Surely every number has an infinite number of factors that are larger than itself - if you allow 1 to be used.

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#11

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 8:36 AM

Hi All,

The factors of 36 are 1,2,2,3,3 which results in the following combinations of possible kid ages and the resulting bus #s.

1+2+18=21

1+3+12=16

1+4+9=14

1+6+6=13

2+2+9=13

2+3+6=11

Tom was told the sum of the kids ages (i.e. the bus number) and he was perplexed. Thus the bus number has to be non-unique and it's 13.

This narrows the ages to 1,6,6 or 2,2,9.

Also, John then added that the oldest plays Piano ... this rules out the answer with two same "oldest ages" and the remaining answer is 2,2,9.

So the kids are ages 2, 2, and 9, and the bus number is 13.

Greg

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#12

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 8:53 AM

Nice.

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#13

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 9:19 AM

And their names?

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#14

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 11:55 AM

Logical, and assuming that Tom is numerate it's the only possible solution; but it would appear you have little experience of families with twins. My wife is half a pair of twins - and the older-younger-sibling rivalry has to be seen to be believed.

Oh, and my children started playing the piano as soon as they could get their sticky little paws on it.

The only conclusion that I could draw is that all the piano teachers in this town refuse to teach children of six and under, and John actually said "... the oldest is taking piano lessons"

But "challenge"? It took about 30 seconds to write down the complete set of possible factorisations - and that included 3,3,4 and 1,1,36 because I was in zombie mode by then.

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#18

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:34 PM

Being in 'zombie mode' can be useful. 38 people fit on a bus easily.

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#24

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 4:05 PM

Rare to see that around here - the buses usually go around practically empty. No wonder Del's negative numbers were needed.

Fyz

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#73

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 8:21 PM

Except at UC Davis. Unitrans can fit 138 people on a bus easily.

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#16

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 11:58 AM

Oh, and what about 11 months' age differential

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#46

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 10:25 AM

I think Greg nailed it. Nice answer.

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#63

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:39 PM

Thanks Troy!

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#50

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:25 AM

It said if you "multiply thier ages" not the "sum of thier ages".

Anonymous Poster
#52

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:35 AM

Look more carefully: he wrote down options that multiply to 36, so he could check which options (for the sum) give ambiguity when you also know the number on the bus.

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#99

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 9:12 AM

BINGO!

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#109

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/07/2007 12:59 AM

Greg's reply in #11 is correct.

If I may add something to the explaination, the trick is that John knows the bus number whereas we don't.

If the puzzle were asked again, and the bus number given as 13, then the trick to the puzzle is revealed, as is the importance of the statement about the oldest child.

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#139

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 4:45 AM

I am quite new to all this and found your calculation quite interesting but on thing I could not work out is how you got to the factors of 36 being 1.2.2.3.3 which formula did you use to arrive at this answer?

Has I have said I am quite new to all this.

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#142

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 7:39 AM

This is a basic maths tool (I think I was taught it at age 11 or 12) whereby one reduces a number to the product of its primes (a prime number number being one that can not be divided by anything other than itself and 1).

for example, the first few prime numbers are (ignoring 1 coz it's special!):

2, 3, 5, 7, 9, 11, 13....

4 = 2x2, so it's not a prime. I hope you can see where this is going.

To find the prime factors of 36, just pick two factors:

36 = 6x6 (probably the most obvious pair)

but 6 = 3x2 (both prime)

so 36 = 2x2x3x3 - the full prime factors.

1 is usually left out since all numbers are divisible by 1, so it's rather stating the obvious.

Why don't you try starting from other factors of 36, such as 2x18, to show that you always end up with the same set of prime factors.

There probably is a formula for doing this - I would imagine that it would involve solving for 2^a x 3^b x 5^c.....

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#144

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 8:25 AM

On this subject, an interesting thing is mathematicians take the trouble to prove that prime factorisation is unique, though it seems fairly obvious.

However, it's possible to define primes in the field of complex numbers (I don't know the details) and in this case prime factorisation is not unique i.e. some complex numbers can be broken down into more than one set of prime factors. In the 18th century I think it was somebody claimed a proof of Fermat's Last Theorem assuming unique complex prime factorisation. Another mathematician (a German, Kummer, if I remember right) pointed out the mistake and the proof was withdrawn.

Cheers....Codey

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#147

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/14/2007 1:52 AM

If the method you were shown is good ER , could you factor this for me ;

25195908475657893494027183240048398571429282126204
03202777713783604366202070759555626401852588078440
69182906412495150821892985591491761845028084891200
72844992687392807287776735971418347270261896375014
97182469116507761337985909570009733045974880842840
17974291006424586918171951187461215151726546322822
16869987549182422433637259085141865462043576798423
38718477444792073993423658482382428119816381501067
48104516603773060562016196762561338441436038339044
14952634432190114657544454178424020924616515723350
77870774981712577246796292638635637328991215483143
81678998850404453640235273819513786365643912120103
97122822120720357

I'll give you a small % of any money I make from you telling me.

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#152

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/27/2007 3:51 PM

ER, I do not know of a formula to find the prime factors of a number, but I do know about a formula that finds prime numbers. It is the Sieve of Erosthenes. Years ago, while a friend was tutoring me in Fortran, I wrote a program using the fromula.

BTW, 9 is not prime.

A few tips to find prime factors:

Even numbers have 2 at least once as a prime factor.

Numbers ending in 0 or 5 have 5 at least once as a prime factor.

Any number that has all digits that are divisible by 3, or the sum of the digits renders a number divisible by 3, will have 3 at least once as a prime factor.

I'm not aware of any tips for 7, 11, etc.

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#153

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/27/2007 4:24 PM

Checking for 7 is not simple, but 11 is straightforward for a reason equivalent to 3.

The rule for 3 results from the fact that 9 = 10-1. So, when you add 9 to a number, you reduce the units by one and increase the remainder by one - so you do not change the sum of the digits - except when the number is divisible by 10, when the sum increases by 9. As 9 is divisible by itself, you can use "proof by induction" to show that if the sum is divisible by 9, then the number itself will also be divisible by 9. Division by three relies on the remainders when dividing by 9.

To check for divisibility by 11, add alternate digits of the number and take the difference between the two sums. If this difference is zero or divisible by 11, so is the original number. (11 is a factor of 99, which in base 100 behaves like 9 does in base 10)

37 has similar special properties...

Regarding prime factorisation, there is a method rather than a formula - starting from 2, divide by each prime until you can do so no longer, and repeat for each next larger prime. Stop when the square of the prime exceeds the residue. Clumsy, but effective.

Because of the application to cryptography, a great deal of attention has been paid to finding rapid methods for finding large factors, but with rather limited success so far.

Fyz

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#154

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/28/2007 1:48 AM

There's a great little online magic trick based partly on some of these observations: just do it before you try to work out how it's done:-

http://www.exstatica.net/flash/psychic.swf

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#155

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/28/2007 4:53 AM

Having established the 3-rule works because of equal remainders...

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#156

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/29/2007 6:36 AM

Quote: BTW, 9 is not prime

oops - finger trouble

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#17

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:29 PM

6i , 6i, 6 . Some children are just unreal ,and as for bus service...

(to save me time later Geordies have yi children)

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#53

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:44 AM

Hi Kris

I think you are on the correct lines. The options are:

-3i 3i 4, and
-1 -4 9

So the number on the bus is 4, and the oldest child is 9
4 really is a bit young to be regarded as playing the piano...

(The four-year old would have imaginary siblings - not uncommon for an only child, and John has been at the Tarot cards again)

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#56

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 1:47 PM

I get so sloppy with the horrid i thingy ! Your clarification and elaboration is perfect Fyz. It's the best answer so far.

Attn all :Since nobody has asked about my pathetic joke yet : yi man

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#59

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:49 PM

Glad you enjoyed it - it couldn't have happened without your inspiration. Very many moons ago, I was at college with a Scottish Geordie (ran into him again only the other century), so explanation not required.

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#91

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:45 AM

Didn't want to spoil others' chance of guessing.

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#94

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:54 AM

Yas see clivvor pet.

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#19

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:38 PM

Well, if you multiply their ages together you get 36, and, if you add their ages you get the number on that bus."

I read the word "and" to be a continuation of the original thought ..... therefore,

1 x 1 x 36 = 36 + 1 + 1 + 36 = 74

1 x 2 x 18 = 36 + 1 + 2 + 18 = 57

1 x 3 x 12 = 36 + 1 + 3 + 12 = 52

1 x 4 x 9 = 36 + 1 + 4 + 9 = 50

etc........

I guess 1,4,9 and the bus is 50

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#21

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:53 PM

If the piano is key to the solution , perhaps you can extend this method to reach 85 88 ? This may not be totally mad , since John's manner of speech to is ex-teacher indicates some age where such familiarity is OK. The children in question must have been around for at least the few years that John has been out of school. I had plenty of teachers who were fine with use of their first name , but plenty who I think do not even have a first name !

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#20

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:46 PM

What if the kids are minus 2, minus 2 and plus 9.

So the bus is a number five. And they will have twins 2 years from now!

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#22

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 1:57 PM

Del , you are a genius ! I nod my horns to you in respect. Some people plan their lives to the smallest detail(s). I no longer care to hear the 'answer' - yours is purrfect !

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#23

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/04/2007 4:02 PM

In the thread of the previous challenge, you said the problems were of no interest without numbers. Would you care to add some other constraints?

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#26

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 1:28 AM

I shall cite you later Fyz. For now I will give space for others to have a go at padding this thread. For now ,I am content to watch your increasing diversions to humour. Watching somebody being drawn into the murky world of frivolity that I inhabit is pleasure enough , until I can risk outright nonsense as the thread decays. You too could have fun by mapping the time related gibberish function of various Challenge Questions. Results could be encrypted in a proposed KFE cypher for Blaine to puzzle over. In due course constraints will be suggested , but as subtlety is not my finest quality you may have to use your imagination. Talking of which , much as I admire Del's proposition it followed my unreal child theorem. I could split the difference with him but it would be imaginary. For now I leave it as fun for non-Brits to figure out what I meant by yi children. They are not at all imaginary.

Now look what you made me do Fyz - I waffled ! Confusing indeed.

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#27

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 1:58 AM

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#32

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:11 AM

I think my #17 covers that (in a sort of way)

I have to ask , what is LARRICAT ?

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#33

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:30 AM

Could that be a method of herding 'ats (have you noticed how they are breeding on CatRoom4)?

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#35

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:40 AM

I shall go off in a huff and change my avatar.

A huff ? is that a bit smaller than a Landau?

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#39

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 7:04 AM

You have long toes Del ! Can you use them to nip round the North Circular and spy on Randall ? Maybe there's a bus.

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#47

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 10:27 AM

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#36

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:42 AM

Plagiarized wisdom I say. Payback may occur.

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#71

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 6:15 PM

its a larrican

an Australian term

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#77

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 1:28 AM

I geddit , sort of. You just have various spellings. A bit like Ocker but with some brains and humour perhaps. Or better still like Clive James ?

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Anonymous Poster
#82

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 5:33 AM

Is this derived from Brummie "Larrikin"?

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#88

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:08 AM

No need to be coy. They stole it , as we did Clive James.

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Anonymous Poster
#25

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 1:05 AM

hi all,

36, 1, 1? this is impossible and unrealistic!!! imagine you have 36-year old child (which cannot be considered a child anymore) and 1 year old twins? the gap is too unreal. where do you think John was during these years?

my bet is 6,3,2 because Tom was surprised to hear that the oldest at the age of 6 plays the piano...

Anonymous Poster
#29

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:43 AM

It never stated all the children had the same mother. I know a 60 yr old with kids that just turned 1 a few months ago.

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#28

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 2:20 AM

Yes, it would seem unrealistic... one of my employees will be 25 next month, his brother will be 20 in September, his other brother is 8 and his sister is 4. That is a 21 year difference. I suppose there are 36 year old "kids" out there with 1 year old brother/sisters.

Interesting thread, seems to have many answers... and only one correct one. I too, don't see the importance of the "plays the piano".

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#37

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:39 AM

The playing of the piano must be significant.

A normal child of 6 will only be tinkering and not playing. The hands are definitely to small)

To rule out a 6 year I need some information.

Did not one of the masters (Bach , Chopin etc ) play at 6?

What was his fathers name and occupation?

Can a 9 year old play?

The local music teacher only accept 10 years up. The mother may have been a music teacher.

The spacing of the ages seems to be consistent with the thinking of a math teacher

Can a Â± 39year old described as old?

Can a 12 year old play? - yes.

Is Â±42 old? Yes (wife 35 to 40)

Is the spacing of 1,2,12 ok ? - not really but he may have calculated that it was his last chance.

Is 1,2,18 and 1,1,36 possible? - yes. Tom would have known them and the piano clue may not have been of any help.

Tom knew the Â±age of the teacher and the number of years he last saw the teacher. As this was not given the answer must be obtained from what we have.

if Tom left school more than 12 years ago he would have known of the birth of at least the oldest child and the tip about the piano would have been useless.

My guess is therefore that he was away for less than 12 years and that the tip about the piano was supposed to help him by excluding the two - 9 year old options.

My guess is therefore 1,2,12 and 15

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Anonymous Poster
#38

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 7:01 AM

The point of the question is that the answer could not be established from the product of the ages and the sum of the ages (the number on the bus). Unless you are concerned about the ambiguity of "the number on the bus", the ONLY possibility is that the sum of the ages is 13.

Many child prodigies were already playing the piano before 6. Hand size is a limitation for wider chords, but there is a lot of good music that can be played with a spread of a sixth. (N.B. my hands are not particularly large, but my comfortable spread is a tenth).

Many piano teachers will accept pupils for the piano regardless of age if they have a reasonable level of accomplishment on another instrument.

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#40

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 7:38 AM

Wasn't Mozart composing at age 6?

Cue the famous quote: "By the time Mozart was my age, he'd been dead for 3 years!"

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Anonymous Poster
#42

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 8:48 AM

Composing at 6 and decomposing at 36

I think Mozart was touring & performing piano in public at the age of 7 - maybe the publicity said 6.

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#90

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:37 AM

The question doesn't imply that the oldest child is proficient at the piano, just "playing". That could mean that the child is old enough to reach the keys and plunk simple songs (i.e. mary had a little lamb). My grandson is only 3, and I'm pretty sure I can show him to do that.

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Anonymous Poster
#92

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:47 AM

My point exactly

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#43

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 9:45 AM

You may want to do your math again! By my math 1*2*12=24, not 36.

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#70

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:55 PM

yes it must be 1,3,12 = 36 and 16 thanks!!

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Anonymous Poster
#81

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 5:03 AM

"The local music teacher only accept 10 years up
"Either that teacher doesn't accept beginners (not uncommon at the top level), or he/she/it is starting them far too late. (Nervous and muscular systems develop according to the use to which they are put)Mozart (not uniquely) was performing in public at 7. Music scholars at private schools are generally required to have skills equivalent to a distinction at Associated Board grade8 by age 13 - most of them would have been playing since at least 7. etc.

This completely loses any point if it is guess work - even if you get the arithmetic right

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#41

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 8:16 AM

Just a thought .....

Could "bus" be a reference to a "data channel"?

in which case I believe 16 is a number of interest (sorry to not know much about "data channels")

Therefore, the ages of 1, 3 and 12 with the "bus" being 16 may hold the answer....

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#44

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 10:04 AM

Or, the twins may be retarded . . . -2, -2, 9

-2 X -2 X 9 = 36

-2 + -2 + 9 = 5 => number on bus

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Anonymous Poster
#45

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 10:25 AM

What help would the additional information be in this case? (Del_the_cat already made the not-born-yet joke, so you need a number that works)

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#48

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 10:39 AM

Missed Del_the_cat's post.

-2, -2, 9 works . . .

I only noticed a couple (may be off by as many as +/- 7) of serious attempts to figure this out (I did it several years ago), so lighten up.

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Anonymous Poster
#49

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:20 AM

Post 11 (not mine) is comprehensive and numerically correct. There was no need to repeat it or elaborate on the maths of the solution - so no-one did.

The fact that Tom needs additional information tells you there are two solutions at that point - which gives the number on the bus as 13. Given that John and Tom believe that the oldest playing the piano gave additional information, that extra data must relate to a six-year-old not being old enough (in their opinions) to play the piano.

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#51

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:32 AM

I must be slipping . . . I try to write something funny, and get blasted for missing Del's post or thinking that I have solved the problem . . .

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Anonymous Poster
#54

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 11:47 AM

Probably me slipping - it was the second answer got to me...

Anonymous Poster
#64

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:03 PM

But, assuming integer values for the bus and the ages, 5 is a unique, unambiguous solution.

If you allow the children to have negative ages (negative integers) and bus numbers to have negative numbers, you still only have unique sums for each set of 3 numbers except for where the sum is 13.

I didn't check using imaginary numbers or non-integral solutions. I suspect you might find that the sums of those are unique.

Anonymous Poster
#67

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:26 PM

As there was a single non-unique case using imaginary and negative integers (see #54), I drew the line at complex kids (I have enough psychological problems of my own, thank you very much).

But there are also far too many possibilities for me to go through if you allow fractional solutions - particularly if you have fractions on the bus. [We could be in HarryPotterLand, or perhaps the number on the bus was the number of fares - senior citizens counting as 2/3, children as 1/2?]

Fyz

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#55

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 12:15 PM

I would say Tom must have been in school to know when John had his youngest child because Tom asked "how old are your 3 kids now". 2X3X6=36, Bus number =2+3+6=11. If his youngest child were only 1 year old Tom must have been in school last year and would surely have known that the oldest was taking piano lessons and if there were twins he would not have not have specified "3 Kids"

Anonymous Poster
#58

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:42 PM

2 2 9 fits as well (no need for twins). Even if there were twins, what would you say if you couldn't remember the name of the eldest? Mainly though, if the bus number was 11, Tom wouldn't have needed any additional information. Go figure.

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#57

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:36 PM

the answer is 2,2 and 9

the possible combinations are

1,1,36=38/1,2,18=21/1,3,12=16/1,4,9=14/1,6,6=13/2,2,9=13/2,3,6=11/3,3,4=10

tom was perlexed because there were two possible combinations with 13 as the sum so john clarified that the oldest plays piano that means eldest is not a twin so we can assume the combi to be 2,2 and 9 and the no. on bus is 13

Anonymous Poster
#60

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 3:52 PM

Fine completion of #11 - but see #14 for reservations that apply to both "solutions".

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#61

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:09 PM

Does the 'the number on that bus' mean the route number or maybe how many passengers?

Never mind, either suggests a whole number.

The factors of 36 lead to a maximum age of 36 for one child.

The original Teacher /Pupil relationship between John and Tom is thought to suggest an age gap of nearly a generation. And if so, Tom would have been at school roughly the same time with one of John's children and thus already familiar with their ages. As tom did not know their ages it suggests an age gap that makes Johns eldest somewhat younger than Tom himself.

The (now) familiarity between John and Tom (on christian name terms) reduces the formality of their own age gap, and maybe with an acquaintance that embodies the knowledge of the existence of three children (but not necessarily their names) that engenders an interest in their ages.

Playing the piano, suggests an age that equates to a physical 'size' gap that implies the two youngest children are not yet big enough.

I plonk for 1, 4 and 9 and a No14 bus. Tom is 23 and John is 41

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Anonymous Poster
#65

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:11 PM

If the bus number is 14: what alternative solution can there have been? So why was Tom confused? Why did John think extra information would help? Look at ##11, 58, and now 63, and weep. It is likely that Tom knew of the children when at school, which according to your solution must be less than a year ago - so 23 is a bit rich on your assumptions (unless you think this was adult education for those who missed out the first time around?)

Anonymous Poster
#62

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 4:36 PM

Tom can see the number on the bus. Adding sets of 3 non-negative integers whose product is 36 yields unique answers for each set except 13 (2,2,9) (6,6,1). If it were anything but 13, there would be no question as to the ages of the children. The final clue is distinguishing whether the twins are older or younger than the third child.

Anonymous Poster
#66

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:15 PM

There are eight combinations of three numbers that equal 36 when multiplied:

1x6x6 sum=13

1x4x9 sum=15

1x3x12 sum=16

1x2x18 sum=21

1x1x36 sum=38

2x2x9 sum=13

2x3x6 sum=11

3x3x4 sum=10

If Tom was perplexed when he noted the number on the bus, then he must have realized there were more than one possible answer. The only sum that is duplicated is 13. When the old teacher tells him "the oldest," his riddle is solved since 2,2,9 provides the only answer clearly demonstrating an oldest child. So, the ages must be 2,2,and 9, and the bus number 13.

I loved the diversion to piano playing! It makes the thought processes dance to a different drum.

Rob

Anonymous Poster
#68

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:31 PM

There is always an oldest child - even (especially?) with twins. Just listen to some (and their parents). See also #14. Of course, the same numerical age doesn't even guarantee twinship.

Anonymous Poster
#69

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 5:39 PM

1 is old, 2 is older, 3 is oldest!

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#87

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 6:45 AM

I agree with #73 but rather change it to descending order.

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#72

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/05/2007 7:02 PM

I do believe that you clearly have it! I cannot see the need for further discussion.

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#95

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/06/2007 7:59 AM

That is the best reasoning thus far. I'll even say that because the teacher identified them by their nominal ages, you have to dismiss the idea of an "older" twin. To make myself feel worthy I will point out a math correction that has no bearing in this answer: 1+4+9=14.

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#138

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 4:16 AM

Am I missing something obvious? Why did the bus have to be the duplicated number?

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#140

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 5:17 AM

If the bus had not been the duplicated number: then Tom would not have been perplexed, and, John would not have needed to supply any more information.

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#141

### Re: My Teachers Three Kids: Newsletter Challenge (06/05/07)

06/13/2007 6:56 AM

There should have been at least three interpretations - the number marked on the bus (usually referred to as the number of the bus where I live), the number of passengers on the bus, or the number of people on the bus including staff. That would have been perplexing. If we allow all three, we would have had no way of finding a solution. If we eliminate the number of the bus (because the maths teacher appears to be a pedant), then we could assume the two possible numbers differ by one...

Fyz

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