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The thing seems totally impossible, and so it is for ordinary regions. The breakthrough is to think of using infinitely iterated processes to define your subsets. The shaded regions in the following are C1 and S1, while their comlements - the unshaded regions - are C2 and S2. The apparent paradox is similar to that in the ``Infinite Hotel'', also known as Hilbert's Hotel, after the great mathematician David Hilbert:
A hotel in Heaven numbers its rooms 1,2,3, etc. ad infinitum. A weary traveller arrives, and is told the hotel is full. But the compassionate Manager says, ``Don't worry, Madame! I will ask the occupant of room 1 to move into room 2, whose occupant will move into room 3, ..., and the occupant of room k having moved into room k+1, for each natural number k, you can have room 1 without anybody being out on the street.''
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Neither salt nor sugar is in the tea jar, nor is tea in
either of their jars. The rice jar holds no rice, barley or currants. The
barley jar holds no rice and the currant jar holds no barley. The
contents of the sugar jar should be where the barley is and the currant
jar holds what ought to be where the flour is, which is not the tea jar.
Sugar is not in the salt jar, nor salt in the sugar jar. Neither barley
nor salt is in the flour jar. The label on the jar holding sugar
describes the contents of the raisin jar. The rice is neither in the tea
jar nor the salt jar. No two items are in each other's jar.
This was solved by Owen Sibanda of N.U.S.T., using the standard
sensible procedure for such problems - make a table with ticks and
crosses.
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This one was solved by Donald T Hove (ex-Kutama, now heading for University).
In division by 3, the integers are divided up into three classes, determined by the three possible remainders: 0,1,2. If, among our five given numbers, all three classes are represented, then the sum of the three numbers giving the three different remainders will itself give remainder 0+1+2=3, so will be divisible by 3. If this doesn't happen, then clearly one of the less-than-three remainders must occur at least three times amongst our five numbers. (Pigeonhole Principle: distribute five letters between less than three pigeonholes and you can be sure that somebody has got at least three letters!) And then of course the sum of these three numbers will surely be divisible by 3, as the remainder will be one of 3×0, 3×1, 3×2.
In considering divisibility by 5, similarly, there are five classes of numbers, or, using the congruence notation invented by Gauss, every number is in one of the five congruence classes modulo 5: 0 (mod 5), 1 (mod 5), 2 (mod 5), 3 (mod 5), 4 (mod 5). If you ask how many numbers you can collect together without any five of them having sum divisible by 5, you must avoid having five of any one class (for their remainders will sum to a multiple of 5), and you must avoid having a representative of each class (for their remainders will sum to 0+1+2+3+4=10). Try choosing four of each of the four classes 1 (mod 5), 2 (mod 5), 3 (mod 5), 4 (mod 5) - that's 16 numbers. But if you start with 17 numbers, and some remainder is not represented, then one of the remaining four remainders must occur at least 5 times, by the Pigeonhole Principle. Therefore, given 17 numbers, it is always possible to select 5 numbers whose sum is divisible by 5.
Generalizing: For divisibility by 7, there are seven congruence classes, and we can choose six from each of the six classes 1 (mod 7), 2 (mod 7), 3 (mod 7), 4 (mod 7), 5 (mod 7), 6 (mod 7), and be sure no seven of them will sum to a multiple of 7. However, if we are given 62+1=37 numbers, and you'll always be able to choose a successful seven - either by taking a representative from each congruence class, or if they aren't all represented, by the Pigeonhole Principle, bcause one class must have at least seven representatives. The extension to any prime number n is now clear.
Donald Hove sent in an answer to this, but he answered a different question! He gave the prime factors first: 2004=1×22×3×167, and then said that all divisors of 2004 are included in the breakdown (twelve terms) of (1+2+22)(1+3)(1+167). Which is true - there are exactly 12 divisors of 2004, making 1992 non-divisors. But that is not the same as the much smaller number of relatively prime numbers less that 2004! For example all even numbers except 2, 4 are non-divisors, but none are relatively prime ( which means havng NO common factors). This problem remains open.
Kennedy Kundanayi of Muchekayaora School and Donald Hove each supplied a correct answer to this one.
For 444 ¼(M digits)¼4=4×111¼(M digits)¼ 1=4K, to be divisible by
333¼(100 digits)¼3=3×111¼(100 digits)¼1=3k, it is necessary for 3 to divide
K, which is easily seen to be equivalent to
M being divisible by 3. It is also necessary for k to divide K
(like 3, k is relatively prime to 4). We only need to show
that this demands that M be a multiple of 100, and all will be clear.
Trial and error will soon convince you, but here is a more rigorous proof
(due to the setter, Munyaradzi Mukachana):
Suppose that M=100q+r,
where 0 < r < 100, and let p=111¼(r digits)¼1.
Then
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Since we know k divides K, we have p=0 and r=0. Thus, finally, M must be divisible by 3 and 100, so the smallest value is M=300 digits.
Normal Section
1. Let A be the sum of three consecutive natural numbers, and B be the sum of the next three natural numbers. Is it possible that the product A×B be equal to 111111111? Explain.
The impossibility was fully demonstrated by our prizewinner
Rudo, as well as by Innocent Warirai and Austin Maregera of
Hippo Valley High School, and by Sifundo Moyo of Kutama. All their
solutions amounted to the routine method of denoting the numbers
n,n+1,n+2, followed by n+3,n+4,n+5, and expressing the given
condition as a quadratic equation in n, simplifying from
A×B=(3n+3)(3n+12)=111111111 to n2+5n+4=12345679, and finally to
n2+5n-12345675=0. While this has real solutions, the discriminant is
not a complete square, so there are no integral solutions n.
There is, however, a much neater and quicker way of showing impossibility.
If n is odd, then A (as sum of two odds and one even) will be even,
and
B (as sum of two evens and one odd) will be odd; while if n is even,
A will be odd and B will be even. Either way, the product A×B
will be even, which the given number is not! Moral: Always look for
the easiest, most elegant solution.
2. During the second term, Temba and Farai each got 64 items marked (or graded). The number of Temba's A's equals the number of Farai's B's; the number of Temba's B's equals the number of Farai's C's; the number of Temba's C's equals the number of Farai's D's; and, finally, the number of Temba's D's equals the number of Farai's A's. Moreover, the boys have each got the same average results, if we weight them as follows: A=5, B=4, C=3, D=2. How many D's did Temba get?
Rudo and Innocent and Austin laid out the given conditions in useful tables with useful notation. Perhaps the best notation to use is this:
| Temba's A's,B's,C's,D's | T5 | T4 | T3 | T2 |
| Farai's A's,B's,C's,D's | F5 | F4 | F3 | F2 |
| & given matchings: | T2 | T5 | T4 | T3 |
Now equal averages can be expressed as follows:
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3. At the Millionaire's Club, the Number One Millionaire remarked that in the current charity campaign to support the Zimbabwean Mathematical Olympiad Team, he had pledged to match the total of all other contributions. Upon hearing this, millionaire Number Two fainted. Can you guess why?
Of course, if Mill.Two has a much smaller fortune than Mill.One, he may well feel nervous about having to contribute at least as much as Mill.One, but he is unlikely to faint, since he knew the Club was full of millionaires when he made the pledge. The reason he fainted must be because he reasoned a bit further, on the basis of TWO of them having made the same pledge, and found himself in what logicians call an `infinite regress':
Even if 1 dollar was contributed by the rest put together, Mill.Two is
pledged to give 1+M1, where M1 is Mill.One's contribution.
But then:
Mill.One is pledged to make his up to 1+M2=1+(1+M1)=2+M1,
Mill.Two is pledged to make his up to 1+(1+(1+M1))=3+M1,
and the process never ends. In fact, each of them will have to donate
their entire fortune trying to match the other's donation plus $1!!
Another way of putting it: the two equations M1=1+M2, M2=1+M1 have no finite solutions, but each is approximated increasingly accurately as M1,M2® ¥.
1. In this 3×3 table, all entries are zero's to begin with, and one may operate on the table as follows: in any 2×2 square, you may increase by 1 all four numbers in it. Is it possible to get the final table as shown on the right? Explain.
This problem was well solved by Shelton Nyakudya of ZRP PPU, Sifundo Moyo of Kutama, Kennedy Kundanayi of Muchekayaora School, Donald Hove ex-Kutama, as well as our prizewinner.
The key is to see that the number of operations performed on each of the four 2×2 squares is recorded by its corner number, and the total number of operations is given by the centre square. Hence the number in the centre must be equal to the sum of the four corner numbers. Since 18 ¹ 22=4+5+7+6, the given square is impossible to achieve. Suppose we changed the middle number to 22 - would the square be achievable now? A few of our respondents observed a further law to be found in all achievable squares: the sum of the outer numbers in any horizontal or vertical row must be equal to the middle number. The new square is consistent with this, and is certainly achievable.
2. Show that f(x)=4cos2 x +7sinxcosx +2sin2 x-Ö[102] is negative for any real x.
In addition to our two prizewinners, Donald Hove and Shelton Nyakudya producd good solutions to this one.
Overall strategy: try to combine the three trig.expressions into one or two cosines/sines whose bounds are obvious. The first good idea, reducing three to two expressions, is to bring the cos2 and sin2 terms together:
4cos2 x +2sin2 x=2(cos2 x +sin2 x)+2cos2x=2+2cos2x.
The next good idea is to express this 2+2cos2x, and also
7sinxcosx, in terms of 2x: the first is cos2x +3 and the
second is [7/2]sin2x. This gives us:
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A slightly different route but notable route was taken at the end by Sifundo Moyo and Donald Hove, to get a rather better bound. True, in this particular problem it wasn't necessary, but it might have been in case we'd had Ö[50] in place of Ö[102]. Call that expression g(x) - it's larger than f(x), but still negative! Hove and Moyo expressed cos2x +[7/2]sin2x in the form Rsin(2x+a), finding R=Ö{1+([7/2])2}=[1/2]Ö[53]. Hence
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3. Pirates decided to hide their treasure on a lonely island. They remember that if one goes 70m to the left from the landing spot, along a straight shoreline, and then from there 110m in a straight line toward a certain tree, one arrives at the place where the treasure is buried. Alternatively, one can go 130m directly towards the tree from the landing spot, then turn left and go another 10m. Unfortunately, when the pirates re-visited the island, they discovered that the tree had been removed by a storm. How do they find the treasure?
This problem was fully solved by Kennedy Kundanayi of Muchekayaora School and by Donald Hove ex-Kutama, as well as our prizewinner. Also, Innocent Warirai and Austin Maregere of Hippo valley High approached it correctly, although they got a bit bogged down with inessential numerical calculation and decimal points.
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